Solution :

(a) From Fig. 2(a); we have

distance traveled $=\frac{2 \pi \mathrm{r}}{4}=\frac{\pi \mathrm{r}}{2}$

Magnitude of displacement

$=$ Length of chord $\mathrm{AB}=\sqrt{2} \mathrm{r}$

(b) From Fig. 2(b), we have

(a)

(b)

Fig. 2

Distance traveled $=\frac{3}{4}(2 \pi \mathrm{r})=\frac{3 \pi \mathrm{r}}{2}$

Magnitude of displacement $=$ Length of chord $A B=\sqrt{2} r$

Solution :

(a) The total distance traveled $=\mathrm{s}$. Obviously

$$ \begin{align*} & \mathrm{s}=\mathrm{AB}+\mathrm{BC}=2 \mathrm{a} \\ & \text { Average speed }=v _{\mathrm{av}}=\frac{\mathrm{s}}{\mathrm{t}}=\frac{2 \mathrm{a}}{\mathrm{t}} \tag{1} \end{align*} $$

(b) The net displacement $=\mathbf{S}=\mathbf{A B}$

Magnitude of net displacement $=|\mathbf{s}|=\mathrm{AC}=\sqrt{2} \mathrm{a}$

The magnitude of the average velocity $=\left|\mathbf{v} _{\mathrm{av}}\right|=\frac{\sqrt{2} \mathrm{a}}{\mathrm{t}}$

Fig. 3

Solution :

The total distance travelled $=\mathrm{S}=\mathrm{S} _{1}+\mathrm{S} _{2}+\ldots \ldots . .=\sum _{\mathrm{i}=1}^{\mathrm{n}} \mathrm{S} _{\mathrm{i}}$

The total time taken $=\mathrm{T}=\mathrm{t} _{1}+\mathrm{t} _{2}+\ldots \ldots . .=\sum _{\mathrm{i}=1}^{\mathrm{n}} \mathrm{t} _{\mathrm{i}}$

$$ \text { or } \mathrm{T}=\frac{\mathrm{S} _{1}}{\mathrm{~V} _{1}}+\frac{\mathrm{S} _{2}}{\mathrm{~V} _{2}}+\frac{\mathrm{S} _{3}}{\mathrm{~V} _{3}} \ldots \ldots . .=\sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\frac{\mathrm{S} _{\mathrm{i}}}{\mathrm{V} _{\mathrm{i}}}\right) $$

The averages speed, $v _{\mathrm{av}}$ is

$$ \mathrm{V} _{\mathrm{av}}=\frac{\mathrm{S}}{\mathrm{T}}=\frac{\sum _{\mathrm{i}=1}^{\mathrm{n}} \mathrm{S} _{\mathrm{i}}}{\sum _{\mathrm{i}=1}^{\mathrm{n}}\left(\frac{\mathrm{S} _{\mathrm{t}}}{\mathrm{V} _{\mathrm{i}}}\right)} $$

Let $\mathrm{S} _{1}=\mathrm{S} _{2}=\mathrm{S}$. The body covers two equal distances with different speeds $\mathrm{V} _{1}$ and $\mathrm{V} _{2}$. Then

$$ \mathrm{V} _{\mathrm{av}}=\frac{2 \mathrm{~S}}{\left(\frac{\mathrm{S}}{\mathrm{V} _{1}}+\frac{\mathrm{S}}{\mathrm{V} _{2}}\right)}=\frac{2 \mathrm{~V} _{1} \mathrm{~V} _{2}}{\mathrm{~V} _{1}+\mathrm{V} _{2}} $$

The average speed equal the harmonic mean of individual speeds.

Solution :

Let $\mathrm{S} _{1}, \mathrm{~S} _{2}, \ldots \ldots \mathrm{S} _{\mathrm{n}}$ be distances traveled in $1^{\text {st }}, 2^{\text {nd }} \ldots . . . \mathrm{n}^{\text {th }}$ time interval, each equal to $t$. Obviously

$$ S _{1}=V _{1} t ; S _{2}=V _{2} t, \ldots \ldots S _{n}=V _{n} t $$

The total distance traveled $\mathrm{S}=\mathrm{S} _{1}+\mathrm{S} _{2}+\ldots \ldots+\mathrm{S} _{\mathrm{n}}$

$$ =\left(\mathrm{V} _{1}+\mathrm{V} _{2}+\ldots . . \mathrm{V} _{\mathrm{n}}\right) \mathrm{t} $$

The total time taken $=\mathrm{T}=\mathrm{nt}$

$\therefore$ Average speed $=\frac{\mathrm{S}}{\mathrm{T}}=\frac{\mathrm{V} _{1}+\mathrm{V} _{2}+\ldots . \mathrm{V} _{\mathrm{N}}}{\mathrm{n}}$

$=$ Arithmetic mean of the individual speeds.

Solution :

(a) (i) $\quad v=$ slope of $x$ vs t graph at $\mathrm{t}=2 \mathrm{~s}$

$$ =\frac{8-2}{4-0}=1.5 \mathrm{~ms}^{-1} $$

Fig. 10

(ii) $v(\mathrm{at} \quad \mathrm{t}=5 \mathrm{~s})=$ Zero

(iii) $\quad \mathrm{v}(\mathrm{at}=10 \mathrm{~s})=\frac{-8-8}{12-8}=-4 \mathrm{~ms}^{-1}$

(b) The total distance traveled in time interval $0-12 \mathrm{~s}$

$$ =6+8+8=22 \mathrm{~m} $$

$\mathrm{V} _{\mathrm{av}}=\frac{\text { Total distance traveled }}{\text { Total time taken }}=\frac{22}{12} \simeq 1.83 \mathrm{~ms}^{-1}$

(c) For time interval

(i) $0 \leq \mathrm{t} \leq 4 \mathrm{~s} ; \quad \mathrm{v}=1.5 \mathrm{~ms}^{-1}$

(ii) $4 \leq \mathrm{t} \leq 8 \mathrm{~s} ; \quad \mathrm{v}=0$

(ii) $8 \leq \mathrm{t} \leq 12 \mathrm{~s} ; \quad \mathrm{v}=-4 \mathrm{~ms}^{-1}$

$v$ vs $t$ graph for motion is shown in Fig. 11

Fig. 11

Solution:

(a) $\mathrm{S}=$ the total distance traveled in $10 \mathrm{~s}$

$$ \begin{aligned} & =\left(\frac{1}{2} \times 4 \times 6\right)+(2 \times 6)+\left(\frac{1}{2} \times 2 \times 6\right)+(6 \times 2) \\ & =12+12+6+12=42 \mathrm{~m} \end{aligned} $$

$\mathrm{T}=$ Total time taken $=10 \mathrm{~s}$

$\therefore$ Average speed $=\mathrm{V} _{\mathrm{av}}=\frac{\mathrm{S}}{\mathrm{T}}=\frac{42}{10}=4.2 \mathrm{~ms}^{-1}$

[Note: In calculating S we have NOT assigned negative sign to area in time interval 8 to 10s.]

(b) The magnitude of total displacement $=|\mathbf{s}|$

$=12+12+6+(-12)=18 \mathrm{~m}$

[Note: Negative sign assigned to area under v vs t graph in time interval 8 to 10s.]

The magnitude of average velocity $=\frac{18}{10}=1.8 \mathrm{~ms}^{-1}$

Solution :

By define the instantaneous velocity $v$ is

$$ \begin{equation*} v=\frac{\mathrm{d} x}{\mathrm{dt}}=4-4 \mathrm{t}+3 \mathrm{t}^{2} \tag{i} \end{equation*} $$

and instantaneous acceleration, $a$, is

$$ \begin{equation*} a=\frac{\mathrm{d} v}{\mathrm{dt}}=-4+6 \mathrm{t} \tag{ii} \end{equation*} $$

At $t=1 s ;$

$$ \begin{aligned} & v(\mathrm{t}=1 \mathrm{~s})=4-4+3=3 \mathrm{~ms}^{-1} \\ & \mathrm{a}(\mathrm{t}=1 \mathrm{~s})=-4+6=2 \mathrm{~ms}^{-2} \end{aligned} $$

Since acceleration varies with time, motion is NOT uniformly accelerated.

Solution :

(a) Given $v=\mathrm{t}^{2}-4$

$\therefore \mathrm{a}=\frac{\mathrm{d} v}{\mathrm{dt}}=2 \mathrm{t}$

$\mathrm{a}(\mathrm{t}=2 \mathrm{~s})=4 \mathrm{~ms}^{-2}$

(b) Let particle move along $x$-axis. Then

$$ v=\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{t}^{2}-4 $$

Displacement $=\int \mathrm{d} x=\int _{0}^{4}\left(\mathrm{t}^{2}-4\right) \mathrm{dt}$

$=\left|\frac{\mathrm{t}^{3}}{3}\right| _{0}^{4}-4|\mathrm{t}| _{0}^{4}$

$=\frac{64}{3}-16=\frac{16}{3} \mathrm{~m}$

(c) Note $v$ is negative in time interval $0 \leq \mathrm{t} \leq 2 \mathrm{~s}$ and positive in time interval $2 \leq \mathrm{t} \leq 4 \mathrm{~s}$. Distance traveled $\mathrm{S}$ is

$$ \begin{aligned} & \mathrm{S}=\left|\int _{0}^{2}\left(\mathrm{t}^{2}-4\right) \mathrm{dt}\right|+\int _{2}^{4}\left|\mathrm{t}^{2}-4\right| \mathrm{dt} \\ & =\frac{16}{3}+\frac{16}{3}=\frac{32}{3} \mathrm{~m} \end{aligned} $$

Solution :

Given,

(a) $\quad v _{0}=108 \mathrm{kmhr}^{-1}=108 \times \frac{5}{18} \mathrm{~ms}^{-1}=30 \mathrm{~ms}^{-1}$ $v=36 \mathrm{kmhr}^{-1}=36 \times \frac{5}{18} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-2}$

$\mathrm{S}=200 \mathrm{~m} ; \mathrm{a}=?$

From, $v^{2}-v _{0}^{2}=2 \mathrm{aS}$, we have

$(10)^{2}-(30)^{2}=2 \mathrm{a} \times 200$

$\therefore \mathrm{a}=-2 \mathrm{~ms}^{-2}$

Negative sign of a indicates retardation

(b) Let t be the time taken. Using $\mathrm{v}=\mathrm{v} _{0}+$ at, we have

$$ 10=30+(-2) \mathrm{t}, \text { or } \mathrm{t}=10 \mathrm{~s} $$

Solution :

Given,

(1) $v _{0}=0 ; \mathrm{a}=+2 \mathrm{~ms}^{-2} ; \mathrm{t} _{1}=5 \mathrm{~s}$

$\therefore v _{1}=0+2 \times 5=10 \mathrm{~ms}^{-1}$

$\mathrm{S} _{1}=0+\frac{1}{2} \times 2 \times(5)^{2}=25 \mathrm{~m}$

The particle is brought to rest in $2 \mathrm{~s}$ having an initial speed of $10 \mathrm{~ms}^{-1}$. The retardation $\mathrm{a} _{2}$ is $0=10+\mathrm{a} _{2} \times 2 \quad, \quad \therefore \mathrm{a} _{2}=-5 \mathrm{~ms}^{-2}$

$\mathrm{S} _{2}=$ distance traveled in time interval $\mathrm{t}=5 \mathrm{~s}$ to $\mathrm{t}=7 \mathrm{~s}$

$=10 \times 2+\frac{1}{2}(-5)(2)^{2}=10 \mathrm{~m}$

$\mathrm{S}=$ The total distance traveled $=\mathrm{S} _{1}+\mathrm{S} _{2}=25+10=35 \mathrm{~m}$

(2) For time interval $0 \leq \mathrm{t} \leq 5 \mathrm{~s}$; the instantaneous speed is

$$ v=0+2 \times t=2 t $$

For time interval $5 \leq \mathrm{t} \leq 7 \mathrm{~s}$; the instantaneous speed is

$v=10+(-5) \mathrm{t}=10-5 \mathrm{t}$

Taking these factors into account $v$ vs $t$ graph is as shown in Fig. 14.

Let $x$ be the instantaneous position co-ordinate of particle starting from origin at $t=0$. For time interval $0 \leq \mathrm{t} \leq 5$ s.

Fig. 14

$x($ at $\mathrm{t}=5 \mathrm{~s})=25 \mathrm{~m}$. For time interval $5 \leq \mathrm{t} \leq 7 \mathrm{~s}$; the instantaneous position coordinate is

$$ \begin{align*} & x=25+10(\mathrm{t}-5)+\frac{1}{2}(-5)(\mathrm{t}-5)^{2} \\ & =25+10(\mathrm{t}-5)-2.5(\mathrm{t}-5)^{2} \tag{2} \\ & x(\mathrm{t}=7 \mathrm{~s})=25+10(7-5)-2.5(7-5)^{2} \\ & =25+20-10=35 \mathrm{~m} \end{align*} $$

These characteristics are shown in $x$ vs $\mathrm{t}$ graph in Fig. 15.

Fig. 15

Solution :

We know, the distance $\mathrm{s} _{\mathrm{n}}$; traveled in $\mathrm{n}^{\text {th }}$ second is

$$ \mathrm{s} _{\mathrm{n}}=v _{0}+\frac{\mathrm{a}}{2}[2 \mathrm{n}-1] $$

Given $\mathrm{s} _{7}=17 \mathrm{~m}$ and $\mathrm{s} _{9}=21 \mathrm{~m}$. Therefore

$$ \begin{equation*} 17=v _{0}+\frac{\mathrm{a}}{2}[2 \times 7-1]=v _{0}+\frac{13}{2} \mathrm{a} \tag{1} \end{equation*} $$

and $\quad 21=v _{0}+\frac{\mathrm{a}}{2}[2 \times 9-1]=v _{0}+\frac{17}{2} \mathrm{a}$

From equations (1) and (2)

$$ \mathrm{a}=2 \mathrm{~ms}^{-1} \text { and } \mathrm{v} _{0}=4 \mathrm{~ms}^{-1} $$

Let $v _{10}$ be the speed acquired by car at $\mathrm{t}=10 \mathrm{~s}$

$$ \begin{aligned} & v _{0}=v _{0}+\mathrm{a} \times 10 \\ & =4+2 \times 10=24 \mathrm{~ms}^{-1} \end{aligned} $$

Let $\mathrm{S}$ be the total distance traveled in $\mathrm{t}=10 \mathrm{~s}$. We have

$$ \mathrm{S}=4 \times 10+\frac{1}{2} \times 2 \times(10)^{2}=140 \mathrm{~m} $$

Solution :

Given,

$$ v=\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{k} x^{1 / 2} \text { or } \frac{\mathrm{d} x}{x^{1 / 2}}=\mathrm{kdt} $$

Integrate

$$ \int _{0}^{x} x^{-1 / 2} \mathrm{~d} x=\mathrm{k} \int _{0}^{\mathrm{t}} \mathrm{dt} \quad \text { or } \quad 2 x^{1 / 2}=\mathrm{kt} \quad \text { or } \quad x=\left(\frac{\mathrm{k}^{2}}{4}\right) \mathrm{t}^{2} $$

The instantaneous velocity, $v$, is

$$ v=\frac{\mathrm{d} x}{\mathrm{dt}}=2\left(\frac{\mathrm{k}^{2}}{4}\right) \mathrm{t}=\frac{\mathrm{k}^{2}}{2} \mathrm{t} $$

Obviously $v$ is directly proportional to t.

Solution :

Choose point of projection $\mathrm{O}$ as origin and vertically upward direction as positive $\mathrm{z}$-axis. as shown in Fig. 17. Given

$$ v _{0}=20 \mathrm{~ms}^{-1}, \mathrm{a}=-10 \mathrm{~ms}^{-2}, \mathrm{z} _{0}=0 $$

Note carefully + and $-\mathrm{ve}$ sign of $v _{0}$ and a. Let $\mathrm{v}$ and $\mathrm{z}$ be the instantaneous velocity and position coordinate of ball. Then

$$ \begin{align*} & \mathrm{v}=\mathrm{v} _{0}+\mathrm{at}=20-10 \mathrm{t} \tag{i} \\ & \mathrm{z}=\mathrm{v} _{0}+\frac{1}{2} a \mathrm{t}^{2}=20 \mathrm{t}-5 \mathrm{t}^{2} \tag{ii} \end{align*} $$

(a) Let particle attain maximum height from ground at $t=t _{1}$. This is shown as point $\mathrm{A}$ in Fig. 17(a). At maximum height instantaneous velocity is zero. For point $A ; v _{A}=0 ; t=t _{1}$. FromEqn. (i)

$0=20-10 \mathrm{t} _{1} \quad \therefore \mathrm{t} _{1}=2 \mathrm{~s}$

The position co-ordinate $\mathrm{z} _{\mathrm{A}}$ of point A gives maximum height of ball. From Eqn. (ii) $\mathrm{H}=20 \times 2-5(2)^{2}=20 \mathrm{~m}$

(b) (i) $\mathrm{At} \quad t=1.5 \mathrm{~s}$; from Eqn. (i) and (ii) we have

$$ \begin{aligned} & \mathrm{v}(\mathrm{t}=1.5 \mathrm{~s})=20-15=5 \mathrm{~ms}^{-1} \\ & \mathrm{z}(\mathrm{t}=1.5 \mathrm{~s})=20 \times 1.5-5 \times(1.5)^{2}=18.75 \mathrm{~m} \end{aligned} $$

The +ve sign of $v$ shows that ball is moving vertically upwards. The $+v e$ sign of $z$ means ball at this moment is above ground at a height of $18.75 \mathrm{~m}$. This is shown in Fig. 17(b).

(ii) $\mathrm{At} \quad t=3 \mathrm{~s}$; from Eqn. (i) and (ii) we have

$$ \begin{aligned} & \mathrm{v}(\mathrm{t}=3 \mathrm{~s})=20-30=-10 \mathrm{~ms}^{-1} \\ & \mathrm{z}(\mathrm{t}=3 \mathrm{~s})=20 \times 3-5(3)^{2}=15 \mathrm{~m} \end{aligned} $$

The negative sign of $v$ indicates the ball is moving vertically downwards. The +ve sign of z indicates ball in above ground at a height of $15 \mathrm{~m}$. This is shown in Fig. 17(c).

Solution :

In Fig. 18(a) $\mathrm{O}$ is point an ground and $\mathrm{O} _{1}$ is the top of tower. Choose $\mathrm{O}$ as origin and vertically upward direction as positive $\mathrm{z}$-axis.

For ball A;

$\mathrm{v} _{\mathrm{AO}} 1=0 ; \mathrm{a} _{\mathrm{A}}=-10 \mathrm{~ms}^{-2} ; \mathrm{z} _{\mathrm{AO}}=+100 \mathrm{~m}$

$$ \therefore \mathrm{v} _{\mathrm{A}}=-10 \mathrm{t} $$

$$ \begin{equation*} \mathrm{z} _{\mathrm{A}}=100+\left(-5 \mathrm{t}^{2}\right) \tag{i} \end{equation*} $$

For ball B,

$$ \begin{align*} & \mathrm{v} _{\mathrm{B}}=+40 \mathrm{~ms}^{-1}, \mathrm{a} _{\mathrm{B}}=-10 \mathrm{~ms}^{-2}, \mathrm{z} _{\mathrm{BO}}=0 \\ & \therefore \mathrm{v} _{\mathrm{B}}=40-10 \mathrm{t} \\ & \mathrm{z} _{\mathrm{B}}=40 \mathrm{t}-\mathrm{st}^{2} \tag{ii} \end{align*} $$

When the two balls cross one another, their instantaneous position co-ordinate is same, i.e. $z _{A}=z _{B}$. From Eqns. (i) and (ii)

$$ \begin{aligned} & 100-5 t^{2}=40 t-5 t^{2} \\ & \text { or } t=2.5 \mathrm{~s} \end{aligned} $$

The two balls cross one another $2.5 \mathrm{~s}$ after projection. The instantaneous position co-ordinate of a from Eqn. (i) is

$$ \mathrm{z} _{\mathrm{A}}=100-5(2.5)^{2}=68.75 \mathrm{~m} $$

The two balls cross one another at a point $68.75 \mathrm{~m}$ above ground. Let $\mathrm{v} _{\mathrm{A}}$ and $\mathrm{v} _{\mathrm{B}}$ be their instantaneous velocity. Obviously

$$ \begin{aligned} & \mathrm{v} _{\mathrm{A}}=-10 \times 2.5=-25 \mathrm{~ms}^{-1} \\ & \mathrm{v} _{\mathrm{B}}=40-20 \times 2.5=+15 \mathrm{~ms}^{-1} \end{aligned} $$

Negative sign of $\mathrm{v} _{\mathrm{A}}$ indicates that ball $\mathrm{A}$ is moving vertically downwards. The positive sign of $\mathrm{v} _{\mathrm{B}}$ indicates that ball B is moving vertically upwards. This is shown in Fig. 18(b).

Solution:

In Fig. 19(a). $\mathrm{O}$ is point on ground and $\mathrm{A}$ is the position of helicopter at $\mathrm{t}=0$ when packet is dropped. Choose $\mathrm{O}$ is origin and vertically upward direction as $+\mathrm{z}-\mathrm{axis}$.

$$ \begin{aligned} & \mathrm{v} _{0}=\text { The initial velocity of packet }=\text { Velocity of helicopter at this moment of time } \\ & =1080 \times \frac{8}{18} \mathrm{~ms}^{-1}=300 \mathrm{~ms}^{-1} \\ & \mathrm{a}=\text { acceleration of packet }=-10 \mathrm{~ms}^{-2} ; \mathrm{z} _{0}=2000 \mathrm{~m} \end{aligned} $$

Let $\mathrm{v}$ and $\mathrm{z}$ denote the instantaneous velocity and position co-ordinate of packet. We have

$$ \begin{align*} & v=300-10 t \tag{1} \\ & z=2000+\left(300 t-5 t^{2}\right) \tag{2} \end{align*} $$

Let packet attain maximum height from ground at $\mathrm{t}=\mathrm{t} _{1}$. In Fig. 19(b); $\mathrm{P}$ is instantaneous position of packet. The instantaneous velocity of packet at $\mathrm{P}$ is zero. From Eqn. (1)

$$ 0=300-10 \mathrm{t} _{1} \quad \therefore \mathrm{t} _{1}=30 \mathrm{~s} $$

$\mathrm{z}\left(\right.$ at $\left. _{1}=30 \mathrm{~s}\right)=\mathrm{H}=$ maximum height of packet above ground. From Eqn. (2)

$$ \mathrm{H}=2000+300 \times 30-5 \times(30)^{2}=6500 \mathrm{~m}=6.5 \mathrm{~km} $$

(b) Let the packet hit ground at $\mathrm{t}=\mathrm{t} _{2}$ and after being dropped. The instantaneous $\mathrm{z} - \mathrm{co}-$ordinate of packet is zero. From Eqn. (2) we have

$$ \begin{aligned} & 0=2000+300 \mathrm{t} _{2}-5 \mathrm{t} _{2}^{2} \\ & \text { or } \mathrm{t} _{2}^{2}-60 \mathrm{t} _{2}-400=0 \\ & \therefore \mathrm{t} _{2}=\frac{60 \pm \sqrt{3600+1600}}{2} \\ & \quad=\frac{60 \pm 72.1}{2}=66.5 \mathrm{~s} \quad \text { [Negative value to } \mathrm{t} _{2} \text { is not possible.] } \end{aligned} $$

The packet hits ground $66.5 \mathrm{~s}$ after being dropped. The instantaneous velocity of packet is

$$ \mathrm{v}=300-660.5=-360.5 \mathrm{~ms}^{-1} $$

Negative sign of $v$ indicates packet is moving vertically downwards.

Solution :

Let $A$ and $B$ be the magnitude of the two given vectors $(A>B)$. $C$ is the magnitude of their resultant. Given $\mathrm{C} _{\max }=\mathrm{A}+\mathrm{B}=14 \quad$ and $\mathrm{C} _{\text {min }}=\mathrm{A}-\mathrm{B}=6$

$$ \therefore A=10 \text { and } B=4 $$

Let $\theta$ be angle between $\mathbf{A}$ and $\mathbf{B}$, so that the magnitude of their resultant, $\mathbf{C}$, is 12.5. From triangle law

$$ \begin{aligned} & (12.5)^{2}=(10)^{2}+(4)^{2}+2 \times 4 \times 10 \cos \theta \\ & \text { or } \cos \theta \simeq 0.867 \quad \therefore \theta=60^{\circ} \end{aligned} $$

Solution :

(a)

(b)

Fig. 31(a) shows river flowing at velocity $\mathbf{v} _{\mathbf{R}}$. The man is at point $A$ and wants to swim to point B. If man aims towards point $B$ the flow of river will make him reach opposite end at a point to right of $B$. Let man swim in a direction making an angle $\theta$ with line $\mathrm{AB}$ in such a manner that his resultant velocity is directed along $\mathrm{AB}$. This is shown in Fig. 31(b). Obviously.

$$ \sin \theta=\frac{\mathrm{v} _{\mathrm{R}}}{\mathrm{v} _{\mathrm{M}}}=\frac{2}{4}=0.5 \quad \therefore \theta=30^{\circ} $$

The magnitude, $\mathrm{v}$, of the resultant velocity is

$$ \mathrm{v}=\sqrt{\mathrm{v} _{\mathrm{M}}^{2}-\mathrm{v} _{\mathrm{R}}^{2}}=\sqrt{(4)^{2}-(2)^{2}}=2 \sqrt{3} \mathrm{kmhr}^{-1} $$

$t=$ The time taken by man to move from A to B

$$ =\frac{4 \mathrm{~km}}{2 \sqrt{3} \mathrm{kmhr}^{-1}}=\frac{2}{\sqrt{3}} \mathrm{hr} $$

(b) When man swims in a direction making an angle $\theta$ with the direction of flow of river, his resultant velocity v, has a magnitude;

$$ v=\sqrt{v _{\mathrm{M}}^{2}+v _{\mathrm{R}}^{2}+2 v _{\mathrm{m}} v _{\mathrm{R}} \cos \theta} $$

The man swims across river in minium time if $v$ is maximum. Obviously $v$ is maximum if $\theta=0^{0}$. Therefore The direction of $\mathrm{v} _{\max }$ is shown in Fig. 31. Let $\mathrm{t}$ be the time taken. Obviously

$$ 4=\mathrm{v} _{\mathrm{m}} \mathrm{t} \text { or } \mathrm{t}=\frac{4}{4}=1 \text { hour } $$

Fig. 32

The man will reach opposite end at point $C$, such that

$$ \mathrm{BC}=\mathrm{v} _{\mathrm{R}} \mathrm{t}=2 \times 1=2 \mathrm{~km} $$

Solution :

Fig. 33(a) and (b) show the velocity $\mathbf{v} _{\mathbf{M}}$ of man and river $\mathbf{v} _{\mathbf{R}}$ during forward and backward journey. The resultant $v _{1}$ and $v _{2}^{\prime}$ in the two cases have magnitude of

$$ \begin{aligned} & \mathrm{v} _{1}=\mathrm{v} _{\mathrm{M}}+\mathrm{v} _{\mathrm{R}}=5+3=8 \mathrm{kmhr}^{-1} \\ & \mathrm{v} _{\mathrm{i}}^{\prime}=\mathrm{v} _{\mathrm{M}}-\mathrm{v} _{\mathrm{R}}=2 \mathrm{kmhr}^{-1} \end{aligned} $$

$\mathrm{T} _{\mathrm{A}}=$ The total time taken by $\operatorname{man} \mathrm{A}=\frac{4}{8}+\frac{4}{2}=2.5 \mathrm{hr}$

(2) For Man B

Fig. 34(a) and 34(b) show $\mathbf{v} _{\mathbf{M}}$ and $\mathbf{v} _{\mathbf{R}}$ during forward and backward journey. The resultant velocity $\mathbf{v} _{2}$ and $\mathbf{v} _{2}^{\prime}$ have same magnitude and it is:

$$ \mathrm{v} _{2}=\sqrt{(5)^{2}-(3)^{2}}=4 \mathrm{kmhr}^{-1} $$

$\mathrm{T} _{\mathrm{B}}=$ Total time taken by man $\mathrm{B}=2\left[\frac{4}{4}\right]=2 \mathrm{~h}$

$\therefore \quad \frac{\mathrm{T} _{\mathrm{A}}}{\mathrm{T} _{\mathrm{B}}}=\frac{2.5}{2}=1.25$

Solution :

Choosing $x-\mathrm{y}$ axis as show in Fig. 36(a); A and B represent position of two cars at $\mathrm{t}=0$. Given $\mathrm{AB}=10 \mathrm{~km}$.

Also, $\mathbf{v} _{\text {BA }}=-20 \hat{\mathrm{i}} ; \quad \mathbf{v} _{\mathbf{B}}=20 \hat{\mathrm{j}}$

The relative velocity of $\mathrm{B}$ with respect to $\mathrm{A}$.

$$ \mathbf{v} _{\mathbf{B A}}=\mathbf{v} _{\mathbf{B}}-\mathbf{v} _{\mathbf{A}}=20(\hat{\mathrm{i}}+\hat{\mathrm{j}}) ; \mathrm{kmhr}^{-1} $$

$\mathbf{v} _{\text {BA }}$ is shown in Fig. 36(b). To driver of car A, car B appears to move along $\mathbf{v} _{\text {BA }}$. The minimum distance between to cars is AC; as shown in Fig. 36(c). Obviously

$$ \mathrm{AC}=\mathrm{AB} \sin 45^{\circ}=\frac{10}{\sqrt{2}} \mathrm{~km} \simeq 7.07 \mathrm{~km} $$

$\mathrm{T}=$ Time taken to acquire minimum distance

$$ =\frac{\mathrm{BC}}{\mathrm{v} _{\mathrm{BA}}}=\frac{\frac{10}{\sqrt{2}}}{20 \sqrt{2}}=\frac{1}{4} \mathrm{hr}=15 \text { mintue } $$

Solution :

Choose a Cartesian co-ordinate system as shown in Fig. 37(a). Let the velocity of wind be

$$ \mathbf{v} _{\mathbf{w}}=\mathrm{v} _{x} \hat{\mathrm{i}}+\mathrm{v} _{\mathrm{y}} \hat{\mathrm{j}} ; \mathrm{ms}^{-1} $$

$\mathrm{v} _{x}$ and $\mathrm{v} _{\mathrm{y}}$ are the $\mathrm{x}-$ and $\mathrm{y}$ - components of velocity of wind.

Case I: Given

$$ \begin{equation*} \mathbf{v} _{\mathbf{M}}=5 \hat{\mathrm{i}} ; \mathrm{ms}^{-1}, \mathbf{v} _{\mathbf{W M}}=\mathrm{v} _{\mathrm{RM}}(-\hat{\mathrm{j}}) \tag{i} \end{equation*} $$

We know,

$$ \begin{equation*} \mathbf{v} _{\mathbf{w M}}=\mathbf{v} _{\mathbf{w}}-\mathbf{v} _{\mathbf{m}}=\left(\mathrm{v} _{x}-5\right) \hat{\mathrm{i}}+\mathrm{v} _{y} \hat{\mathrm{j}} \tag{ii} \end{equation*} $$

From Eqns. (i) and (ii)

$$ \begin{equation*} \mathrm{v} _{x}-5=0 \quad \text { or } \quad \mathrm{v} _{x}=5 \mathrm{~ms}^{-1} \tag{iv} \end{equation*} $$

Case II:

$$ \begin{equation*} \mathbf{v} _{\mathrm{M}}^{\prime}=10 \hat{\mathrm{i}}, \mathbf{v} _{\mathrm{wM}}^{\prime}=\mathbf{v} _{\mathrm{wM}}\left[\frac{1}{\sqrt{2}}(-\hat{\mathrm{i}})+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}\right] \tag{v} \end{equation*} $$

Also, $\mathbf{v} _{\mathbf{M}}{ }^{\prime}=(5-10) \hat{\mathrm{i}}+\mathrm{v} _{\mathrm{y}} \hat{\mathrm{j}} \mathrm{\hspace{70mm}}(vi)$

From Eqns (v) and (vi); we have

$$ \mathrm{v} _{\mathrm{y}}=5-10 \text { or } \mathrm{v} _{\mathrm{y}}=-5 \mathrm{~ms}^{-1} $$

The velocity of wind, $\mathrm{v} _{\mathrm{w}}$; is

$$ \mathbf{v} _{\mathbf{w}}=(5 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}) \mathrm{ms}^{-1} $$

The speed of wind is $5 \sqrt{2} \mathrm{~ms}^{-1}$; and it is blowing in south-east direction.

Scalar or Dot Product of Vectors

Consider two vectors $\mathbf{A}$ and $\mathbf{B}$ making an angle $\theta$ with one another. The dot product $\mathbf{A} \cdot \mathbf{B}$ by definition is $\mathbf{A} \cdot \mathbf{B}=\mathrm{ab} \cos \theta$

It is a scalar quantity. In words, dot product of $\mathbf{A}$ and $\mathbf{B}$ is product of magnitude of $\mathbf{A}$ and component of $\mathbf{B}$ in the direction of $\mathbf{A}$ or vice-versa. The dot product of two non zero vectors, i.e. $A \neq 0, B \neq 0$ can be a positive or -ve number depending on $\theta$ whether is an acute or obtuse angle. If $\mathrm{A} \neq 0, \mathrm{~B} \neq 0$, but $\mathbf{A} \cdot \mathbf{B}=0 ; \cos \theta=0$ or $\theta=\frac{\pi}{2}$. Dot product of perpendicular vectors is zero. Dot product is commutative, i.e. $\mathbf{A} \cdot \mathbf{B}=\mathbf{B} \cdot \mathbf{A}$. Dot product is also distributive i.e.

$$ \mathbf{A}(\mathbf{B}+\mathbf{C})=\mathbf{A} \cdot \mathbf{B}+\mathbf{A} \cdot \mathbf{C} $$

Scalar product of a vector with itself equals square of magnitude of vector. i.e. $\mathbf{A} \cdot \mathbf{A}=A^{2}$

$\text { Let } \hat{\mathrm{i}}, \hat{\mathrm{j}} \text { and } \hat{\mathrm{k}} \text { be unit vectors along } x-, \mathrm{y} \text {-and } \mathrm{z} \text {-axis of a Cartesian co-ordinate system. Then }$

$\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}=0 ; \hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}=1$

In a Cartesian co-ordinate system; let

$$ \mathbf{A}=\mathrm{A} _{x} \hat{\mathrm{i}}+\mathrm{A} _{\mathrm{y}} \hat{\mathrm{j}}+\mathrm{A} _{z} \hat{\mathrm{k}} \quad \text { and } \quad \mathbf{B}=\mathrm{B} _{x} \hat{\mathrm{i}}+\mathrm{B} _{y} \hat{\mathrm{j}}+\mathrm{B} _{z} \hat{\mathrm{k}} $$

Then, $\mathbf{A} \cdot \mathbf{B}=\mathrm{A} _{x} \mathrm{~B} _{x}+\mathrm{A} _{\mathrm{y}} \mathrm{B} _{\mathrm{y}}+\mathrm{A} _{\mathrm{z}} \mathrm{B} _{\mathrm{z}}$

Solution :

$\mathbf{A}=|\mathbf{A}|=\sqrt{(2)^{2}+(4)^{2}+(-1)^{2}}=\sqrt{21}, \mathrm{~B}=|\mathbf{B}|=\sqrt{(-3)^{2}+(1)^{2}+(1)^{2}}=\sqrt{11}$

$\mathbf{A} \cdot \mathbf{B}=2 \times(-3)+4 \times 1+(-1) \times 1=-3$

Also, $\mathbf{A} \cdot \mathbf{B}=\mathrm{AB} \cos \theta$, therefore

$$ -3=\sqrt{21} \times \sqrt{11} \cos \theta $$

or $\cos \theta=-\left[\frac{3}{\sqrt{231}}\right] \simeq-\left[\frac{3}{15.1}\right] \simeq 0.2 \quad \theta=\cos ^{-1}(-0.2) \simeq 101.5^{0}$

Solution :

Let $\mathbf{B}=\hat{i}+\hat{j}$

$$ |\mathbf{A}|=\mathrm{A}=\sqrt{(3)^{2}+(2)^{2}}=\sqrt{13},|\mathbf{B}|=\mathrm{B}=\sqrt{(1)^{2}+(1)^{2}}=\sqrt{2} $$

$\mathbf{A} \cdot \mathbf{B}=3 \times 1+2 \times 1=5$

$=($ Components of $\mathbf{A}$ along $\mathbf{B}) \times B$

$\therefore$ Magnitude of component of $\mathbf{A}$ along $(\hat{i}+\hat{j})=\frac{5}{\sqrt{2}}$. Also, $\hat{B}=\frac{\mathbf{B}}{|\mathbf{B}|}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}$

In terms of $\hat{i}$ and $\hat{j}$; component of $\mathbf{A}$ along $(\hat{i}+\hat{j})$ is

$$ \frac{5}{\sqrt{2}} \hat{\mathrm{B}}=\frac{5}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}}) $$

Solution :

We know vector product of two parallel vector is zero. Now

$$ \begin{aligned} & \mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ -6 & 9 & 3 \end{array}\right| \\ & =\hat{i}(-9+9)+\hat{j}(6-6)+\hat{k}(18-18)=\text { zero } \end{aligned} $$

Hence $\mathbf{A}$ and $\mathbf{B}$ are parallel vectors.

Solution:

Let $\mathbf{C}=\mathbf{A} \times \mathbf{B}$. We know $\mathbf{C}$ is perpendicular to both $\mathbf{A}$ and $\mathbf{B}$. Therefore $\hat{\mathrm{c}}$ is a unit vector perpendicular to both $\mathbf{A}$ and $\mathbf{B}$. Now

$$ \begin{aligned} & \mathbf{C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right| \\ & =\hat{i}(2+1)+\hat{j}(1-4)+\hat{k}(-2-1)=3 \hat{i}-3 \hat{j}-3 \hat{k} \end{aligned} $$

Also, $|\mathbf{C}|=\sqrt{(3)^{2}+(-3)^{2}+(-3)^{2}}=3 \sqrt{3}$

$$ \therefore \hat{\mathrm{c}}=\frac{\mathbf{C}}{|\mathbf{C}|}=\frac{1}{\sqrt{3}}[\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}] $$

Solution :

Given $\mathrm{v} _{0}=100 \mathrm{~ms}^{-1}, \theta=30^{0}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$

(a) The time of flight, $T$, is

$$ \mathrm{T}=\frac{2 \mathrm{v} _{0} \sin \theta}{\mathrm{g}}=\frac{2 \times 100 \times \sin 30}{10}=10 \mathrm{~s} $$

(b) The horizontal range, $\mathrm{R}$, is

$$ \mathrm{R}=\frac{\mathrm{v} _{0}^{2} \sin 2 \theta}{\mathrm{g}}=\frac{(100)^{2} \times \sin 60}{10} \mathrm{~m} \simeq 867 \mathrm{~m} $$

(c) Let $\mathrm{v} _{x}$ and $\mathrm{v} _{\mathrm{y}}$ be the instantaneous $x$ - and $\mathrm{y}$-component of velocity at $\mathrm{t}=2 \mathrm{~s}$. Obviously

$$ \mathrm{v} _{x}=\mathrm{v} _{x 0}=100 \cos 30=86.7 \mathrm{~ms}^{-1} $$

$\mathrm{v} _{\mathrm{y}}=\mathrm{v} _{\mathrm{yo}}-\mathrm{gt}=\left(100 \sin 30^{\circ}\right)-10 \times 2$

$=50-20=30 \mathrm{~ms}^{-1}$

$\mathrm{v}=$ The instantaneous speed of projectile.

$=\sqrt{\mathrm{v} _{\mathrm{x}}^{2}+\mathrm{y} _{\mathrm{y}}^{2}}=\sqrt{(86.7)^{2}+(30)^{2}} \simeq 91.6 \mathrm{~ms}^{-1}$

Let $\alpha$ be the angle the instantaneous velocity $\mathbf{v}$ makes with horizontal direction. We know

$$ \tan \alpha=\frac{\mathrm{v} _{\mathrm{y}}}{\mathrm{v} _{\mathrm{x}}}=\frac{30}{86.7} \simeq 0.346 ; \quad \text { or } \quad \alpha=\tan ^{-1}(0.346) \simeq 10.9^{0} $$

Solution :

In Fig. 44, $O$ is point of projection of ball and $A B$ is the vertical wall. Choose $x$ - and $y$-axis at shown in Fig. 44. Obviously

$\mathrm{v} _{x 0}=\mathrm{v} _{0} \cos 60=\frac{\mathrm{v} _{0}}{2}$ and $\mathrm{v} _{\mathrm{y} 0}=\mathrm{v} _{0} \sin 60 \simeq 0.867 \mathrm{v} _{0}$

The co-ordinates of point $\mathrm{A}$ of wall are:

$$ x _{\mathrm{A}}=3 \mathrm{~m} ; \quad \mathrm{y} _{\mathrm{A}}=3 \tan 60^{\circ}=4.96 \mathrm{~m} $$

Let the ball graze past point $A$ at $t=0.2 \mathrm{~s}$. Obviously

$$ x _{\mathrm{A}}=3 \mathrm{~m}=\frac{\mathrm{v} _{0}}{2} \times 0.2, \quad \text { or } \quad \mathrm{v} _{0}=30 \mathrm{~ms}^{-1} $$

The vertical height, $h$, of wall equals y co-ordinate of projectile at $5=0.2 \mathrm{~s}$. Therefore,

$$ \begin{aligned} & \mathrm{h}=\left(\mathrm{v} _{0} \sin 60\right) \times 0.2-5(0.2)^{2} \\ & =30 \times 0.867 \times 0.2-0.2 \\ & =5.2-0.2 \simeq 5 \mathrm{~m} \end{aligned} $$

The maximum vertical height attained by the ball, $\mathrm{H}$, is

$$ \mathrm{H}=\frac{\mathrm{v} _{0}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{(30)^{2} \times(0.857)^{2}}{2 \times 10} \mathrm{~m} \simeq 33.8 \mathrm{~m} $$

Solution :

$\mathrm{v} _{x 0}=40 \cos 30^{\circ}=34.7 \mathrm{~ms}^{-1}$ and $\mathrm{v} _{\mathrm{yo}}=40 \sin 30^{\circ}=20 \mathrm{~ms}^{-1}$

The instantaneous y co-ordinate of stone, $y$, is

$$ \mathrm{y}=\left(\mathrm{v} _{0} \sin \theta\right) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}=20 \mathrm{t}-5 \mathrm{t}^{2} $$

Given $\mathrm{y}=15 \mathrm{~m}$. Therefore,

$$ \begin{aligned} & 15=20 \mathrm{t}-5 \mathrm{t}^{2} \text { or } \mathrm{t}^{2}-4 \mathrm{t}+3=0 \\ & \therefore \mathrm{t}=\frac{4 \pm \sqrt{16-12}}{2}=\frac{4+2}{2}=18 \text { or } 3 \mathrm{~s} \end{aligned} $$

The particle attains given vertical height at $t=1 \mathrm{~s}$, while going up towards position of maximum vertical height and at $\mathrm{t}=3 \mathrm{~s}$ while moving downwards. Let $x _{1}$ and $x _{2}$ be the horizontal co-ordinates of particle at $\mathrm{t}=1 \mathrm{~s}$ and $\mathrm{t}=3 \mathrm{~s}$ respectively. Obviously.

$x _{1}=\mathrm{v} _{x 0} \times 1=34.7 \mathrm{~m}$ and $x _{2}=34.7 \times 3=104.1 \mathrm{~m}$

Solution :

In Fig. 45, O is the position of particle on edge of table. Choose a co-ordinate system as shown in Fig. 45. For the ball

Fig. 45

$$ \begin{align*} & \mathrm{v} _{x 0}=\mathrm{v} _{0}= 10 \mathrm{~ms}^{-1}, \mathrm{v} _{\mathrm{yo}}=0 ; \text { and } \mathrm{a} _{\mathrm{y}}=+10 \mathrm{~ms}^{-1} \text { Therefore; } \\ & \mathrm{v} _{x}=\mathrm{v} _{x 0}=10 \mathrm{~ms}^{-1} \text { at all times } \\ & \mathrm{v} _{\mathrm{y}}=\mathrm{gt}=10 \mathrm{t \quad ms}^{-1} \tag{i} \\ & x=\mathrm{v} _{x 0} \times \mathrm{t}=10 \mathrm{t}, \text { and } \mathrm{y}=\frac{1}{2} \times 10 \mathrm{t}^{2}=5 \mathrm{t}^{2} \tag{ii} \end{align*} $$

From Eqns (ii) we have

$$ \begin{equation*} \mathrm{y}=5\left(\frac{x}{10}\right)^{2}=0.05 x^{2} \tag{iii} \end{equation*} $$

This is equation of trajectory of ball. Let ball hit ground at point $P$, at $t=t$. For point $P ; y=+2.5 \mathrm{~m}$

$$ \therefore 2.5=5 \mathrm{t}^{2} \quad \text { or } \quad \mathrm{t}=\frac{1}{\sqrt{2}} \simeq 0.7 \mathrm{~s} $$

$\mathrm{O} _{1} \mathrm{P}=$ Horizontal distance of $\mathrm{P}$ from the foot $\mathrm{O} _{1}$ of table.

$$ =\mathrm{v} _{x 0} \times \mathrm{t}=10 \times 0.7 \simeq 7 \mathrm{~m} $$

Let $\mathrm{v} _{x}$ and $\mathrm{v} _{\mathrm{y}}$ be the $x$ - and $\mathrm{y}$-components of velocity of ball as it hits ground. Obviously

$$ \begin{aligned} & \mathrm{v} _{x}=\mathrm{v} _{x 0}=10 \mathrm{~ms}^{-1} ; \quad \mathrm{v} _{\mathrm{y}}=10 \mathrm{x} 0.7 \mathrm{~ms}^{-1}=7 \mathrm{~ms}^{-1} \\ & \therefore \mathrm{v}=\sqrt{\mathrm{v} _{x}^{2}+\mathrm{v} _{\mathrm{y}}^{2}}=\sqrt{149} \mathrm{~ms}^{-1} \simeq 12.21 \mathrm{~ms}^{-1} \end{aligned} $$

Let $\alpha$ be the angle the instantaneous velocity makes with horizontal direction. Then:

$$ \tan \alpha=\frac{\mathrm{v} _{\mathrm{y}}}{\mathrm{v} _{x}}=0.7 \quad \text { or } \quad \alpha=\tan ^{-1}(0.7) \simeq 35^{\circ} $$

Solution :

In Fig. 46, A is position of point where the ball hits bat. Choose co-ordinate axis as shown in Fig. 46. Given,

$\mathrm{AB}=\mathrm{R}=106.7=\frac{\mathrm{v} _{0}^{2} \sin 90^{\circ}}{\mathrm{g}}$

$\therefore \mathrm{v} _{0}=\sqrt{1067} \mathrm{~ms}^{-1}=32 \mathrm{~ms}^{-1}$

Fig. 46

The question of trajectory of ball is

$$ \begin{align*} & \mathrm{y}=1.22+(\tan 45) x-\frac{10 x^{2}}{2\left(\mathrm{v} _{0} \cos 45\right)^{2}} \\ & =1.22+x-\left[\frac{10 x^{2}}{2 \times\left(\frac{1067}{2}\right)}\right] \\ & =1.22+x-\left[\frac{10 x^{2}}{1067}\right] \tag{i} \end{align*} $$

When ball is at $x=97.5 \mathrm{~m}$; its $\mathrm{y}$-co-ordinate is

$$ y=1.22+97.5-\frac{10(97.5)^{2}}{106.7}=9.58 \mathrm{~m} $$

The height of wall at $x=97.5 \mathrm{~m}$ is $7.3 \mathrm{~m}$. Since $\mathrm{y}>7.3 \mathrm{~m}$; the ball will clear the wall.

Solution:

(1) Given $\omega _{0}=0 ; \quad \omega=120 \mathrm{rpm}=\frac{120}{60} \mathrm{rps}=2 \times 2 \pi \mathrm{rad} \quad \mathrm{s}^{-1} \quad \mathrm{t}=\pi, \mathrm{s} ; \quad \alpha=$ ?

Using equation $\omega _{0}=\omega _{0}+\alpha$ t, we have

$4 \pi=\alpha . \pi \quad$ or $\quad \alpha=4 \quad \mathrm{rad \quad s}^{-2}$

$\mathrm{a}=$ The magnitude of linear acceleration

$=\mathrm{R} \alpha=0.2 \times 4 \mathrm{~ms}^{-2}=0.8 \mathrm{~ms}^{-2}$

(2) The total angle described $\theta$, is

$\theta=\omega _{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}=0+\frac{1}{2} \times 4 \times(\pi)^{2}=2 \pi^{2}$ radian

Number of revolution completed $=\frac{\theta}{2 \pi}=\frac{2 \pi^{2}}{2 \pi}=\pi$

The linear distance traveled, $\mathrm{S}$; is

$$ \mathrm{S}=\mathrm{R} \theta=0.2 \times 2 \pi^{2} \simeq 4 \mathrm{~m} \quad\left[\because \pi^{2} \simeq 10\right] $$

Solution :

$\omega$ =Angular speed of second’s hand of clock

$$ =\frac{2 \pi}{60} \mathrm{rad} \mathrm{s}^{-1}=\frac{\pi}{30} \mathrm{rad} \mathrm{s}^{-1} $$

$v=$ The linear speed of tip of second’s hand of clock

$$ =\ell \omega ;=0.2 \times \frac{\pi}{30} \mathrm{~ms}^{-1} \quad[\text { Given, } \ell=20 \mathrm{~cm}=0.2 \mathrm{~m}] $$

Let the seconds hand of clock rotate through an angle $\theta$ in time t. We have $\theta=\omega t$. In Fig. 52(a) A and B denote initial and final position of tip of second’s hand. $\mathbf{v} _{\mathbf{A}}$ and $\mathbf{v} _{\mathbf{B}}$ is the linear velocity of tip of hand in position A and B.

$\Delta \mathbf{v}=$ The change in linear velocity $=\mathbf{v} _{\mathbf{B}}-\mathbf{v} _{\mathbf{A}}$

From Fig. 52(b); $|\Delta \mathbf{v}|=M N=2 \mathrm{v} \sin \left(\frac{\theta}{2}\right)$. Therefore

$$ \begin{aligned} & 0.67 \pi \times 10^{-2}=0.2 \times \frac{\pi}{30} \sin \left(\frac{\theta}{2}\right), \text { or } \sin \left(\frac{\theta}{2}\right)=\frac{1}{2} \\ & \therefore \theta=60^{\circ}=\frac{\pi}{3} \text { radian } \end{aligned} $$

Now, $\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi \times 30}{3 \times \pi}=10 \mathrm{~s}$

Solution :

Let $\mathrm{v} _{\max }$ be the maximum linear speed of particle so that string does not break. The tension in string equals to breaking strength of string. The tension in string provides necessary centripetal force on particle. Therefore

$$ \mathrm{T} _{\max }=\frac{\mathrm{mv} _{\max }^{2}}{\ell} $$

Given, $\mathrm{T} _{\text {max }}=1.2 \mathrm{kgf}=12 \mathrm{~N} ; \quad \ell=1.5 \mathrm{~m} ; \quad \mathrm{m}=0.5 \mathrm{~kg}$

$\therefore \mathrm{v} _{\text {max }}=\sqrt{\frac{\mathrm{T} _{\text {max }} \ell}{\mathrm{m}}}=\sqrt{\frac{12 \times 1.5}{0.5}}=6 \mathrm{~ms}^{-1}$

$\omega _{\max }=$ The maximum angular speed of particle $=\frac{\mathrm{v} _{\max }}{\ell}=\frac{6}{1.5}=4 \mathrm{rad} / \mathrm{s}$

$\therefore \mathrm{n} _{\max }=$ The maximum number of revolutions $=\frac{\omega _{\max }}{2 \pi}=\frac{4}{2 \pi} \mathrm{rps}=\frac{4 \times 60}{2 \pi} \mathrm{rpm}$

$$ =\frac{120}{\pi} \mathrm{rpm} $$

Solution :

$\omega=$ The instantaneous angular speed of particle

$$ \begin{equation*} =\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{kt} \tag{1} \end{equation*} $$

As $\omega$ varies with time; the particle has a non-uniform circular motion. The instantaneous linear acceleration, $\mathrm{a}$; has a radial component, $\mathrm{a} _{\mathrm{r}}$; and tangential component $\mathrm{a} _{\mathrm{t}}$. We know

$$ \mathrm{a} _{\mathrm{r}}=\mathrm{R} \omega^{2}=4 \mathrm{k}^{2} \mathrm{Rt}^{2}, \quad \text { and } \quad \mathrm{a} _{\mathrm{t}}=\mathrm{R} \alpha=\mathrm{R}\left(\frac{\mathrm{d} \omega}{\mathrm{dt}}\right)=2 \mathrm{kR} $$

The magnitude of instantaneous linear acceleration, $a$, is

$$ \begin{aligned} & \mathrm{a}=\sqrt{\mathrm{a} _{\mathrm{r}}^{2}+\mathrm{a} _{\mathrm{t}}^{2}}=\sqrt{\left(4 \mathrm{k}^{2} \mathrm{Rt}^{2}\right)^{2}+(2 \mathrm{kR})^{2}} \\ & =2 \mathrm{kR} \sqrt{1+4 \mathrm{k}^{2} \mathrm{t}^{2}} \end{aligned} $$

Correct option is (1)

(IIT 2014)

Solution:

We know that the particle will take a time $\mathrm{t} _{1}\left(=\frac{\mathrm{u}}{\mathrm{g}}\right)$ to reach the highest point $($ for which $\mathrm{v}=0$ ). Let $\mathrm{t} _{2}$ be the time taken by the particle to hit the ground. Then

$$ -\mathrm{H}=\mathrm{ut} _{2}+\frac{1}{2}(-\mathrm{g}) \mathrm{t} _{2}^{2} $$

We are given that $\mathrm{t} _{2}=\mathrm{nt} _{1}$. This gives

$$ 2 \mathrm{gH}=\mathrm{nu}^{2}(\mathrm{n}-2) $$

Correct option is (1)

(IIT 2016)

Solution:

There is no acceleration along the horizontal direction. Therefore

$$ x=\mathrm{v} _{o x} \mathrm{t}=1 \times \mathrm{t}=\mathrm{t} $$

Also, $y=v _{\text {oy }} \mathrm{t}+\frac{1}{2}(-\mathrm{g}) \mathrm{t}^{2}=2 \mathrm{t}-5 \mathrm{t}^{2}$

Eliminating $\mathrm{t}$ we get

$$ y=2 x-5 x^{2} $$

This is the equation of the trajectory.

Correct option is (1)

(IIT 2015)

Solution:

For the first object, we have

$\mathrm{y} _{1}=10 \mathrm{t}+\frac{1}{2}(-10) \mathrm{t}^{2}$

If it takes a time $t _{1}$ to reach the ground, we have

$-240=10 \mathrm{t} _{1}=5 \mathrm{t} _{1}^{2} \quad$ or $\quad \mathrm{t} _{1}=8 \mathrm{~s}$

For the second object,

$\mathrm{y} _{2}=40 \mathrm{t}-5 \mathrm{t}^{2}$

Time $t _{2}$, taken by it to reach the ground, comes out to be 12 s. The relative displacement of the second object, with respect to the first, is

$$ \mathrm{y} _{21}=\mathrm{y} _{2}-\mathrm{y} _{1}=30 \mathrm{t} \text { for }(0<\mathrm{t}<8) $$

This is equation of a straight line.

For $8<\mathrm{t}<12$, we would have

$\mathrm{y} _{21}=40 \mathrm{t}-5 \mathrm{t}^{2}$

This is equation of a parabola. The graph would have the shape shown in (1).

Correct answer: (2)

Solution:

In the time interval considered acceleration is not uniform. Eqn (iii) is equation of a uniformly accelerated motion.

Correct answer: (3)

Solution:

Let $\mathrm{S}$ and $\mathrm{T}$ denote the total distance and the total time of journey. $\mathrm{S} _{1}=0.4 \mathrm{~S}=\frac{25}{5}$ is the distance travelled with constant speed $v$. The time taken, $t _{1}$, is

$$ \begin{equation*} \mathrm{t} _{1}=\frac{2}{5}\left(\frac{\mathrm{S}}{\mathrm{v}}\right) \tag{i} \end{equation*} $$

During the second part of the journey, the time taken $=\mathrm{t} _{2}=\frac{\mathrm{T}}{2}$

The distance travelled $\mathrm{S} _{2}$ is

$$ \begin{equation*} \mathrm{S} _{2}=2 \mathrm{vt} _{2}=\mathrm{vT} \tag{ii} \end{equation*} $$

Let $\mathrm{S} _{3}$ be the distance traveled during the remaining journey and $\mathrm{t} _{3}$ be time taken. Obviously

$$ S _{3}=S-\left[\frac{2 S}{5}+v T\right]=\frac{3 S}{5}-v T $$

and $\quad \mathrm{t} _{3}=\frac{\mathrm{S} _{3}}{3 \mathrm{v}}=\frac{\mathrm{S}}{5 \mathrm{v}}-\frac{\mathrm{T}}{3}$

Now, $\mathrm{T}=\mathrm{t} _{1}+\frac{\mathrm{T}}{2}+\mathrm{t} _{3}$

or $\frac{\mathrm{T}}{2}=\mathrm{t} _{1}+\mathrm{t} _{3}=\frac{2 \mathrm{~S}}{5 \mathrm{v}}+\left(\frac{\mathrm{S}}{5 \mathrm{v}}-\frac{\mathrm{T}}{3}\right)$

or $\frac{5 \mathrm{~T}}{6}=\frac{3 \mathrm{~S}}{5 \mathrm{v}}$

The average speed of the complete journey is

$$ \overline{\mathrm{v}}=\frac{\mathrm{S}}{\mathrm{T}}=\left(\frac{25}{15}\right) \mathrm{v} $$

Correct answer: (1)

Solution:

Let $\mathrm{V}$ and $\mathrm{V} _{\mathrm{R}}$ denote speed of motor boat and river respectively. When motor boat moves down the stream its resultant velocity is $\left(\mathrm{V}+\mathrm{V} _{\mathrm{R}}\right)$ and when it moves up the stream resultant velocity is $\left(\mathrm{V}-\mathrm{V} _{\mathrm{R}}\right)$. Let $\mathrm{t}$ be time taken by motor boat, after it turns back to meet raft at $\mathrm{Q}$. The total distance $\mathrm{S}$ traveled by boat is:

$$ \begin{equation*} \mathrm{S}=\left(\mathrm{V}+\mathrm{V} _{\mathrm{R}}\right) \frac{1}{2}+\left(\mathrm{V}-\mathrm{V} _{\mathrm{R}}\right) \mathrm{t} \tag{i} \end{equation*} $$

where $\mathrm{V}$ and $\mathrm{V} _{\mathrm{R}}$ are expressed in $\mathrm{kmhr}^{-1}$. Also

$$ \begin{align*} & \mathrm{S}=\text { average speed of boat } \times \text { total time taken } \\ & =\left[\frac{\left(\mathrm{V}+\mathrm{V} _{\mathrm{R}}\right)+\left(\mathrm{V}-\mathrm{V} _{\mathrm{R}}\right)}{2}\right]\left[\frac{1}{2}+\mathrm{t}\right] \\ & =\mathrm{V}\left(\frac{1}{2}+\mathrm{t}\right) \tag{ii} \end{align*} $$

From Eqns (i) and (ii) we have

$$ \begin{aligned} & \left(V+V _{R}\right) \frac{1}{2}+\left(V-V _{R}\right) t=V\left(\frac{1}{2}+t\right) \\ & \text { or } V _{R}+2\left(V-V _{R}\right) t=2 V t \\ & \text { or } V _{R}(1-2 t)=0 \end{aligned} $$

A program to give wings to girl students $\therefore \mathrm{t}=\frac{1}{2} \mathrm{hr}$ $\left[\because \mathrm{V} _{\mathrm{R}} \neq 0\right]$

The total time taken by motor boat $=\frac{1}{2}+\frac{1}{2}=1 \mathrm{hr}$. In this time - raft has traveled a distance $=\mathrm{PQ}$ $=4 \mathrm{~km}$. The raft moves with same speed as river. Hence

$$ \begin{aligned} & V _{R}=\text { Speed of river }=\frac{\text { Total distance travelled }}{\text { Total time taken }} \\ & =4 \mathrm{kmhr}^{-1} \end{aligned} $$

Correct answer: (1)

Solution:

$\mathrm{s} _{\mathrm{n}}=$ distance travelled in $\mathrm{n}^{\text {th }}$ second by a particle starting from rest having uniform acceleration a

$$ =\frac{\mathrm{a}}{2}[2 \mathrm{n}-1] $$

Time interval $t=4 \mathrm{~s}$ to $\mathrm{t}=5 \mathrm{~s}$ is $5^{\text {th }}$ second of motion. Therefore

$$ s=\frac{a}{2}[2 \times 5-1]=\frac{9 \mathrm{a}}{2} $$

Similarly time interval $\mathrm{t}=\mathrm{n}$ and $\mathrm{t}=\mathrm{n}+1$ is $(\mathrm{n}+1)^{\mathrm{th}}$ second of motion. Therefore

$$ \mathrm{s} _{1}=\frac{\mathrm{a}}{2}[2(\mathrm{n}+1)-1]=\frac{\mathrm{a}}{2}[2 \mathrm{n}+1] $$

Given, $\frac{\mathrm{s} _{1}}{\mathrm{~s}}=\frac{15}{9}=\frac{2 \mathrm{n}+1}{9}$

$\therefore 15=2 \mathrm{n}+1 \quad$ or $\quad \mathrm{n}=7$

Correct answer : (4)

Solution:

Let ’ $\mathrm{g}$ ’ denote acceleration due to gravity. We know,

$$ \mathrm{T}=\frac{2 \mathrm{U}}{\mathrm{g}} \text { and } \mathrm{H}=\frac{\mathrm{U}^{2}}{2 \mathrm{~g}} $$

The new speed of projection $U^{\prime}$ is

$$ \mathrm{U}^{\prime}=(1+\mathrm{n}) \mathrm{U} $$

Therefore, $\mathrm{T}^{\prime}=\frac{2 \mathrm{U}^{\prime}}{\mathrm{g}}=(1+\mathrm{n}) \mathrm{T} \quad$ and $\quad \mathrm{H}^{\prime}=\frac{\left(\mathrm{U}^{\prime}\right)^{2}}{2 \mathrm{~g}}=(1+\mathrm{n})^{2} \mathrm{H}$

The percentage change in $\mathrm{H}=\alpha=\left(\frac{\mathrm{H}^{\prime}-\mathrm{H}}{\mathrm{H}}\right) \times 100=\left[(1+n)^{2}-1\right] \times 100=\left[\mathrm{n}^{2}+2 \mathrm{n}\right] \times 100$

The percentage change in $\mathrm{T}=\beta=\frac{\mathrm{T}^{\prime}-\mathrm{T}}{\mathrm{T}} \times 100=\mathrm{n} \times 100$

Obviously $\frac{\alpha}{\beta}=\frac{n(1+2 n)}{n}=1+2 n$

Correct answer: (4)

Solution:

Given,

$$ \begin{equation*} \mathrm{v}=\frac{\mathrm{dz}}{\mathrm{dt}}=\mathrm{kz}^{1 / 2} \tag{i} \end{equation*} $$

or $\mathrm{z}^{-1 / 2} \mathrm{dz}=\mathrm{kdt}$

Integrate,

$$ \begin{aligned} & \int _{0}^{z} z^{-1 / 2} d z=k \int _{0}^{t} d t \\ & 2 z^{1 / 2}=k t \end{aligned} $$

or $\mathrm{z}=\frac{\mathrm{k}^{2} \mathrm{t}^{2}}{4} \hspace{50mm} . .. . . . .(ii)$

$t=$ The time taken to move from $z=0$ to $z=z$

$=\frac{2}{\mathrm{k}} \sqrt{\mathrm{z}}$

$\mathrm{v} _{\mathrm{av}}=$ The average velocity $=\frac{\mathrm{z}}{\mathrm{t}}$

$$ \begin{equation*} =\frac{1}{2} \mathrm{k} \sqrt{\mathrm{z}} \tag{iii} \end{equation*} $$

$\mathrm{v}=$ The instantaneous velocity at $\mathrm{z}=\mathrm{z}$ is given by Eqn (i). Therefore,

$$ \frac{\mathrm{v}}{\mathrm{v} _{\mathrm{av}}}=\frac{\mathrm{k} \sqrt{\mathrm{z}}}{\frac{1}{2} \mathrm{k} \sqrt{\mathrm{z}}}=2 $$

Correct answer: (3)

Solution:

The equation of given straight line is

$$ \ln (v)=\ln \left(v _{0}\right)+\left(\tan 120^{\circ}\right) \mathrm{t} $$

$$ =\ln \left(v _{0}\right)-\sqrt{3} t $$

or $\quad \ln \left(\frac{\mathrm{v}}{\mathrm{v} _{0}}\right)=-\sqrt{3} \mathrm{t}$

$$ \begin{equation*} \therefore \mathrm{v}=\mathrm{v} _{0} \mathrm{e}^{-\sqrt{3} \mathrm{t}} \tag{i} \end{equation*} $$

At $\quad \mathrm{t}=\frac{1}{\sqrt{3}} \mathrm{~s} ; \quad \mathrm{v}=\mathrm{v} _{0} \mathrm{e}^{-1}=\frac{\mathrm{v} _{0}}{\mathrm{e}}$

Solution:

The instantaneous position co-ordiante $x _{1}$ and $x _{2}$ of the two particles are

$$ \begin{align*} & x _{1}=\frac{1}{2} \times 2\left(\mathrm{t}^{2}\right)=\mathrm{t}^{2} \tag{i} \\ & x _{2}=2+1 \times \mathrm{t}=2+\mathrm{t} \tag{ii} \end{align*} $$

Obviously,

$$ \begin{equation*} x=x _{1}-x _{2}=\mathrm{t}^{2}-2-\mathrm{t} \tag{iii} \end{equation*} $$

$x$ vs $\mathrm{t}$ graph is parabolic in nature. $x$ is a negative number at $\mathrm{t}=0 . x$ is maximum when

$$ \begin{aligned} & \frac{\mathrm{d} x}{\mathrm{dt}}=0=2 \mathrm{t}-1 \\ & \text { or } \mathrm{t}=0.5 \mathrm{~s} \\ & x _{\max }=0.25-2-0.5=-2.25 \mathrm{~m} \end{aligned} $$

Also, $x=0$ if

$\mathrm{t}^{2}-\mathrm{t}-2=0$

or $\quad \mathrm{t}=\frac{1 \pm \sqrt{1+8}}{2}=\frac{1+3}{2}=2 \mathrm{~s}$

For $\mathrm{t}>2 \mathrm{~s} ; x$ is $\mathrm{a}+\mathrm{ve}$ number and varies as a parabolic function of t. These characteristics are shown in (4).

Correct answer: (2)

Solution:

Choosing the point of projection of balls as origin of co-ordinates and vertically upward direction as direction of $+\mathrm{z}-$ axis the instantaneous position co-ordinates $\mathrm{z} _{1}$ and $\mathrm{z} _{2}$ of the two balls is

$$ \begin{aligned} & \mathrm{z} _{1}=\mathrm{vt}-\frac{1}{2} g \mathrm{t}^{2} \\ & \mathrm{z} _{2}=\mathrm{v}(\mathrm{t}-\mathrm{n})-\frac{1}{2} \mathrm{~g}(\mathrm{t}-\mathrm{n})^{2} \end{aligned} $$

When the two balls meet one another $\mathrm{z} _{1}=\mathrm{z} _{2}$

$$ \begin{aligned} & \therefore \mathrm{vt}-\frac{1}{2} \mathrm{gt}^{2}=\mathrm{v}(\mathrm{t}-\mathrm{n})-\frac{1}{2} \mathrm{~g}(\mathrm{t}-\mathrm{n})^{2} \\ & 0=-\mathrm{nv}-\frac{1}{2} \mathrm{~g}\left(\mathrm{n}^{2}-2 \mathrm{nt}\right) \end{aligned} $$

or $\mathrm{t}=\frac{\mathrm{n}}{2}+\frac{\mathrm{v}}{\mathrm{g}}$

The first ball will hit ground at $t=\frac{2 v}{g}$. The maximum value of $t$ is $\frac{2 v}{g}$. Hence

$$ \begin{aligned} & \frac{2 \mathrm{v}}{\mathrm{g}}=\frac{\mathrm{n} _{\max }}{2}+\frac{\mathrm{v}}{\mathrm{g}} \\ & \therefore \mathrm{n} _{\max }=\frac{2 \mathrm{v}}{\mathrm{g}} \end{aligned} $$

Correct answer: (2)

Solution:

Let $t _{1}$ be time taken by the ball to move freely under gravity from $\mathrm{O}$ to $\mathrm{P}$. Obviously

$$ \begin{align*} & \mathrm{H}-\mathrm{h}=\frac{1}{2} \mathrm{gt} _{1}^{2} \\ & \therefore \mathrm{t} _{1}=\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}} \tag{i} \end{align*} $$

After impact at $\mathrm{P}$, velocity of particle becomes horizontal, i.e. its vertical component is zero. Let it take a time $\mathrm{t} _{2}$ to hit ground. Obviously

$$ \begin{align*} & \mathrm{h}=\frac{1}{2} \mathrm{gt} _{2}{ }^{2} \\ & \therefore \mathrm{t} _{1}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \tag{ii} \end{align*} $$

Now, $\mathrm{t}=\mathrm{t} _{1}+\mathrm{t} _{2}=\sqrt{\frac{2(\mathrm{H}-\mathrm{h})}{\mathrm{g}}}+\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}$

$\mathrm{t}$ is maximum if $\frac{\mathrm{dt}}{\mathrm{dh}}=0$. This gives

$$ \frac{\mathrm{h}}{\mathrm{H}}=\frac{1}{2}=0.5 $$

Correct answer: (3)

Solution:

Since body has uniformly accelerated motion; starting from rest.

$$ \begin{equation*} S=\frac{1}{2} \alpha \tau^{2} \tag{i} \end{equation*} $$

Let $\mathrm{t}^{\prime}$ be time taken by body to come to rest due to retardation $2 \alpha$. Obviously $\mathrm{t}^{\prime}=\frac{\tau}{2}$. The distance travelled $\mathrm{S}^{\prime}$, during this time is

$$ \begin{align*} & S^{\prime}=(\alpha \tau) \frac{\tau}{2}+\frac{1}{2}(-2 \alpha)\left(\frac{\tau}{2}\right)^{2} \\ & =\frac{\alpha \tau^{2}}{2}-\frac{\alpha \tau^{2}}{4}=\frac{\alpha \tau^{2}}{4}=\frac{S}{2} \tag{ii} \end{align*} $$

Let $S _{1}$, be the distance travelled with constant speed $v=\alpha \tau$. Then $S _{1}=(\alpha \tau) t$. Given,

$$ \begin{aligned} & 3 \mathrm{~S}=\mathrm{S}+\mathrm{S} _{1}+\mathrm{S}^{\prime}=\mathrm{S}+\mathrm{S} _{1}+\frac{\mathrm{S}}{2} \\ & \therefore \mathrm{S} _{1}=\frac{3}{2} \mathrm{~S} \\ & \therefore \mathrm{t}=\frac{\mathrm{S} _{1}}{\alpha \tau}=\frac{3}{2}\left(\frac{\mathrm{S}}{\alpha \tau}\right)=\frac{3}{4} \tau \end{aligned} $$

or $\frac{\mathrm{t}}{\tau}=\frac{3}{4}$

Correct answer: (2)

Solution:

1. For time interval $0 \leq \mathrm{t} \leq 1 \mathrm{~s}$; the motion is uniformly accelerated with an acceleration $\mathrm{a} _{1}=2 \mathrm{~ms}^{-2}$. The instantaneous positive co-ordinate $x$ is

$$ \begin{equation*} x=\frac{2}{2} \times 2 \times \mathrm{t}^{2}=\mathrm{t}^{2} \tag{1} \end{equation*} $$

Eqn (1) is equation of parabola. At $\mathrm{t} _{1}=1 \mathrm{~s} ; x=1 \mathrm{~m}$.

2. For time interval $1 \leq \mathrm{t} \leq 2 \mathrm{~s}$; the particle has uniform motion with speed $\mathrm{v}=\mathrm{v} _{1}=2 \mathrm{~ms}^{-1}$. The instantaneous position co-ordinate $x$; in this time interval is given by:

$$ \begin{align*} & x=x _{1}+v _{1}(\mathrm{t}-1) \\ & =1+2(\mathrm{t}-1) \tag{2} \end{align*} $$

Eqn (2) is equation of a straight line. At $\mathrm{t} _{2}=2 \mathrm{~s}, x _{2}=1+2(2-1)=3 \mathrm{~m}$.

3. For time interval $2 \leq \mathrm{t} \leq 4 \mathrm{~s}$; the motion is uniformly retarded with retardation $\mathrm{a} _{2}=2 \mathrm{~ms}^{-2}$. The instantaneous position co-ordinate of particle in this time interval is given by

$$ \begin{align*} & x=x _{2}+\left[2(\mathrm{t}-2)+\frac{1}{2}(-2)(\mathrm{t}-2)^{2}\right] \\ & =3+2(\mathrm{t}-2)-(\mathrm{t}-2)^{2} \tag{3} \end{align*} $$

Eqn (3) is equation of parable. At $\mathrm{t} _{3}=3 \mathrm{~s}, \mathrm{v} _{3}=0$ and

$$ x _{3}=3+2(3-2)-(3-2)^{2}=4 $$

At $\mathrm{t} _{4}=4 \mathrm{~s} ; \mathrm{v} _{4}=-2 \mathrm{~ms}^{-1}$. The position co-ordinate $x _{4}$ is

$x _{4}=3+2(4-2)-(4-2)^{2}=3+4-4=3 \mathrm{~m}$

The characteristics are correctly shown in (2).

Correct answer: (2)

Solution:

The equation representing given $\mathrm{v}$ vs $x$ graph is

$$ \begin{align*} & \mathrm{v}=\mathrm{v} _{0}-\left(\frac{\mathrm{v} _{0}}{\alpha}\right) x \\ & =\mathrm{v} _{0}\left[1-\frac{x}{\alpha}\right] \tag{1} \end{align*} $$

The instantaneous acceleration; $a$, is

$$ \begin{align*} & \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\left(\frac{\mathrm{dv}}{\mathrm{d} x}\right)\left(\frac{\mathrm{d} x}{\mathrm{dt}}\right)=\mathrm{v}\left(\frac{\mathrm{dv}}{\mathrm{d} x}\right) \\ & =\mathrm{v} _{0}\left[1-\frac{x}{\alpha}\right]\left[-\frac{\mathrm{v} _{0}}{\alpha}\right] \\ & =\left(\frac{\mathrm{v} _{0}^{2}}{\alpha}\right)+\left(\frac{\mathrm{v} _{0}^{2}}{\alpha^{2}}\right) x \tag{2} \end{align*} $$

Eqn (2) is equation of a straight line with positive slope equal to $\frac{v _{0}^{2}}{\alpha^{2}}$ and a negative intercept on acceleration axis. Also at $x=\alpha ; \mathrm{a}=0$ and at $x=2 \alpha ; \mathrm{a}=\frac{\mathrm{v} _{0}^{2}}{\alpha}$. These characteristics are shown in (2).

Correct answer:

Solution:

For the co-ordinate system shown, let $\mathrm{z}$ be the instantaneous position co-ordinate of particle. Obviously.

$$ \begin{equation*} \mathrm{z}=30 \mathrm{t}-5 \mathrm{t}^{2} \tag{1} \end{equation*} $$

Let the particle be at point $\mathrm{P}$ at $\mathrm{t}=\mathrm{t} _{\mathrm{P}}$. At this point $\mathrm{z}=\mathrm{z} _{\mathrm{P}}=+25 \mathrm{~m}$

$$ \begin{aligned} & \therefore 25=30 \mathrm{t} _{\mathrm{p}}-5 \mathrm{t} _{\mathrm{p}}^{2} \\ & \text { or } \mathrm{t} _{\mathrm{p}}^{2}-6 \mathrm{t} _{\mathrm{p}}+5=0 \\ & \therefore \mathrm{t} _{\mathrm{p}}=\frac{6 \pm \sqrt{36-20}}{2}=\frac{6 \pm 4}{2}=1 \text { or } 5 \mathrm{~s} \end{aligned} $$

Let particle be at point $\mathrm{Q}$ at $\mathrm{t}=\mathrm{t} _{\mathrm{Q}}$. Then $\mathrm{z}=\mathrm{z} _{\mathrm{Q}}=40 \mathrm{~m}$.

$$ \therefore 40=30 \mathrm{t} _{\mathrm{Q}}-5 \mathrm{t} _{\mathrm{Q}}^{2} \text { or } \mathrm{t} _{\mathrm{Q}}^{2}-6 \mathrm{t} _{\mathrm{Q}}+8=0 $$

or $\quad \mathrm{t} _{\mathrm{Q}}=\frac{6 \pm \sqrt{36-32}}{2}=\frac{6 \pm 2}{2}=2$ or $5 \mathrm{~s}$

Similarly for point $\mathrm{R} ; \mathrm{t}=\mathrm{t} _{\mathrm{R}}$ and $\mathrm{z}=\mathrm{z} _{\mathrm{R}}=+10 \mathrm{~m}$. Therefore

$$ \begin{aligned} & 10=30 \mathrm{t} _{\mathrm{R}}-5 \mathrm{t} _{\mathrm{R}}^{2} \text { or } \mathrm{t} _{\mathrm{R}}^{2}-6 \mathrm{t} _{\mathrm{R}}+2=0 \\ & \therefore \mathrm{t} _{\mathrm{R}}=\frac{6 \pm \sqrt{36-8}}{2} \approx \frac{6 \pm 5.3}{2}=0.35 \text { or } 5.665 \mathrm{~s} \end{aligned} $$

Obviously during upward journey

$$ \mathrm{t} _{1}=\mathrm{t} _{\mathrm{Q}}-\mathrm{t} _{\mathrm{p}}=2-1=1 \mathrm{~s} $$

and during downward journey

$$ \begin{aligned} & \mathrm{t} _{2}=\mathrm{t} _{\mathrm{R}}-\mathrm{t} _{\mathrm{p}} \\ & =5.675-5=0.665 \mathrm{~s} \\ & \therefore \frac{\mathrm{t} _{1}}{\mathrm{t} _{2}}=\frac{1}{0.665} \simeq \frac{3}{2} \end{aligned} $$

Correct answer: (1)

Solution:

The instantaneous velocity $v$ of particle is

$$ \begin{equation*} \mathrm{v}=\mathrm{v} _{0}+\mathrm{at}=\mathrm{v} _{0}+\left(\mathrm{a} _{0}-\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right) \mathrm{t} \tag{1} \end{equation*} $$

The instantaneous velocity v is zero when particle is momentarily at rest. From Eqn. (1) we get

$$ 0=\mathrm{v} _{0}+\mathrm{a} _{0} \mathrm{t}-\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}} \mathrm{t}^{2} $$

or $\quad\left(\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right) \mathrm{t}^{2}-\mathrm{a} _{0} \mathrm{t}-\mathrm{v} _{0}=0$

$$ \therefore \quad \mathrm{t}=\frac{\mathrm{a} _{0} \pm \sqrt{\mathrm{a} _{0}^{2}+4\left(\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right) \mathrm{v} _{0}}}{\frac{4 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}}=\frac{\mathrm{a} _{0} \pm 3 \mathrm{a} _{0}}{\left(\frac{4 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right)} $$

or $\quad \mathrm{t}=\frac{4 \mathrm{a} _{0}}{\left(\frac{4 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right)}=\frac{\mathrm{v} _{0}}{\mathrm{a} _{0}}$

To obtain the instantaneous position co-ordinate $x$ of particle we have,

$$ \begin{align*} & \mathrm{v}=\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{v} _{0}+\mathrm{a} _{0} \mathrm{t}-\left(\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right) \mathrm{t}^{2} \\ & \text { or } \int _{0}^{x} \mathrm{~d} x=\int _{0}^{\mathrm{t}}\left[\mathrm{v} _{0}+\mathrm{a} _{0} \mathrm{t}-\left(\frac{2 \mathrm{a} _{0}^{2}}{\mathrm{v} _{0}}\right) \mathrm{t}^{2}\right] \mathrm{dt} \\ & \text { or } \quad x=\mathrm{v} _{0} \mathrm{t}+\mathrm{a} _{0} \frac{\mathrm{t}^{2}}{2}-\frac{2 \mathrm{a} _{0}^{2}}{3 \mathrm{v} _{0}} \mathrm{t}^{3} \tag{3} \end{align*} $$

At $t=\frac{\mathrm{v} _{0}}{\mathrm{a} _{0}} ;$ the instantaneous position co-ordinate is

$$ \begin{aligned} & x=\mathrm{v} _{0}\left(\frac{\mathrm{v} _{0}}{\mathrm{a} _{0}}\right)+\frac{\mathrm{a} _{0}}{2}\left(\frac{\mathrm{v} _{0}}{\mathrm{a} _{0}}\right)^{2}-\frac{2 \mathrm{a} _{0}^{2}}{3 \mathrm{v} _{0}}\left(\frac{\mathrm{v} _{0}}{\mathrm{a} _{0}}\right)^{3} \\ & =\frac{\mathrm{v} _{0}^{2}}{\mathrm{a} _{0}}+\frac{\mathrm{v} _{0}^{2}}{2 \mathrm{a} _{0}}-\frac{2 \mathrm{v} _{0}^{2}}{3 \mathrm{a} _{0}}=\frac{5 \mathrm{v} _{0}^{2}}{6 \mathrm{a} _{0}} \end{aligned} $$

Correct answer: (3)

Solution:

Taking ground as origin of co-ordinates and vertically upward direction as direction of $+z-$ axis Fig. (a) shows position of two particles at $t=0$.

$$ \begin{gathered} \mathrm{U} _{\mathrm{A} 0}=+10 \mathrm{~ms}^{-1} ; \quad \mathrm{U} _{\mathrm{B} \sigma}+20 \mathrm{~ms}^{-1} ; \mathrm{a} _{\mathrm{A}}=\mathrm{a} _{\mathrm{B}}=-10 \mathrm{~ms}^{-2} \\ \mathrm{U} _{0}=\mathrm{U} _{\mathrm{B} 0}-\mathrm{U} _{\mathrm{A} 0}=10 \mathrm{~ms}^{-1} \end{gathered} $$

The instantaneous velocity $U _{A}$ and $U _{B}$ of $A$ and $B$, is given by

$$ \begin{aligned} & \mathrm{U} _{\mathrm{A}}=10-10 \mathrm{t} \text { and } \mathrm{U} _{\mathrm{B}}=20-10 \mathrm{t} \\ & \therefore \mathrm{U}=\mathrm{U} _{\mathrm{B}}-\mathrm{U} _{\mathrm{A}}=10 \mathrm{~ms}^{-1} \end{aligned} $$

The above result is valid for time interval $0 \leq \mathrm{t} \leq 2 \mathrm{~s}$. At $\mathrm{t}=2 \mathrm{~s}$; particle A comes back to ground and subsequently is at rest. At $\mathrm{t}=2 \mathrm{~s}$; particle $\mathrm{B}$ is at its highest position $\mathrm{B} _{1}$ above $\mathrm{B}$ and its instantaneous veloicty is zero. Also $\mathrm{BB} _{1}=\frac{(20)^{2}}{2 \times 10}=20 \mathrm{~m}$. Now B starts falling freely under gravity and it hits ground at $\mathrm{t}=\mathrm{t} _{1}$. Now $\mathrm{z} _{\mathrm{B} 0}=+60 \mathrm{~m}$ and $\mathrm{z} _{\mathrm{B}}=0$ at $\mathrm{t}=\mathrm{t} _{1}$. Therefore

$$ 0=60=20 \mathrm{t} _{1}-5 \mathrm{t} _{1}^{2} $$

or $\quad \mathrm{t} _{1}^{2}-4 \mathrm{t} _{1}-12=0$

$$ \therefore \mathrm{t} _{1}=\frac{4 \pm \sqrt{16+48}}{2}=\frac{4 \pm 8}{2}=6 \mathrm{~s} $$

In time interval $2 \leq \mathrm{t} \leq 6 \mathrm{~s}$; the instantaneous speed of $\mathrm{B}$ is

$$ \mathrm{U} _{\mathrm{B}}=20-10 \mathrm{t} $$

Since $U _{A}=0 ; U=U _{B}-U _{A}=20-10 t$. In time interval $2 \leq t \leq 6 \mathrm{~s} \quad U$ is a negative number. U vs t graph is a straight line. At $t=6 \mathrm{~s} ; \mathrm{U}=20-10 \times 6=-40 \mathrm{~ms}^{-1}$. For $\mathrm{t}>6 \mathrm{~s}$; both $A$ and $\mathrm{B}$ at rest, hence $\mathrm{U}=0$. These characteristics are correctly shown in (3).

Correct answer: (1)

Solution:

Let $\mathrm{z} _{\mathrm{A}}$ and $\mathrm{z} _{\mathrm{B}}$ be the instantaneous position co-ordinate of $\mathrm{A}$ and $\mathrm{B}$. Obvisouly

$$ \begin{equation*} \mathrm{z} _{\mathrm{A}}=30 \mathrm{t}-5 \mathrm{t}^{2} \tag{i} \end{equation*} $$

and $\mathrm{z} _{\mathrm{B}}=75+\left(10 \mathrm{t}-5 \mathrm{t}^{2}\right) \hspace{50mm} . . . . . . . . (ii) $

$\therefore \mathrm{z}=\mathrm{z} _{\mathrm{B}}-\mathrm{z} _{\mathrm{A}}=75-20 \mathrm{t}$

Let $A$ and $B$ hit ground at $t _{A}$ and $t _{B}$ after projection. Then for particle $A, z _{A}=0, t=t _{A}$

$$ \therefore 0=30 \mathrm{t} _{\mathrm{A}}-5 \mathrm{t} _{\mathrm{A}}^{2}=5 \mathrm{t} _{\mathrm{A}}\left(6-\mathrm{t} _{\mathrm{A}}\right) \quad \text { or } \quad \mathrm{t} _{\mathrm{A}}=6 \mathrm{~s} $$

For particle $\mathrm{B} ; \mathrm{z} _{\mathrm{B}}=0$; $\mathrm{at} \mathrm{t}=\mathrm{t} _{\mathrm{B}}$. FromEqn. (ii) we have

$$ \begin{aligned} & 0=75+10 t _{B}-5 t _{B}^{2} \\ & \text { or } t _{B}^{2}-2 t _{B}-15=0 \\ & \text { or } t _{B}=\frac{2 \pm \sqrt{4+60}}{2}=\frac{2+8}{2}=5 \mathrm{~s} \end{aligned} $$

For time interval $0 \leq \mathrm{t} \leq 5 \mathrm{~s}$; both $\mathrm{A}$ and $\mathrm{B}$ are in motion $\mathrm{z}$ is given by Eqn. (iii). $\mathrm{z}$ varies linearly with time $\mathrm{t}$. At $\mathrm{t}=5 \mathrm{~s}$.

$$ \mathrm{z} _{\mathrm{A}}=75-20 \times 5=-25 \mathrm{~m} $$

Since $\mathrm{z}=\mathrm{z} _{\mathrm{B}}-\mathrm{z} _{\mathrm{A}}$ and $\mathrm{z} _{\mathrm{B}}=0 ; \mathrm{z} _{\mathrm{A}}=25 \mathrm{~m}$. At this time A is $25 \mathrm{~m}$ above ground. However $\mathrm{z}$ is a negative number. For time interval $5 \leq \mathrm{t} \leq 6 \mathrm{~s} ; \mathrm{z}=-\mathrm{z} _{\mathrm{A}}\left[\because \mathrm{z} _{\mathrm{B}}=0\right]$. Therefore

$$ z=30 t-5 t^{2} $$

In this time interval $z$ is a parabolic function of $t$. These characteristics all shown correctly in (1).

Solution:

Particle A moves along $x$-axis and $\mathrm{B}$ along $\mathrm{y}$-axis. The instantaneous position co-ordinates of $\mathrm{A}$ and $\mathrm{B}$ are $(x, 0)$ and $(0, y)$. Obviously

$$ x=\alpha-\frac{1}{2} \mathrm{a} _{1} \mathrm{t}^{2} \quad \text { and } \quad \mathrm{y}=\beta-\frac{1}{2} \mathrm{a} _{2} \mathrm{t}^{2} $$

The instantaneous distance and between the two particle is $\mathrm{s}=\sqrt{x^{2}+\mathrm{y}^{2}}$ For $\mathrm{s}$ to be minimum

$$ \frac{\mathrm{ds}}{\mathrm{dt}}=0 \text { i.e. } \frac{1}{2}\left(x^{2}+\mathrm{y}^{2}\right)^{-1 / 2}\left[2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right]=0 $$

This condition is satisfied if

$$ x \frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dt}}=0 $$

or $\quad\left(\alpha-\frac{1}{2} a _{1} t^{2}\right) a _{1} t+\left(\beta-\frac{1}{2} a _{2} t^{2}\right) a _{2} t=0$ or

$\mathrm{a} _{1} \alpha+\mathrm{a} _{2} \beta=\frac{1}{2}\left(\mathrm{a} _{1}^{2}+\mathrm{a} _{2}^{2}\right) \mathrm{t}^{2}$

$\therefore \mathrm{t}=\sqrt{\frac{2\left(\mathrm{a} _{1} \alpha+\mathrm{a} _{2} \beta\right)}{\mathrm{a} _{1}^{2}+\mathrm{a} _{2}^{2}}}$

Correct answer:

Solution:

Since the resultant of given vectors is zero, they form a closed polygon $A B C D A$ as shown in Fig. We know the sum of the two sides of a triangle is always less than the third side. In triangle $B C D$.

$$ \mathrm{BD}<\mathrm{q}+\mathrm{r} $$

Similarly from triangle $\mathrm{ABD}$

$$ \begin{array}{ll} & \mathrm{AB}=\mathrm{p}<\mathrm{BD}+\mathrm{DA} \\ \text { or } \quad \mathrm{p}<\mathrm{q}+\mathrm{r}+\mathrm{s} \end{array} $$

Correct answer: (1)

Solution:

Let $F _{1}$ and $F _{2}$ be the magnitude of the two forces. Given,

$$ \begin{aligned} & \mathrm{F} _{\max }=\mathrm{F} _{1}+\mathrm{F} _{2}=11 \text { and } \mathrm{F} _{\min }=\mathrm{F} _{1}-\mathrm{F} _{2}=5 \\ & \therefore \mathrm{F} _{1}=8 \mathrm{~N} \text { and } \mathrm{F} _{2}=3 \mathrm{~N} \end{aligned} $$

When the forces are increased by $2 \mathrm{~N}$ each; the magnitude $F _{1}^{\prime}$ and $F _{2}^{\prime}$ of the two forces is

$$ \mathrm{F} _{1}^{\prime}=10 \mathrm{~N} \text { and } \mathrm{F} _{2}^{\prime}=5 \mathrm{~N} $$

Now $F _{1}^{\prime}$ and $F _{2}^{\prime}$ act at right angles to one another as shown in Fig. Their resultant

$$ \mathrm{F}^{\prime}=\sqrt{(10)^{2}+(5)^{2}}=\sqrt{125}=5 \sqrt{5} \mathrm{~N} $$

$\tan \alpha=\frac{\mathrm{F} _{2}^{\prime}}{\mathrm{F} _{1}^{\prime}}=0.5 \quad$ or $\quad \alpha=\tan ^{-1}(0.5)$

Correct answer:

Solution:

Number of given vectors form a closed polygon because total angle between them is $2 \pi$. Their resultant is zero. Therefore sum of all $(n+2)$ vectors is sum of two vectors of same magnitude $\mathrm{F}$ making an angle $\frac{2 \pi}{\mathrm{n}}$ with one other. The resultant $\mathrm{R}$ is shown in Fig. Obviously

$$ \mathrm{R}=2 \mathrm{~F} \cos \alpha ; \text { and } 2 \alpha=\theta=\frac{2 \pi}{\mathrm{n}} \quad \therefore \alpha=\frac{\pi}{\mathrm{n}} $$

or $R=2 F \cos \left(\frac{\pi}{n}\right)$

R makes an angle $\alpha=\frac{\pi}{\mathrm{n}}$ with the first vector.

Correct answer: (1)

Solution:

Obviously,

$$ \frac{P _{x}}{P _{x}^{\prime}}=\frac{p \cos \theta}{p \cos (\theta-30)}=\frac{1}{\sqrt{3}} $$

or $\quad \sqrt{3} \cos \theta=\cos (\theta-30)$ $\therefore \theta=60^{\circ}$

Now $\mathbf{P}$ and $\mathbf{Q}$ have same magnitude and angle between them is also $30^{\circ}$. The magnitude of the resultant $\mathbf{R}$ is

$$ \begin{aligned} & \mathrm{R}=\sqrt{\mathrm{p}^{2}+\mathrm{p}^{2}+2 \mathrm{p}^{2} \cos 30^{0}}=\mathrm{p} \sqrt{(2+\sqrt{3})} \\ & \therefore \frac{\mathrm{R}}{\mathrm{P}}=(2+\sqrt{3})^{1 / 2} \end{aligned} $$

Correct answer:

Solution:

Choose a co-ordinate system on the floor with intial position of insect as origin of co-ordinates. The three displacement vectors $\mathbf{s} _{1}, \mathbf{s} _{2}$ and $\mathbf{s} _{3}$ are as shown in Fig. (a). Expressing the vectors in terms of their rectangular components we have

$$ \begin{aligned} & \mathbf{s} _{1}=5 \hat{i} ; \quad \mathbf{s} _{2}=2 \hat{j} \\ & \mathbf{s} _{3}=(2 \sqrt{2} \cos 45) \hat{i}+\left(2 \sqrt{2} \sin 45^{0}\right)(-\hat{j})=-2 \hat{i}-2 \hat{j} \end{aligned} $$

The resultant displacement $\mathrm{s}$ is

$$ \mathbf{s}=\mathbf{s} _{1}+\mathbf{s} _{2}+\mathbf{s} _{3}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}} $$

This is shown in Fig. (b). Magnitude of $\mathbf{s}$ is

$$ \mathrm{s}=\sqrt{(3)^{2}+(-4)^{2}}=5 \mathrm{~m} ; \quad \text { and } \tan \alpha=\frac{4}{3} \quad \therefore \alpha=\tan ^{-1}\left(\frac{4}{3}\right) $$

Correct answer :(2)

Solution:

Choose a co-ordinate system as shown in Fig. Given,

$$ \begin{aligned} & \mathbf{v} _{\mathbf{A}}=(30 \cos 60) \hat{\mathrm{i}}+(30 \sin 60) \hat{\mathrm{j}}=15 \hat{\mathrm{i}}+26 \hat{\mathrm{j}} \\ & \mathbf{v} _{\mathbf{A B}}=41 \hat{\mathrm{j}} \end{aligned} $$

Let velocity of car $B$ be

$$ \mathbf{v} _{\mathbf{B}}=x \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}} $$

We know,

$$ \begin{aligned} & \mathbf{v} _{\mathbf{A B}}=\mathbf{v} _{\mathbf{A}}-\mathbf{v} _{\mathbf{B}} \\ & =(15-x) \hat{\mathrm{i}}+(26-\mathrm{y}) \hat{\mathrm{j}}=41 \hat{\mathrm{j}} \end{aligned} $$

$\therefore x=15$ and $\mathrm{y}=15$

We have,

$$ \mathbf{v} _{\mathbf{B}}=(15 \hat{\mathrm{i}}+15 \hat{\mathrm{j}}) \mathrm{kmhr}^{-1} $$

Obviously $\mathrm{v} _{\mathrm{B}}=15 \sqrt{2} \mathrm{kmhr}^{-1}$ in north-east direction.

Correct answer: (1)

Solution:

Choosing the co-ordinate systemas shown in Fig. let $\mathbf{v} _{\mathbf{A}}=x _{1} \hat{\mathrm{i}}+\mathrm{y} _{1} \hat{\mathrm{j}}$; kmhr ${ }^{-1}$ and $\mathbf{v} _{\mathbf{B}}=x _{2} \hat{\mathrm{i}}+\mathrm{y} _{2} \hat{\mathrm{j}} ; \mathrm{kmhr}^{-1}$

Given, $\mathbf{v} _{\mathbf{C}}=30 \hat{\mathrm{i}}+30 \sqrt{3} \hat{\mathrm{j}}$. Also, $\mathbf{v} _{\mathbf{C A}}=30 \sqrt{3} \hat{\mathrm{j}}$

Therefore,

$$ \mathbf{v} _{\mathbf{C A}}=\mathbf{v} _{\mathbf{C}}-\mathbf{v} _{\mathbf{A}} \text { or } 15 \sqrt{3} \hat{\mathrm{j}}=\left(30-x _{1}\right) \hat{\mathrm{i}}+\left(30 \sqrt{3}-\mathrm{y} _{1}\right) \hat{\mathrm{j}} $$

Comparing coefficents of $\hat{i}$ and $\hat{j}$, we get

$$ \begin{aligned} & 30-x _{1}=0 \quad \text { or } x _{1}=30 \text { and } 15 \sqrt{3}=30 \sqrt{3}-\mathrm{y} _{1} \quad \text { or } \mathrm{y} _{1}=15 \sqrt{3} \\ & \therefore \mathrm{v} _{\mathrm{A}}=30 \hat{\mathrm{i}}+15 \sqrt{3} \hat{\mathrm{j}} ; \mathrm{kmhr}^{-1} \end{aligned} $$

Similarly,

$$ \begin{gathered} \mathbf{v} _{\mathbf{C B}}=20 \hat{\mathrm{i}}=\mathbf{v} _{\mathbf{C}}-\mathbf{v} _{\mathbf{B}} \\ \text { or } \quad 20 \hat{\mathrm{i}}=\left(30-x _{2}\right) \hat{\mathrm{i}}-\left(30 \sqrt{3}-\mathrm{y} _{2}\right) \hat{\mathrm{j}} \end{gathered} $$

Comparing coefficients of $\hat{i}$ and $\hat{j}$

$$ \begin{aligned} & 20=30-x _{2} \quad \text { or } \quad x _{2}=10 ; \text { and } \\ & 30 \sqrt{3}-\mathrm{y} _{2}=0 \text { or } \mathrm{y} _{2}=30 \sqrt{3} \\ & \therefore \quad \mathbf{v} _{\mathbf{B}}=10 \hat{\mathrm{i}}+30 \sqrt{3} \hat{\mathrm{j}} \end{aligned} $$

These are expression for $\mathbf{v} _{\mathbf{A}}$ and $\mathbf{v} _{\mathbf{B}}$ in (1)

Correct answer: (4)

Solution:

Let swimmer 1 swim in a direction to left of point $\mathrm{Q}$ in such a manner that his resultant velocity is along PQ. This is shown in Fig. (a). The resultant speed $v _{1}$ of swimmer 1 is

$\mathrm{v} _{1}=\sqrt{\mathrm{v} _{\mathrm{S}}^{2}-\mathrm{v} _{\mathrm{R}}^{2}}=\sqrt{(5)^{2}-(4)^{2}}=3 \mathrm{kmhr}^{-1}$

Let $\omega$ be width of stream in kilometer.

$\mathrm{t} _{1}=$ Time taken by swimmer 1 to move from $P$ to $Q$

$$ \begin{equation*} =\frac{\omega}{\mathrm{v} _{1}}=\frac{\omega}{3} \mathrm{hr} \tag{1} \end{equation*} $$

For swimmer 2; the velocity vectors are as shown in Fig. (b). Let $\mathrm{t} _{2}$ be the total time taken by swimmer 2 to reach Q. Obviously

$$ \begin{aligned} & \mathrm{t} _{2}=\mathrm{t} _{2}^{\prime}+\mathrm{t} _{2}^{\prime} \\ & \mathrm{t} _{2}^{\prime}=\frac{\mathrm{PQ}}{\mathrm{V} _{\mathrm{S}}}=\frac{\omega}{\mathrm{V} _{\mathrm{S}}}=\frac{\omega}{5} ; \text { and } \mathrm{t} _{2}^{\prime}=\frac{\mathrm{QR}}{\mathrm{V}}=\frac{\mathrm{V} _{\mathrm{R}} \mathrm{t} _{2}^{\prime}}{\mathrm{V}} \end{aligned} $$

$$ \text { or } \mathrm{t} _{2}=\frac{\omega}{5}+\frac{4}{\mathrm{~V}} \times\left(\frac{\omega}{5}\right) $$

Given, $\mathrm{t} _{1}=\mathrm{t} _{2}$; therefore

$$ \begin{aligned} & \frac{\omega}{3}=\frac{\omega}{5}+\frac{4}{5} \frac{\omega}{\mathrm{V}} \quad \text { or } \quad \frac{1}{3}-\frac{1}{5}=\frac{4}{5 \mathrm{~V}} \\ & \therefore \mathrm{V}=6 \mathrm{kmhr}^{-1} \end{aligned} $$

Correct answer: (2)

Solution:

$\mathbf{A}+\mathbf{B}=(a+c) \hat{i}+3 b \hat{j} ; \mathbf{A}-\mathbf{B}=(a-c) \hat{i}-b \hat{j}$

The vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$ are perpendicular to one another; therefore

$$ \begin{aligned} & (\mathbf{A}+\mathbf{B}) \cdot(\mathbf{A}-\mathbf{B})=0 \\ & \left(\mathrm{a}^{2}-\mathrm{c}^{2}\right)-3 \mathrm{~b}^{2}=0 \quad \text { or } \quad\left(\mathrm{c}^{2}+n b^{2}\right)-\mathrm{c}^{2}-3 \mathrm{~b}^{2}=0 \\ & \therefore \mathrm{n}=3 \end{aligned} $$

Correct answer: (1)

Solution:

$\mathrm{A} _{11}=$ Component of $\mathbf{A}$ along $\mathbf{B}$ has a magnitude of $\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}$.

Now $\mathbf{A} \cdot \mathbf{B}=10 \times 3-2.5 \times 4=20$

and $|\mathbf{B}|=\sqrt{(3)^{2}+(4)^{2}}=5$ $\therefore \mathrm{A} _{11}=\frac{20}{5}=4$

In terms of vectors

$$ \begin{aligned} & \mathbf{A} _{11}=4 \hat{B}=4 \frac{[3 \hat{i}+4 \hat{j}]}{5} \\ & =2.4 \hat{i}+3.2 \hat{j} \end{aligned} $$

Correct answer: (1)

Solution:

F the instantaneous force on charged particle is

$$ \mathbf{F}=q(\mathbf{v} \times \mathbf{B})=q\left[\left(v _{1} \hat{i}+v _{2} \hat{j}\right) \times B \hat{k}\right] $$

$$ =\operatorname{Bqv} _{1}(-\hat{\mathrm{j}})+\operatorname{Bqv} _{2}(\hat{\mathrm{i}}) $$

Let $\mathbf{P}=x \hat{i}+y \hat{j}+z \hat{k}$. Since $\mathbf{P}$ is perpendicular to $\mathbf{F}, \mathbf{P} \cdot \mathbf{F}=0$

$\therefore \quad(x \hat{i}+y \hat{j}+z \hat{k}) \cdot B q\left(v _{2} \hat{i}-v _{1} \hat{j}\right)=0$

or $\operatorname{Bqv} _{2} x-\operatorname{Bqv} _{1} \mathrm{y}=0 \quad$ or $\quad \frac{x}{\mathrm{y}}=\frac{\mathrm{v} _{1}}{\mathrm{v} _{2}} \hspace{40mm} . . . . . . . . .(i)$

P.F is zero if $x$ and y satisfy Eqn (i); irrespective of value of $z$.

Correct answer: (4)

Solution:

The position vector of particle at $B$ with respect to point $A$ is

$$ \begin{aligned} \mathbf{r} _{B A}= & \mathbf{r} _{B}-\mathbf{r} _{A}=(5 \hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}+3 \hat{j}-\hat{k}) \\ = & 3 \hat{i}-\hat{j}+4 \hat{k} \end{aligned} $$

$\ell _{\mathrm{BA}}$; the angular momentum of particle at $\mathrm{B}$ about point $\mathrm{A}$ is

$$ \begin{aligned} \ell _{\mathbf{B A}} & =\mathbf{r} _{\mathbf{B A}} \times \mathbf{P} \\ & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 4 \\ 3 & 2 & 4 \end{array}\right|=\hat{\mathrm{i}}[-4-8]+\hat{\mathrm{j}}[12-12]+\hat{\mathrm{k}}[6+3] \\ & =-12 \hat{\mathrm{i}}+9 \hat{\mathrm{k}} \end{aligned} $$

Correct answer: (2)

Solution:

Let $(x, y)$ be the Cartesian co-ordinates of instantaneous position $\mathrm{P}$ of particle. The instantaneous position vector is:

$$ \mathbf{r}=(R \cos \theta) \hat{i}+(R \sin \theta) \hat{j} $$

The angular velocity $\omega$ is

$$ \boldsymbol{\omega}=\omega \hat{\mathrm{k}} $$

The instantaneous linear velocity $\mathrm{v}$ is

$$ \mathbf{v}=\boldsymbol{\omega} \times \mathbf{r} $$

$$ \begin{aligned} & =(\omega \hat{k}) \times[(R \cos \theta) \hat{i}+(R \sin \theta) \hat{j}] \\ & =(+R \omega \cos \theta) \hat{j}+(-R \omega \sin \theta) \hat{i} \end{aligned} $$

Correct answer: (1)

Solution:

Choosing $x^{\prime}$ and $y^{\prime}$ axis as shown in Fig.

$$ \begin{aligned} & \mathrm{v} _{0 \mathrm{x}}^{\prime}=\mathrm{u} \cos (\theta-30) \\ & \mathrm{v} _{0 y}^{\prime}=\mathrm{u} \sin (\theta-30) \end{aligned} $$

Resolving acceleration due to gravity into its components along $x^{\prime}$ and $y^{\prime}$ axis,

$$ \begin{array}{r} \mathrm{a} _{x}^{\prime}=-\mathrm{g} \sin 30=-\left(\frac{\mathrm{g}}{2}\right) \\ \mathrm{a} _{y}^{\prime}=-\mathrm{g} \cos 30=-\left(\frac{\sqrt{3}}{2} \mathrm{~g}\right) \end{array} $$

Let $\mathrm{v} _{x}^{\prime}$ and $\mathrm{v} _{y}^{\prime}$ be the $x^{\prime}$ and $\mathrm{y}^{\prime}$ component of instantaneous velocity of particle at $\mathrm{P}$.

$$ \begin{aligned} & \mathrm{v} _{x}^{\prime}=\mathrm{v} _{\mathrm{o} x}^{\prime}+\mathrm{a} _{x}^{\prime} \mathrm{t}=\mathrm{u} \cos (\theta-30)-\frac{\mathrm{g}}{2} \mathrm{t} \\ & \mathrm{v} _{y}^{\prime}=\mathrm{v} _{\mathrm{o} y}^{\prime}+\mathrm{a} _{y}^{\prime} \mathrm{t}=\mathrm{u} \sin (\theta-30)-\frac{\sqrt{3}}{2} \mathrm{gt} \end{aligned} $$

As $\mathrm{v} _{x}^{\prime}=0$; when particle hits $x^{\prime}-$ axis (i.e. the inclined plane)

$$ \begin{equation*} \therefore \quad \mathrm{t}=\frac{2 \mathrm{u} \cos (\theta-30)}{\mathrm{g}} \tag{1} \end{equation*} $$

Also for piont $\mathrm{P}, \mathrm{y}^{\prime}=0$. Therefore

$$ \begin{align*} & 0=\mathrm{u} \sin (\theta-30) \mathrm{t}-\frac{\sqrt{3}}{4} \mathrm{gt}^{2} \\ & \text { or } \mathrm{t}=\frac{4 \mathrm{u} \sin (\theta-30)}{\sqrt{3} \mathrm{~g}} \tag{2} \end{align*} $$

From Eqns (1) and (2) we have

$$ \frac{2 \mathrm{u} \cos (\theta-30)}{\mathrm{g}}=\frac{4 \mathrm{u} \sin (\theta-30)}{\sqrt{3} \mathrm{~g}} \text { or } \tan \left(\theta-30^{\circ}\right)=\frac{\sqrt{3}}{2} $$

$\therefore \theta=30^{0}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Correct answer:(3)

Solution:

Fig. shows a particle projected at an angle $\theta$ with horizontal. The particle passes through point $\mathrm{P}, \mathrm{t}$ second after projection. Co-ordinates of $\mathrm{P}$ are

$$ x=r \cos \alpha, y=r \sin \alpha $$

Therefore,

$$ \begin{align*} & x=(u \cos \theta) t=r \cos \alpha \\ & \text { or } t=\frac{r \cos \alpha}{u \cos \theta} \tag{1} \\ & y=(u \sin \theta) t-\frac{1}{2} g^{2} \end{align*} $$

or $\quad r \sin \alpha=u \sin \theta\left(\frac{r \cos \alpha}{u \cos \theta}\right)-\frac{1}{2} g t^{2}$

or $\quad \frac{1}{2} \mathrm{gt}^{2}=\mathrm{r}\left[\frac{\sin \theta \cos \alpha}{\cos \theta}-\sin \alpha\right]$

$$ \begin{align*} & =\mathrm{r}\left[\frac{\sin \theta \cos \alpha-\cos \theta \sin \alpha}{\cos \theta}\right] \\ & \therefore \quad \mathrm{t}^{2}=\frac{2 \mathrm{r} \sin (\theta-\alpha)}{\mathrm{g} \cos \theta} \tag{2} \end{align*} $$

From Enq (2) $t$ is directly proportional to $\sqrt{\mathrm{r}}$, therefore

$\mathrm{t} _{\mathrm{A}} \mathrm{t} _{\mathrm{B}}$ is directly proportional to $(\sqrt{\mathrm{r}} \cdot \sqrt{\mathrm{r}})=\mathrm{r}$

Correct answer:(4)

Solution:

The equation of trajectory of a projectile is:

$$ \begin{equation*} y=(\tan \theta) x-\frac{g x^{2}}{2 v _{0}^{2} \sin ^{2} \theta} \tag{1} \end{equation*} $$

Comaparing given equation of trajectory with Eqn. (1) we have, $\tan \theta=2$.

$\therefore \sin \theta=\frac{\tan \theta}{\sqrt{1+\tan ^{2} \theta}}=\frac{2}{\sqrt{5}} \quad$ or $\quad \theta=\sin ^{-1} \frac{2}{\sqrt{5}}$

Also,

$$ \begin{gathered} \frac{\mathrm{g}}{2 \mathrm{v} _{0}^{2} \sin ^{2} \theta}=\frac{1}{5} \text { or } \mathrm{v} _{0}^{2} \sin ^{2} \theta=\frac{10 \times 5}{2}=25 \\ \therefore \mathrm{v} _{0} \sin \theta=5 \quad \text { or } \quad \mathrm{v} _{0}=\frac{5 \sqrt{5}}{2} \mathrm{~ms}^{-1} \\ \mathrm{H}=\text { The maximum vertical height }=\frac{2 \mathrm{v} _{0}^{2} \sin 2 \theta}{\mathrm{g}}=\frac{2 \times(5)^{2}}{10}=5 \mathrm{~m} \\ \mathrm{~T}=\text { The time of flight }=\frac{2 \mathrm{v} _{0} \sin \theta}{\mathrm{g}}=\frac{2 \times 5}{10}=1 \mathrm{~s} \end{gathered} $$

Correct answer:(2)

Solution:

In Fig. $\mathrm{O}$ is top of tower. Choosing co-ordinate axis as shown in Fig. and $\mathrm{t}=0$ when ball $\mathrm{A}$ is projected; the instantaneous co-ordinates $\left(x _{1}, y _{1}\right)$ of ball A are:

$$ \begin{align*} & x _{1}=\left(\mathrm{v} _{0} \cos 60\right) \mathrm{t}=\frac{\mathrm{v} _{0}}{2} \mathrm{t} \tag{1} \\ & y _{1}=\left(\mathrm{v} _{0} \sin 60\right) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \\ & =\frac{\sqrt{3} \mathrm{v} _{0}}{2} \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \tag{2} \end{align*} $$

The instantaneous co-ordinates $\left(x _{2}, \mathrm{y} _{2}\right)$ at $\mathrm{t}=\mathrm{t}$ $(t>n)$ of ball $B$ are

$$ \begin{align*} & x _{2}=\mathrm{v} _{0}(\mathrm{t}-\mathrm{n}) \tag{3} \\ & \mathrm{y} _{2}=-\frac{1}{2} \mathrm{~g}(\mathrm{t}-\mathrm{n})^{2} \tag{4} \end{align*} $$

Let $\mathrm{P}$ be the point where the two balls collide with one another. Co-ordinates of ball $\mathrm{A}$ and $\mathrm{B}$ are same at this moment of time. From Eqns. (1) and (3).

$$ \begin{align*} & \frac{\mathrm{v} _{0}}{2} \mathrm{t}=\mathrm{v} _{0}(\mathrm{t}-\mathrm{n}) \\ & \therefore \mathrm{n}=\frac{\mathrm{t}}{2} \tag{5} \end{align*} $$

From Eqns (2) and (4); we have,

$$ \begin{align*} & \frac{\sqrt{3}}{2} \mathrm{v} _{0} \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}=-\frac{1}{2} \mathrm{~g}(\mathrm{t}-\mathrm{n})^{2} \\ & \text { or } \sqrt{3} \mathrm{v} _{0} \mathrm{t}-\mathrm{gt}^{2}=-\frac{\mathrm{gt}^{2}}{4} \\ & \text { or } \mathrm{t}=\frac{4 \mathrm{v} _{0}}{\sqrt{3} \mathrm{~g}} \tag{6} \end{align*} $$

Obvisouly, from Eqns. (5) and (6) we have

$$ \mathrm{n}=\frac{\mathrm{t}}{2}=\frac{2 \mathrm{v} _{0}}{\sqrt{3} \mathrm{~g}} $$

Correct answer:(2)

Solution:

Choosing $x$ and $y$-axis as shown in Fig., the instantaneous $x$ and y co-ordinates of ball projected from $\mathrm{O}$ are

$$ \begin{align*} & x=\left(\mathrm{v} _{0} \cos 60\right) \mathrm{t}=\left(\frac{\mathrm{v} _{0}}{2}\right) \mathrm{t} \tag{i} \\ & \mathrm{y}=\left(\mathrm{v} _{0} \cos 60\right) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}=\frac{\sqrt{3}}{2} \mathrm{v} _{0} \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \tag{ii} \end{align*} $$

The equation of straight line $\mathrm{OB}$, giving slant side of wedge is

$$ \begin{equation*} y=-(\tan 30) x=-\frac{x}{\sqrt{3}} \tag{iii} \end{equation*} $$

Let ball hit wedge at $\mathrm{P}, \mathrm{t}$ second after projection. Point $\mathrm{P}$ satisfies Eqns (i) and (ii) as well as Eqn (iii). Therefore,

$$ \begin{align*} & -\left(\frac{1}{\sqrt{3}}\right)\left(\frac{\mathrm{v} _{0}}{2} \mathrm{t}\right)=\frac{\sqrt{3}}{2} \mathrm{v} _{0} \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \\ & \therefore \mathrm{t}=\frac{2 \mathrm{v} _{0}}{\sqrt{3} \mathrm{~g}} \tag{iv} \end{align*} $$

Obvisouly $\mathrm{QP}=x$-co-ordinate of ball at $\quad t=\frac{2 x _{0}}{\sqrt{3} g}$. Therefore

$$ \mathrm{QP}=\left(\frac{\mathrm{v} _{0}}{2}\right) \frac{2 \mathrm{v} _{0}}{\sqrt{3} \mathrm{~g}}=\frac{\mathrm{v} _{0}^{2}}{\sqrt{3} \mathrm{~g}} $$

From right angled $\triangle \mathrm{OPQ} \frac{\mathrm{QP}}{\mathrm{OP}}=\cos 30$, therefore

$\mathrm{OP}=\frac{\mathrm{QP}}{\cos 30}=\frac{2}{\sqrt{3}} \times \frac{\mathrm{v} _{0}^{2}}{\sqrt{3} \mathrm{~g}}=\frac{2 \mathrm{v} _{0}^{2}}{3 \mathrm{~g}}$

Correct answer: (3)

Solution:

In Fig. $A B$ is pole of vertical height $h$. Given $B$ is the position of bird. Given $O C=2 h$ is the maximum vertical height attained by projectile fired from $O$ with speed $v _{0}$ at an angle $\theta$ with the horizontal. Obviously

$$ 2 \mathrm{~h}=\frac{\mathrm{v} _{0}^{2} \sin ^{2} \theta}{2 \mathrm{~g}} $$

$\therefore \mathrm{v} _{0} \sin \theta=2 \sqrt{\mathrm{gh}} \hspace{40mm} . . . . . .(1) $

Let the projectile attain vertical height $h$ at time $t$ after projection. Then

$$ \mathrm{h}=\left(\mathrm{v} _{0} \sin \theta\right) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} $$

or $\mathrm{gt}^{2}-\left(2 \mathrm{v} _{0} \sin \theta\right) \mathrm{t}+2 \mathrm{~h}=0$

$\therefore \mathrm{t}=\frac{2 \mathrm{v} _{0} \sin \theta \pm \sqrt{\left(2 \mathrm{v} _{0} \sin \theta\right)^{2}-8 \mathrm{gh}}}{2 \mathrm{~g}}$

$$ =\frac{4 \sqrt{\mathrm{gh}} \pm \sqrt{16 \mathrm{gh}-8 \mathrm{gh}}}{2 \mathrm{~g}}=\frac{(2 \pm \sqrt{2})}{\mathrm{g}} \sqrt{\mathrm{gh}} $$

$$ =(2 \pm \sqrt{2}) \sqrt{\frac{\mathrm{h}}{\mathrm{g}}} $$

Let $t _{1}=(2-\sqrt{2}) \sqrt{\frac{h}{g}}$ and $t _{2}=(2+\sqrt{2}) \sqrt{\frac{h}{g}}$.

The bird gets hit at point $B _{1}$ i.e. at $t=t _{2}$ after projection. Therefore

$$ \mathrm{BB} _{1}=\mathrm{vt} _{2}=\mathrm{v}(2+\sqrt{2}) \sqrt{\frac{\mathrm{h}}{\mathrm{g}}} $$

Also,

$$ \mathrm{BB} _{1}=\left(\mathrm{v} _{0} \cos \theta\right)\left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)=\left(\mathrm{v} _{0} \cos \theta\right) 2 \sqrt{2} \sqrt{\frac{\mathrm{h}}{\mathrm{g}}} $$

Clearly,

$$ \frac{\mathrm{v}}{\mathrm{v} _{0} \cos \theta}=\frac{2 \sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{\left(1+\frac{1}{\sqrt{2}}\right)} $$

Correct answer: (3)

Solution:

For ball A

In hitting $\mathrm{N}^{\text {th }}$ step from top; the ball travels a horizontal distance of $\mathrm{N} \omega$ and vertical distance $\mathrm{Nh}$. Let the ball hit $\mathrm{N}^{\text {th }}$ step $t$ second after projection. Obviously

$$ \begin{align*} & \mathrm{N} \omega=\mathrm{u} _{1} \mathrm{t} \text { and } \mathrm{Nh}=\frac{1}{2} \mathrm{gt}^{2} \\ & \quad \therefore \mathrm{Nh}=\frac{1}{2} \mathrm{~g}\left(\frac{\mathrm{N} \omega}{\mathrm{u} _{1}}\right)^{2} \quad \text { or } \quad \mathrm{u} _{1}^{2}=\frac{\mathrm{gN} \omega^{2}}{2 \mathrm{~h}} \tag{1} \end{align*} $$

For ball B

Ball $\mathrm{B}$ is $\mathrm{n}$ steps below the top most step. In hits step $(\mathrm{N}-\mathrm{n})^{\text {th }}$ below its point of projection. Therefore

$$ \begin{equation*} \mathrm{u} _{2}^{2}=\frac{\mathrm{g}(\mathrm{N}-\mathrm{n}) \omega^{2}}{2 \mathrm{~h}} \tag{2} \end{equation*} $$

From Eqns. (1) and (2), we have

$$ \begin{aligned} & \left(\frac{\mathrm{u} _{1}}{\mathrm{u} _{2}}\right)^{2}=\frac{\mathrm{N}}{\mathrm{N}-\mathrm{n}} \\ & \therefore \frac{\mathrm{u} _{1}}{\mathrm{u} _{2}}=\sqrt{\frac{\mathrm{N}}{\mathrm{N}-\mathrm{n}}} \end{aligned} $$

Correct answer: (1)

Solution:

Let $x$ be the horizontal distance of wall from point of projection. Let the ball just grazes past top of wall $\mathrm{t}$ second after projection. $\mathrm{At} \quad \mathrm{t}=\mathrm{t} ; \mathrm{y}=\mathrm{h}=1.8 \mathrm{~m}$. Therefore

$1.8=(20 \sin 30) \mathrm{t}-\frac{1}{2} \times 10 \mathrm{t}^{2}=10 \mathrm{t}-5 \mathrm{t}^{2}$

or $5 \mathrm{t}^{2}-10 \mathrm{t}+1.8=0$

$\therefore \mathrm{t}=\frac{10 \pm \sqrt{100-36}}{10}=\frac{10 \pm 8}{10}=2 \mathrm{~s}$ or $1.8 \mathrm{~s}$

Obvisouly,

$$ \begin{aligned} & x=\left(\mathrm{v} _{0} \cos \theta\right) \mathrm{t}=(10 \sqrt{3}) \times 0.2 \text { or }(10 \sqrt{3}) 1.8 \\ & =2 \sqrt{3} \mathrm{~m} \text { or } 18 \sqrt{3} \mathrm{~m} \end{aligned} $$

The distance of the point where ball hits ground from point of projection equals the horizontal range. Therefore,

$$ \mathrm{R}=\frac{\mathrm{v} _{0}^{2} \sin \theta}{\mathrm{g}}=\frac{400 \times \frac{\sqrt{3}}{2}}{10}=20 \sqrt{3} $$

The required ratio is $\frac{x}{\mathrm{R}-x}$. It is

$\frac{2 \sqrt{3}}{18 \sqrt{3}}=\frac{1}{9} \quad$ or $\quad \frac{18 \sqrt{3}}{2 \sqrt{3}}=9$.

The correct choice is (1)

Correct answer: (4)

Solution:

In Fig. $\mathrm{O}$ and $\mathrm{C}$ are position of gun and car at $\mathrm{t}=0$. The bullet hits ground at point $\mathrm{P} . \mathrm{OP}=$ $\mathrm{R}=$ range of bullet. Let bullet hit ground at point $P$. Let $T$ be time of flight of bullet. The car and bullet reach point $\mathrm{P}$ at same time.

Therefore $\mathrm{OP}=\mathrm{R}=\left(\mathrm{v} _{0} \cos \theta\right) \mathrm{T}$.

The maximum vertical height $\mathrm{H}$; attained by bullet is

$\mathrm{H}=\frac{\mathrm{v} _{0}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

$=\frac{\left(\frac{4 \mathrm{~V}}{3 \cos \theta}\right)^{2} \sin ^{2} \theta}{2 \mathrm{~g}}=\frac{8}{9 \mathrm{~g}} \mathrm{~V}^{2} \tan ^{2} \theta$

Correct answer: (3)

Solution:

Let $\mathrm{v} _{0 x}$ and $\mathrm{v} _{\text {oy }}$ denote the initial value of $x$ and $\mathrm{y}$-component of velocity of projectile. Given

$$ \mathrm{v} _{0 x}=20 \cos 30^{\circ}=10 \sqrt{3} \mathrm{~ms}^{-1} ; \text { and } \mathrm{v} _{\mathrm{oy}}=20 \sin 30^{\circ}=10 \mathrm{~ms}^{-1} $$

The instantaneous $x$ and y co-ordinates of projectile are

$$ x=\mathrm{v} _{0 x} \mathrm{t}=10 \sqrt{3} \mathrm{t} \text { and } \mathrm{y}=10 \mathrm{t}-\frac{1}{2}(10) \mathrm{t}^{2}=10 \mathrm{t}-5 \mathrm{t}^{2} $$

The projectile is at a vertical height of $1.8 \mathrm{~m}$, (i.e. $\mathrm{y}=1.8 \mathrm{~m}$ ) at $\mathrm{t}=\mathrm{t}$.

$$ \begin{aligned} & \therefore 1.8=10 \mathrm{t}-5 \mathrm{t}^{2} \text { or } 5 \mathrm{t}^{2}-10 \mathrm{t}+1.8=0 \\ & \mathrm{t}=\frac{10 \pm \sqrt{100-36}}{10}=\frac{10 \pm 8}{10}=0.28 \text { or } 1.8 \mathrm{~s} \end{aligned} $$

The projectile attains $\mathrm{y}=1.8 \mathrm{~m}$ at two times $\mathrm{t} _{1}=0.2 \mathrm{~s}$ or $1.8 \mathrm{~s}$. The corresponding $x$-coordinates are

$$ x=10 \sqrt{3} \times 0.2=2 \sqrt{3} \mathrm{~m} \text { or } x=10 \sqrt{3} \times 1.8=18 \sqrt{3} \mathrm{~m} $$

The position vector of projectile is

$$ \mathbf{r}=2 \sqrt{3} \hat{i}+1.8 \hat{j} \text { or } \mathbf{r}=18 \sqrt{3} \hat{i}+1.8 \hat{j} $$

Correct answer: (2)

Solution:

Let $\mathbf{v}$ be the instantaneous velocity vector of projectile. Obviously

$$ \mathbf{v}=(10 \sqrt{3}) \hat{\mathrm{i}}+(30-10 \mathrm{t}) \hat{\mathrm{j}} \quad \mathrm{ms}^{-1} $$

$(\Delta \mathbf{p}) _{1}=$ Change in linear momentum in $1^{\text {st }}$ second of motion i.e. time interval $t=0$ to $t=1 \mathrm{~s}$.

$$ \begin{aligned} & =0.1[\{(10 \sqrt{3}) \hat{\mathrm{i}}+20 \hat{\mathrm{j}}\}-\{(10 \sqrt{3}) \hat{\mathrm{i}}+30 \hat{\mathrm{j}}\}] \\ & =-1 \hat{\mathrm{j}} ; \mathrm{kgms}^{-1} \end{aligned} $$

Similarly,

$(\Delta \mathbf{p}) _{2}=$ Change in linear momentum in $2^{\text {nd }}$ second of motion i.e. time interval $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$.

$$ \begin{aligned} & =0.1[\{(10 \sqrt{3}) \hat{i}+10 \hat{j}\}-\{(10 \sqrt{3}) \hat{i}+20 \hat{j}\}]=-1 \hat{j} k g m s^{-1} \\ & \therefore \frac{(\Delta \mathbf{p}) _{1}}{(\Delta \mathbf{p}) _{2}}=1 \end{aligned} $$

Correct answer: (3)

Solution:

The initial direction of firing of bullet is at an angle of $45^{\circ}$ with horizontal. The acceleration vector is always in a vertically downward direction. Since angle between $\mathbf{v}$ and $\mathbf{a}$ is $120^{\circ}, \mathbf{v}$ makes an angle of $30^{\circ}$ with the horizontal. Obviously

$$ \tan 30=\frac{\mathrm{v} _{\mathrm{y}}}{\mathrm{v} _{x}} $$

where $\mathrm{v} _{x}$ and $\mathrm{v} _{\mathrm{y}}$ denote the instantaneous $x$ - and $\mathrm{y}$-component of instantaneous velocity vector i.e. $\mathbf{v}$. We know $\mathrm{v} _{x}=\mathrm{v} _{0 x}=20 \mathrm{~ms}^{-1}$ and $\mathrm{v} _{\mathrm{y}}=\mathrm{v} _{0 \mathrm{y}}-\mathrm{gt}=20-10 \mathrm{t}$.

$$ \therefore \frac{1}{\sqrt{3}}=\frac{20-10 \mathrm{t}}{20} \quad \text { or } \quad \mathrm{t}=2(\sqrt{3}-1) $$

Correct answer: (3)

Solution:

Fig. (a) shows the path of the projectile projected from point $\mathrm{O}$. A is the highest point on its parabolic path and $\mathrm{OP}=\mathrm{R}=$ horizontal range of the projectile. The instantaneous speed of projectile at $\mathrm{A} _{\mathrm{v}} \cos \theta$. At A the projectile breaks up into two fragments having mass $\mathrm{m} _{1}=\frac{\mathrm{m}}{3}$ and $\mathrm{m} _{2}=\frac{2 \mathrm{~m}}{3}$ ( $\mathrm{m}$ is original mass of projectile). The breakup of projectile occurs on its own, i.e. there is no external force, therefore, the linear momentum is conserved. Given that fragment of mass $m _{2}$ hits ground at point B. This fragment moves vertically downward. The horizontal component of its velocity is zero. Let $\mathrm{v} _{1}$ be horizontal component of velocity of fragment of mass $m _{1}$. From law of conversation of linear momentum.

$$ \mathrm{mv} _{0} \cos \theta=\frac{\mathrm{m} _{1}}{3} \mathrm{v} _{1}, \quad \mathrm{v} _{1}=3 \mathrm{v} _{0} \cos \theta $$

There is no vertical component of velocity of both fragments. In hitting ground they travel same vertical distance $\mathrm{H}$.

$\therefore \mathrm{t} _{1}=\mathrm{t} _{2}=\frac{\mathrm{T}}{2}$ or $\frac{\mathrm{t} _{1}}{\mathrm{t} _{2}}=1$

Also for the fragment of mass $m _{1}$

$$ \mathrm{BR}=\left(3 \mathrm{v} _{0} \cos \theta\right) \frac{\mathrm{T}}{2}=\frac{3 \mathrm{R}}{2} $$

Obviously,

$$ \mathrm{OR}=\mathrm{OB}+\mathrm{BR}=\frac{\mathrm{R}}{2}+\frac{3 \mathrm{R}}{2}=2 \mathrm{R} $$

Correct answer:

Solution:

$\mathrm{k}=$ Force constant of rubber string $=0.05 \mathrm{kgf} \mathrm{cm}^{-1}=\frac{0.05 \times 10}{10^{-2}} \mathrm{Nm}^{-1}=50 \mathrm{Nm}^{-1}$

$$ \omega=300 \mathrm{r} . \text { p.m. }=5 \mathrm{r} . \mathrm{p} . \mathrm{s} .=10 \pi \mathrm{rad} \quad \mathrm{s}^{-1} $$

Let $x$ be increase in length of rubber string when particle attached to it moves in a horizontal circle. The radius of circle $=\mathrm{r}=(0.5+x) \mathrm{m}$. The tension in string $=\mathrm{T}=\mathrm{mr} \omega^{2}=\mathrm{k} x$

$\therefore 10^{-2}(0.5+x) \times(10 \pi)^{2}=50 x$

or $10^{-2} \times 10^{3}(0.5+x)=50 x \quad\left[\pi^{2} \simeq 10\right]$

$5+10 x=50 x$

$\therefore x=\frac{5}{40} \mathrm{~m}=0.125 \mathrm{~m}$

$\mathrm{T}=\mathrm{k} x=50 \times 0.125=6.25 \mathrm{~N}$

Correct answer: (4)

Solution:

The acceleration has two components.

(a) The radial component of acceleration is due to change direction of the velocity and equals $\frac{\mathrm{v}^{2}}{\mathrm{r}}$.

(b) The tangential component is due to change in magnitude of $v$ and equals $r \alpha$.

The two component of acceleration are mutually perpendicular. The net acceleration is $\mathrm{a}=\left[\left(\frac{\mathrm{v}^{2}}{\mathrm{r}}\right)^{2}+(\mathrm{r} \alpha)^{2}\right]^{1 / 2}$

Correct answer: (3)

Solution:

The instantaneous tension, $\mathrm{T}$, in string is

$$ \mathrm{T}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{k}}{\mathrm{r}^{3}} $$

$$ \text { or } \quad \mathrm{v} \alpha \frac{1}{\mathrm{r}} \text { or } \mathrm{vr}=\mathrm{const} $$

The instantaneous angular momentum of particle is mvr. Obviously this is a constant.

Correct answer: (2)

Solution:

Since $\mathbf{F} \cdot \mathbf{v}=0$; the instantaneous velocity is always perpendicular to the instantaneous direction of force. This condition is satisfied for a particle moving in a horizontal circle with uniform speed. The force acting on particle provides the necessary centeripetal force.

Correct answer: (2)

Solution:

Let $\mathrm{m}, \mathrm{v}$, and $\mathrm{r}$ denote mass of particle; speed of particle and length of string. The breaking strength, $\mathrm{T}$; of string is

$\mathrm{T}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{0.1 \times(2)^{2}}{0.2}=2 \mathrm{~N}$

$\mathrm{T}^{\prime}=$ Breaking strength of second string $=(1.75) \mathrm{T}=3.5 \mathrm{~N}$

$\mathrm{r}^{\prime}=$ Length of second string $=2 \mathrm{r}=0.4 \mathrm{~m}$

Let $\mathrm{m}$ ’ be mass of particle attached to string. Given,

$$ \begin{gathered} \mathrm{T}^{\prime}=3.5=\frac{\mathrm{m}^{\prime} \mathrm{v}^{2}}{\left(\mathrm{r}^{\prime}\right)} \\ \text { or } \quad \mathrm{m}^{\prime}=\frac{3.5 \times 0.4}{4} \mathrm{~kg}=0.35 \mathrm{~kg} \end{gathered} $$

Change in mass of particle $=\mathrm{m}^{\prime}-\mathrm{m}=0.25 \mathrm{~kg}$

Correct answer: (2)

Solution:

In Fig. $\hat{r}$ and $\hat{t}$ are instantaneous unit vector along radially outward and tangential direction. We know,

$\mathbf{v}=(\mathrm{r} \omega) \hat{\mathrm{t}}$ and $\mathbf{a}=\left(-\mathrm{r} \omega^{2}\right) \hat{\mathrm{r}}$. Resolve $\hat{\mathrm{r}}$ and $\hat{\mathrm{t}}$ into their components along $x$ and $y$-axis. Obviously

$$ \hat{\mathrm{r}}=(\cos \theta) \hat{\mathrm{i}}+(\sin \theta) \hat{\mathrm{j}} $$

and $\hat{\mathrm{t}}=(-\cos \theta) \hat{\mathrm{i}}+(\sin \theta) \hat{\mathrm{j}}$

Obviously (2) in the correct choice.

Correct answer: (4)

Solution:

The angular speed, $\omega$, of minutes hand of clock is

$$ \omega=\frac{2 \pi}{60 \times 60} \mathrm{rad} / \mathrm{s} $$

The linear speed $v$ of the tip of the minute’s hands

$$ =\ell \times \omega=\frac{0.3 \times 2 \pi}{60 \times 60} \mathrm{~ms}^{-1}=\frac{\pi}{6} \mathrm{mms}^{-1} $$

Let $\theta$ be the angle through which the minute’s hand of clock has turned in time initial t. The magnitude of change in velocity of the tip of minute’s hand $=2 \mathrm{v} \sin \left(\frac{\theta}{2}\right)$. Therefore

$$ \frac{\pi}{6}=2\left(\frac{\pi}{6}\right) \sin \left(\frac{\theta}{2}\right) \quad \text { or } \quad \sin \left(\frac{\theta}{2}\right)=\frac{\pi}{6} $$

$\therefore \theta=\frac{\pi}{3}=60^{\circ}$

The time interval $=\frac{60}{360} \mathrm{hr}=\frac{1}{6} \mathrm{hr}=10$ minute

Correct answer: (1)

Solution:

Since there is no gravitating matter nearby; whether particle moves is a horizontal or vertical circle; situation is same. In both cases speed can be constant but velocity is not constant. The direction of velocity changes.

Correct answer: (3)

Solution:

For motion in horizontal circle speed is constant but velocity is not. The direction of velocity changes.

For motion in vertical circle neither speed nor velocity is constant. Speed decreases as particle moves from lowest point of circular path to highest point and increases during the other half part of motion along vertical circle.

Correct answer: (1)

Solution:

Let $t$ be time taken by bullet to travel horizontal distance $D$. Obviously $t=\frac{D}{v} \cdot y=$ The vertical distance by which bullet misses hitting bird $=\frac{1}{2} \mathrm{gt}^{2}$

Let bird fly with vertical speed $\mathrm{v}^{\prime}$ when bullet leaves gun. The distance by which bullet misses hitting bird is $\left(v _{1} t-y\right)$. Similarly the bullet misses hitting bird in $2^{\text {nd }}$ case by distance $\left(y-v _{2} t\right)$. Given,

$$ \mathrm{v} _{1} \mathrm{t}-\mathrm{y}=\mathrm{y}-\mathrm{v} _{2} \mathrm{t} $$

or $\mathrm{t}=\frac{2 \mathrm{y}}{\mathrm{v} _{1}+\mathrm{v} _{2}}=\frac{1}{\left(\mathrm{v} _{1}+\mathrm{v} _{2}\right)} \mathrm{gt}^{2}$

$\therefore \mathrm{t}=\frac{\mathrm{v} _{1}+\mathrm{v} _{2}}{\mathrm{~g}} \quad$ or $\frac{\mathrm{D}}{\mathrm{v}}=\frac{\mathrm{v} _{1}+\mathrm{v} _{2}}{\mathrm{~g}}$

or $\quad v=\frac{\mathrm{gD}}{\left(\mathrm{v} _{1}+\mathrm{v} _{2}\right)}$

Correct answer: (2)

Solution:

Fig. shows particles $A$ and $B$ projected from point $O$ horizontally. Given $v _{A}=v _{1}, v _{B}=2 v _{1}$. Let $A _{1}$ and $B _{1}$ be the position of two particles at $t=t$ when their velocities are mutually perpendicular i.e. $\mathrm{A} _{1} \mathrm{O} _{1} \mathrm{~B} _{1}=\frac{\pi}{2}$. Let $\theta _{1}$ and $\theta _{2}$ be the angle the instantaneous velocity vector of $A$ and $B$ make with horizontal. Obviously

$\tan \theta _{1}=\frac{\mathrm{v} _{1}}{\mathrm{gt}} ; \quad \tan \theta _{2}=\frac{2 \mathrm{v} _{1}}{\mathrm{gt}}$

From $\Delta A_1 O B_1 ; \theta_1+\theta_2=\frac{\pi}{2}$ $$ \therefore \tan \theta_2=\tan \left(\frac{\pi}{2}-\theta_1\right)=\cot \theta_1 $$ or $\quad \tan \theta_1 \cdot \tan \theta_2=1$ $$ \therefore \frac{2 \mathrm{v}_1^2}{\mathrm{~g}^2 \mathrm{t}^2}=1 $$ or $\quad t=\sqrt{2}\left(\frac{v_1}{g}\right)$

Correct answer: (4)

Solution:

Given $|\mathbf{A} \times \mathbf{B}|=\sqrt{3}(\mathbf{A} \cdot \mathbf{B})$

$\therefore A B \sin \theta=\sqrt{3} A B \cos \theta \quad$ or $\tan \theta=\sqrt{3}$

$\therefore \theta=60^{\circ}$

Now, $|\mathbf{A}+\mathbf{B}|^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \times \frac{1}{2}$ $\left[\because \cos 60^{\circ}=\frac{1}{2}\right]$

and $|\mathbf{A}-\mathbf{B}|^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \times \frac{1}{2}$

$\therefore|\mathbf{A}+\mathbf{B}|^{2}-|\mathbf{A}-\mathbf{B}|^{2}=2 \mathrm{AB}$

or $\frac{1}{2}\left\{|\mathbf{A}+\mathbf{B}|^{2}-|\mathbf{A}-\mathbf{B}|^{2}\right\}=\mathrm{AB}$

Correct answer (4)

Correct answer: (2)

Solution:

(i) $\mathrm{h}=0+\frac{1}{2} \mathrm{gt} _{1}^{2}$

$\therefore \mathrm{t} _{1}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}$

(ii) $\quad 2 \mathrm{~h}=0+\frac{1}{2} \mathrm{~g}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)^{2}$

$\therefore \mathrm{t} _{1}+\mathrm{t} _{2}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{2}$

(iii) $3 \mathrm{~h}=0+\frac{1}{2} \mathrm{~g}\left(\mathrm{t} _{1}+\mathrm{t} _{2}+\mathrm{t} _{3}\right)^{2} \therefore \mathrm{t} _{1}+\mathrm{t} _{2}+\mathrm{t} _{3}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{3}$

From Eqns. (i), (ii) and (iii), we have

$$ \begin{aligned} & \mathrm{t} _{1}= \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \\ & \mathrm{t} _{2}=\left(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{2}\right)-\mathrm{t} _{1}=\left(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{2}\right)-\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \\ &= \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}[\sqrt{2}-1] \\ & \mathrm{t} _{3}=\left(\mathrm{t} _{1}+\mathrm{t} _{2}+\mathrm{t} _{3}\right)-\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right) \\ &= \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{3}-\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \times \sqrt{2}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}[\sqrt{3}-\sqrt{2}] \\ & \therefore \mathrm{t} _{1}: \mathrm{t} _{2}: \mathrm{t} _{3}=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2}) \end{aligned} $$

Correct answer (2)

Correct answer(3)

Solution:

When the man throws the ball down, he will get a velocity v upwards, such that:

$$ \mathrm{mv}=\mathrm{MV} \quad(\text { Law of conservation of linear momentum) } $$

or $\quad 0.2 \times 2=50 \times \mathrm{V} \quad$ or $\quad \mathrm{V}=\frac{1}{50} \mathrm{~ms}^{-1}$

Time taken by the ball to reach floor is

$\mathrm{t}=\frac{10}{2}=5 \mathrm{~s} \quad$ (The ball will not accelerate in the absence of gravity)

$\therefore$ Distance moved up by the man in this $\mathrm{t}=5 \mathrm{~s}$ is

$$ \mathrm{d}=\mathrm{V} \times \mathrm{t}=\frac{1}{50} \times 5=0.1 \mathrm{~m} ; \text { upwards } $$

Hence the distance of the man from the floor, when the stone hits the floor is

$$ \mathrm{H}=\mathrm{h}+\mathrm{d}=10+0.1=10.1 \mathrm{~m} $$

Correct answer(2)

Solution:

Acceleration of A down the plane is $\mathrm{g} \sin 60 \mathrm{~g} \cos 30^{\circ}$ as shown in Fig. Resolve acceleration into horizontal and vertical components as shown in Fig. Obvisouly

The vertical acceleration of $\mathrm{A}=\left(\mathrm{g} \cos 30^{\circ}\right) \cos 30^{\circ}=\mathrm{g} \times \frac{3}{4}$

Similarly the vertical acceleration of $\mathrm{B}=\left(\mathrm{g} \cos 60^{\circ}\right) \cos 60^{\circ}=\mathrm{g} \times \frac{1}{4}$

$\therefore$ Relative vertical acceleration of $A$ with respect to $\mathrm{B}=\mathrm{g} \times \frac{3}{4}-\mathrm{g} \times \frac{1}{4}$

$$ =\mathrm{g} \times \frac{2}{4}=\frac{\mathrm{g}}{2}=4.9 \mathrm{~ms}^{-2} $$

Correct answer: (1)

Solution:

Given $s=t^{3}+5$. Therefore the instantaneous velocity

$$ \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=3 \mathrm{t}^{2} $$

Since this velocity is not constant in magnitude, the point $\mathrm{P}$ will have a tangential acceleration, $\mathrm{a} _{1}=\frac{\mathrm{dv}}{\mathrm{dt}}=3 \times 2 \mathrm{t}=6 \mathrm{t}$

At $\mathrm{t}=2 \mathrm{~s}, \mathrm{a} _{1}=6 \times 2=12 \mathrm{~ms}^{-2}$.

Now, $\mathrm{P}$ is moving in circular path therefore $\mathrm{P}$ will have centripetal acceleration.

$$ \mathrm{a} _{2}=\frac{\mathrm{v}^{2}}{\mathrm{R}}=\frac{\left(3 \mathrm{t}^{2}\right)^{2}}{\mathrm{R}}=\frac{9 \mathrm{t}^{4}}{20} $$

At $\mathrm{t}=2 \mathrm{~s}, \mathrm{a} _{2} \frac{9 \times 2 \times 2 \times 2 \times 2}{20}=7.2 \mathrm{~ms}^{-2}$

$\therefore \quad \mathrm{a}=\sqrt{\mathrm{a} _{1}^{2}+\mathrm{a} _{2}^{2}}=\sqrt{12^{2}+(7.2)^{2}} \approx 14.5 \mathrm{~ms}^{-2}$

Correct answer: (3)

Solution:

Suppose the bullet loses $\frac{\mathrm{u}}{\mathrm{n}}$ of its initial velocity, after penetrating through distance s. So, final velocity, after traveling distance $\mathrm{s}$ in target, is:

$$ \mathrm{v}=\mathrm{u}-\frac{\mathrm{u}}{\mathrm{n}}=\frac{\mathrm{u}}{\mathrm{n}}(\mathrm{n}-1) $$

If a is the deceleration, then :

$$ \begin{gathered} v^{2}=u^{2}-2 a s \\ \text { or } \quad\left[\frac{u}{n}(n-1)\right]^{2}=u^{2}-2 \text { as } \\ \therefore \quad a=\frac{1}{2 s}\left[u^{2}-\frac{u^{2}}{n^{2}}\left(n^{2}-2 n+1\right)\right] \\ \quad=\frac{1}{2 s}\left[u^{2}-u^{2}+\frac{u^{2}}{n^{2}} \times 2 n-\frac{u^{2}}{n^{2}}\right] \\ \quad=\frac{1}{2 s}\left[\frac{u^{2}}{n^{2}}(2 n-1)\right] \end{gathered} $$

If the bullet goes a further distance $x$, before coming to rest, then

$$ \begin{aligned} & \mathrm{v} _{\mathrm{f}}^{2}=\mathrm{v}^{2}-2 \mathrm{a} x \\ & \text { or } \quad 0=\frac{\mathrm{u}^{2}}{\mathrm{n}^{2}}(\mathrm{n}-1)^{2}-2 \times\left[\frac{1}{2 \mathrm{~s}} \times \frac{\mathrm{u}^{2}}{\mathrm{n}^{2}}(2 \mathrm{n}-1)\right] x \\ & \therefore \quad x=\frac{(\mathrm{n}-1)^{2}}{(2 \mathrm{n}-1)} \mathrm{s} \end{aligned} $$

Given, $s=3 \mathrm{~m}$, therefore

$$ x=\frac{(3-1)^{2}}{(2 \times 3-1)} \times 20=\frac{4}{5} \times 20=16 \mathrm{~m} $$

Therefore, total penetration is

$$ \mathrm{s}+x=20+16=36 $$

Alternatively:

$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \hspace{40mm} . . . . . .(1) $

Here, $v=u=\frac{1}{3} u=\frac{2}{3} u$ and $s=20 m$

Since a is -ve, Eqn. (1) can be written as:

$$ \frac{4}{9} u^{2}=u^{2}-29 \times 20 $$

$\therefore 2 \mathrm{a} \times 20=\frac{5}{9} \mathrm{u}^{2}$

or $2 \mathrm{a}=\left(\frac{5}{9}\right) \frac{\mathrm{u}^{2}}{20} \hspace{40mm} . . . . . .(2)$

Now, if ’ $\mathrm{d}$ ’ is the total penetration, then

$$ \mathrm{v} _{\mathrm{f}}^{2}=\mathrm{u}^{2}-2 \mathrm{a} \times \mathrm{d} \quad(\mathrm{a}=\text { deceleration }) $$

$$ \mathrm{O}^{2}=\mathrm{u}^{2}-\left(\frac{5}{9}\right) \frac{\mathrm{u}^{2}}{20} \times \mathrm{d} $$

or $\quad u^{2}=\frac{5}{9} \times \frac{u^{2} \times d}{20}$

$$ \mathrm{d}=\frac{180}{5}=36 \mathrm{~m} $$

Correct answer(2)

Solution:

Given, $\mathrm{y} _{(0)}=+5 \mathrm{~m}$. From the graph,

Velocity $=-8 \mathrm{~ms}^{-1}$ at $\mathrm{t}=0$

$$ \therefore \mathrm{v} _{(0)}=-8 \mathrm{~ms}^{-1} $$

Now, $\mathrm{v}-\mathrm{t}$ graph is a straight line. Therefore acceleration is a constant $\mathrm{a}=$ slope of $\mathrm{v} \quad vs \quad \mathrm{t}$ graph.

$$ =+\frac{8}{2}=+4 \mathrm{~ms}^{-2} $$

The final position at $t=t$ can be written as: $y _{(t)}=y _{(0)}+v _{(0)} t+\frac{1}{2} a t^{2}$

or $\quad y=5-8 t+\frac{1}{2} \times 4 t^{2}$

or $y=2 t^{2}-8 t+5$

Correct answer(2)

Solution:

The momentum imparted to the two pieces together is:

$\Delta \mathbf{p}=1 \times(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}-\mathrm{s} \quad(\mathrm{m}=1 \mathrm{~kg})$

As the body of $3 \mathrm{~kg}$ is initially at rest the change in momentum for the third piece should be equal and opposite that for the first two taken together.

$\therefore \Delta \mathbf{p}$ for $3^{\text {rd }}$ piece $=-(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{N}-\mathrm{s}$

Time taken for this change, $\Delta \mathrm{t}=10^{-4} \mathrm{~s}$

$\therefore$ Force on the third piece is

$$ \mathbf{F} _{3}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=-\frac{3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}}{10^{-4}} \mathrm{~N} $$

$$ \begin{gathered} \mathbf{F} _{3}=-(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \times 10^{4} \mathrm{~N} \\ \therefore\left|\mathbf{F} _{3}\right|=\mathrm{F} _{3}=\sqrt{3^{2}+4^{2}} \times 10^{4} \mathrm{~N}=5 \times 10^{4} \mathrm{~N} \\ =50 \times 10^{3} \mathrm{~N}=50 \text { Kilo newton } \end{gathered} $$

Correct answer: (1)

Solution:

Given : $x=\mathrm{mt}^{2}-\mathrm{nt}^{3}$. $\therefore \mathrm{v}=\frac{\mathrm{d} x}{\mathrm{dt}}=2 \mathrm{mt}-3 \mathrm{nt}^{2} \hspace{60mm} . . . . . .(i)$

Hence, $\mathrm{a}=\frac{\mathrm{d} v}{\mathrm{dt}}=2 \mathrm{~m}-6 \mathrm{nt} \hspace{40mm} . . . . . .(ii)$

or $\quad a=(-6 n) t+2 m$

Eqn. (ii) is of the form $\mathrm{y}=\mathrm{m} x+\mathrm{c}$ for the equation of a straight line, where $\mathrm{m}$ is slope and $\mathrm{c}$ is $\mathrm{y}$ intercept.

Hence $a-t$ graph is a straight line with

(i) slope of $-6 \mathrm{~m}$.

(ii) of-intercept $=+2 \mathrm{~m}$ and

(iii) a $=0$ at $\mathrm{t}=\frac{2 \mathrm{~m}}{6 \mathrm{n}}=\frac{\mathrm{m}}{3 \mathrm{n}}$

(i), (ii) and (iii) are true only for (1)

Correct answer: (3)

Solution:

Here, $\mathrm{v} _{1}+\mathrm{v} _{2}=6 \mathrm{~ms}^{-1}$, and $\mathrm{v} _{1}-\mathrm{v} _{2}=2.4 \mathrm{~ms}^{-1}$

$\therefore 2 \mathrm{v} _{1}=8.4 \mathrm{~ms}^{-1}$

or $\mathrm{v} _{1}=4.2 \mathrm{~ms}^{-1}$, and $\mathrm{v} _{2}=1.8 \mathrm{~ms}^{-1}$

Correct answer: (2)

Solution:

The particle is in non-uniform motion from $t=20 \mathrm{~s}$ to $\mathrm{t}=40 \mathrm{~s}$. Total displacement in this interval is equal to the area under the graph in this interval.

$\therefore \mathrm{S}=2 \times\left[\frac{1}{2}(1+4) \times 10\right]=50 \mathrm{~m}$

Correct answer: (2)

Solution:

Suppose $x$ is the length of the shadow. When the man is at $\mathrm{d}$ distance from the pole, as shown in Fig. If he takes $t$ seconds to reach the pole, the edge of his shadow will also reach the pole in the same time $\mathrm{t} \frac{\mathrm{h}}{\mathrm{x}}=\frac{\mathrm{H}}{\mathrm{x}+\mathrm{d}}$ and $\mathrm{d}=$ ut

$\therefore x \mathrm{H}=\mathrm{h} x+\mathrm{hd}$

or $\quad x(\mathrm{H}-\mathrm{h})=\mathrm{hut}$

$\therefore x=\frac{\mathrm{hut}}{\mathrm{H}-\mathrm{h}}$

The distance travelled by the edge of the shadow, in t seconds, is equal to $x+\mathrm{d}$. The velocity of the edge, $v$ is

$$ \begin{aligned} & \mathrm{v}=\frac{x+\mathrm{d}}{\mathrm{t}}=\frac{\left[\frac{\mathrm{hut}}{\mathrm{H}-\mathrm{h}}+\mathrm{ut}\right]}{\mathrm{t}} \\ & =\frac{\mathrm{hu}}{\mathrm{H}-\mathrm{h}}+\mathrm{u}=\left(\frac{\mathrm{h}+\mathrm{H}-\mathrm{h}}{\mathrm{H}-\mathrm{h}}\right) \mathrm{u} \\ & \quad=\left(\frac{\mathrm{H}}{\mathrm{H}-\mathrm{h}}\right) \mathrm{u} \end{aligned} $$$ \qquad $

Correct answer: (3)

Solution:

Mass does not influence the time and speed. Now, for a body dropped, $\mathrm{u}=0$

$\therefore \mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ reduces to

$\mathrm{S}=0+\frac{1}{2} \mathrm{gt}^{2}$

$\therefore \mathrm{S} _{1}=\frac{1}{2} \mathrm{gt} _{1}^{2}$ and $\mathrm{S} _{2}=\frac{1}{2} \mathrm{gt} _{2}^{2}$

So, $\mathrm{t} _{1}: \mathrm{t} _{2}=\sqrt{\mathrm{S} _{1}}: \sqrt{\mathrm{S} _{2}}=4: 5$

Now, $\mathrm{v}=\mathrm{u}+\mathrm{at}=0+\mathrm{gt}$

$\therefore \mathrm{v} _{1}=\mathrm{gt} _{1}$ and $\mathrm{v} _{2}=\mathrm{gt} _{2}$

$\therefore \mathrm{v} _{1}: \mathrm{v} _{2}=\mathrm{t} _{1}: \mathrm{t} _{2}=4: 5$

Correct answer: (1)

Solution:

According to the diagram given:

$$ x^{2}+y^{2}=\ell^{2}=\text { constant } $$

Differentiating with respect to time,

$$ 2 x \frac{\mathrm{d} x}{\mathrm{dt}}+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dt}}=0 $$

$\therefore x \mathrm{v} _{x}=-\mathrm{yv} _{\mathrm{y}}$

or $\quad \mathrm{v} _{\mathrm{x}}=-\mathrm{v} _{\mathrm{y}}\left(\frac{\mathrm{y}}{x}\right)$

$\mathrm{v} _{x}=-\mathrm{v} _{y}\left(\frac{\mathrm{y}}{\sqrt{\ell _{2}-\mathrm{y} _{2}}}\right)$

Correct answer: (1)

Correct answer : (3)

Solution:

Given,

$x _{0}=-5 \mathrm{~m}$. The distance $\mathrm{s} _{3}$ travelled in third second is

$\mathrm{s} _{3}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 3-1)$

or $\quad 10=\mathrm{u}+\frac{5}{2} \mathrm{a} \hspace{40mm} . . . . . .(1)$

Similarly the distance travelled in $6^{\text {th }}$ second; $\mathrm{s} _{6}$, is

$\mathrm{s} _{6}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 6-1)$

or $\quad 28=u+\frac{11}{2} \mathrm{a} \hspace{40mm} . . . . . .(2)$

From Eqns. (1) and (2) we have $18=3 \mathrm{a}$ or $\mathrm{a}=6 \mathrm{~ms}^{-2}$

From Eqn(1), we get

$10=\mathrm{u}+\frac{5}{2} \times 6 \quad$ or $\mathrm{u}=-5 \mathrm{~ms}^{-1}$

At $\mathrm{t}=8 \mathrm{~s}$, the position $x _{8}$ of particle is

$x _{8}=x _{0}+\left(\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\right)$

$=-5+(-5) \times 8+\frac{1}{2} \times 6 \times 8^{2}=-45+192=147 \mathrm{~m}$

Correct answer:(3)

Solution:

Let $\mathrm{t} _{\mathrm{A}}$ be the time taken to fall through height $\mathrm{h}=180 \mathrm{~m}$, after the ball is dropped.

$$ \therefore \mathrm{t} _{0 \mathrm{~A}}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=\sqrt{\frac{2 \times 180}{10}}=6 \mathrm{~s} $$

Now, $\mathrm{v} _{\mathrm{A}}=\mathrm{u}+\mathrm{at}=0+\mathrm{gt}$

$$ =10 \times 6=60 \mathrm{~ms}^{-1} $$

$\mathrm{v} _{\mathrm{B}}=\mathrm{v} _{\mathrm{A}}-10 \% \mathrm{v} _{\mathrm{A}}=60-\left(\frac{1}{10} \times 60\right)$

$$ =60-6=54 \mathrm{~ms}^{-1} $$

Let $t _{B D}$ be the time taken to bounce up from the floor, reach the maximum height and then come down to hit the floor again.

$\therefore \mathrm{t} _{\mathrm{BD}}=\frac{2 \mathrm{u}}{\mathrm{g}}=\frac{2 \times 54}{10}=10.8 \mathrm{~s}$

Hence the total time taken is

$\mathrm{t} _{\mathrm{OD}}=\mathrm{t} _{\mathrm{OA}}+\mathrm{t} _{\mathrm{BD}}=16.8 \mathrm{~s}$

Correct answer: (3)

Solution:

The speed-time graph for the motion of the train is shown in the figure. The total distance covered is:

$S=\left(\frac{1}{2} \times t \times 60\right)+(60 \times 8 t)+\left(\frac{1}{2} t \times 60\right)=540 t$

Total time taken $=t+8 t+t=10 t$

$\therefore$ Average speed $=\frac{540 \mathrm{t}}{10 \mathrm{t}}=54 \mathrm{kmh}^{-1}$

Correct answer: (4)

Solution:

Let $A$ take $t _{1}$ seconds to finish the race. Then, according to the data given, $B$ will take $\left(t _{1}+t\right)$ seconds to finish the race. Also, let $\mathrm{v} _{1}$ be the velocity of $B$ at the finishing point. Then the velocity of $A$ at the finishing point will be $\left(\mathrm{v} _{1}+\mathrm{v}\right)$. Therefore, for $\mathrm{A}$,

$$ \begin{equation*} \mathrm{v} _{1}+\mathrm{v}=0+\mathrm{a} _{1} \mathrm{t} _{1} \tag{1} \end{equation*} $$

and for $\mathrm{B}, \mathrm{v} _{1}=0+\mathrm{a} _{2}\left(\mathrm{t} _{1}+\mathrm{t}\right) \hspace{40mm} . . . . . .(2)$

$\therefore \mathrm{v}=\left(\mathrm{a} _{1}-\mathrm{a} _{2}\right) \mathrm{t} _{1}-\mathrm{a} _{2} \mathrm{t} \hspace{40mm} . . . . . .(3) $

Now, total distance travelled by both $A$ and $B$ is same. i.e. $S _{A}=S _{B}$. Using the formula, $S=u t+\frac{1}{2} a t^{2}$, we get;

$S _{A}=0+\frac{1}{2} a _{1} t _{1}^{2} \quad$ and $\quad S _{B}=0+\frac{1}{2} a _{2}\left(t _{1}+t\right)^{2}$

or $\sqrt{\mathrm{a} _{1}^{\prime}} \mathrm{t} _{1}=\sqrt{\mathrm{a} _{2}^{\prime}}\left(\mathrm{t} _{1}+\mathrm{t}\right)$

$\therefore \mathrm{t} _{1}\left(\sqrt{\mathrm{a} _{1}}-\sqrt{\mathrm{a} _{2}}\right)=\sqrt{\mathrm{a} _{2}} \mathrm{t}$

or $\quad \mathrm{t} _{1}=\frac{\left(\sqrt{\mathrm{a} _{2}}\right) \mathrm{t}}{\left(\sqrt{\mathrm{a} _{1}}-\sqrt{\mathrm{a} _{2}}\right)} \hspace{40mm} . . . . . .(4)$

Subsituting value of $t _{1}$ from Eqn (4) in Enq (3) we have

$$ \begin{aligned} v= & \left(a _{1}-a _{2}\right) \cdot \frac{\sqrt{a _{2}} t}{\left(\sqrt{a _{1}}-\sqrt{a _{2}}\right)}-a _{2} t \\ & =\frac{\left[\sqrt{a _{1}} \sqrt{a _{2}}-a _{2} \sqrt{a _{2}}-a _{2} \sqrt{a _{1}}+a _{2} \sqrt{a _{2}}\right]}{\sqrt{a _{1}}-\sqrt{a _{2}}} t \\ & =\frac{\left(a _{1} \sqrt{a 2}-a _{2} \sqrt{a _{1}}\right)}{\sqrt{a _{1}}-\sqrt{a _{2}}} t \\ & =\sqrt{a _{1} a _{2}} \frac{\left(\sqrt{a _{1}}-\sqrt{a _{2}}\right)}{\left(\sqrt{a _{1}}-\sqrt{a _{2}}\right)} t \\ & =\left(a _{1} a _{2}\right) t \end{aligned} $$

Correct answer: (1)

Solution:

When the two particles meet we have for the first particle,

$$ \begin{equation*} \frac{\mathrm{h}}{\mathrm{n}}=\frac{1}{2} \mathrm{gt}^{2} \tag{1} \end{equation*} $$

and for the second particle

$$ \begin{equation*} \mathrm{h}-\frac{\mathrm{h}}{\mathrm{n}}=\mathrm{u} _{2} \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \tag{2} \end{equation*} $$

Adding Eqn, (1) and (2), we have

$$ \begin{equation*} \mathrm{h}=\mathrm{u} _{2} \mathrm{t} \text { or } \mathrm{t}=\frac{\mathrm{h}}{\mathrm{u} _{2}} \tag{3} \end{equation*} $$

Substituting value of t given by Eqn. (3) in Eqn. (1), we get

$$ \begin{align*} & \frac{\mathrm{h}}{\mathrm{n}}=\frac{1}{2} \mathrm{~g} \frac{\mathrm{h}^{2}}{\mathrm{u} _{2}^{2}} \\ & \therefore \mathrm{u} _{2}^{2}=\frac{1}{2} \mathrm{ngh} \tag{4} \end{align*} $$

At time $t$, when the two particle meet for the first particle

$$ \begin{align*} & \mathrm{v} _{1}^{2}=\mathrm{u} _{1}^{2}+2 \mathrm{~g}\left(\frac{\mathrm{h}}{\mathrm{n}}\right) \\ & \therefore \mathrm{v} _{1}^{2}=\mathrm{gh} \frac{2}{\mathrm{~h}} \quad\left(\because \mathrm{u} _{1}=0\right) \tag{5} \end{align*} $$

For the second particle, we have,

$$ \begin{align*} v _{2}^{2}= & u _{2}^{2}-2 g\left(h-\frac{h}{n}\right) \\ & =\frac{1}{2} n g h-g h\left(2-\frac{2}{n}\right) \\ & =\operatorname{gh}\left(\frac{n}{2}-2+\frac{2}{n}\right) \\ & =g h\left[\frac{n^{2}-4 n+4}{2 n}\right] \\ & =g h \frac{(n-2)^{2}}{2 n} \tag{6} \end{align*} $$

From Eqn s. (5) and (6), we have

$$ \frac{v _{1}^{2}}{v _{2}^{2}}=\frac{\operatorname{gh}\left(\frac{2}{n}\right)}{\operatorname{gh} \frac{(n-2)^{2}}{2 n}}=\frac{4}{(n-2)^{2}} $$

$\therefore \frac{\mathrm{v} _{1}}{\mathrm{v} _{2}}=\frac{2}{(\mathrm{n}-2)}$

or $\quad \mathrm{v} _{1}: \mathrm{v} _{2}=2:(\mathrm{n}-2)$

Correct answer: (4)

Solution:

Let ’ $u$ ’ be the velocity of projection of the particle at 0 . The total time taken to reach the ground back is $\mathrm{t} _{1}+\mathrm{t} _{2}$.

From $\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$, we have,

$0=\mathrm{u}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)+\frac{1}{2}(-\mathrm{g})\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right)^{2}$

$\therefore \mathrm{u}=\frac{\mathrm{g}}{2}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right) \hspace{40mm} . . . . . .(1)$

Now, for the first $t$, second of motion, we have,

$$ \begin{aligned} \mathrm{h}=\mathrm{ut} _{1} & -\frac{1}{2} \mathrm{gt} _{1}^{2}=\frac{\mathrm{g}}{2}\left(\mathrm{t} _{1}+\mathrm{t} _{2}\right) \mathrm{t} _{1}-\frac{\mathrm{g}}{2} \mathrm{t} _{1}^{2} \\ & =\frac{\mathrm{g}}{2}\left[\mathrm{t} _{1}^{2}+\mathrm{t} _{1} \mathrm{t} _{2}-\mathrm{t} _{1}^{2}\right]=\frac{1}{2} \mathrm{gt} _{1} \mathrm{t} _{2} \end{aligned} $$

Correct answer: (4)

Solution:

Let the total length of the train be $2 x$. Let the velocity of the middle point of the train passing the pole be $v _{m}$. Then, for the first half of the journey of train,

$$ \begin{equation*} \mathrm{v} _{\mathrm{m}}^{2}=\mathrm{u}^{2}+2 \mathrm{a} x \tag{1} \end{equation*} $$

For the second half part of the journey, we have;

$$ \mathrm{v}^{2}=\mathrm{v} _{\mathrm{m}}^{2}+2 \mathrm{a} x $$

or $\mathrm{v} _{\mathrm{m}}^{2}=\mathrm{v}^{2}-2 \mathrm{a} x \hspace{40mm} . . . . . .(2)$

Adding Eqns. (1) and (2) we have

$$ \begin{aligned} & 2 \mathrm{v} _{\mathrm{m}}^{2}=\mathrm{u}^{2}+\mathrm{v}^{2} \\ & \therefore \mathrm{v} _{\mathrm{m}}=\sqrt{\frac{\mathrm{u}^{2}+\mathrm{v}^{2}}{2}} \end{aligned} $$

Correct answer: (4)

Solution:

Let $\mathrm{v}$ be the velocity of the balloon, when it has risen to the height $\mathrm{H}$. Then,

$$ \mathrm{v}^{2}=0^{2}+2\left(\frac{\mathrm{g}}{8}\right) \mathrm{H} $$

$$ \begin{equation*} \text { or } \quad v=\frac{\sqrt{g H}}{2} \tag{1} \end{equation*} $$

This $\mathrm{v}$ will be the initial velocity of the stone when dropped from the balloon. If it is the time taken by the stone to reach the ground, then:

$-\mathrm{H}=\mathrm{v} \times \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$

or $\quad-\mathrm{H}=\frac{\sqrt{\mathrm{gH}}}{2} \times \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}$ or $\quad \mathrm{gt}^{2}-(\sqrt{\mathrm{gH}}) \mathrm{t}-2 \mathrm{H}=0 \hspace{20mm} . . . . . .(2)$

$\therefore \quad \mathrm{t}=\frac{\sqrt{\mathrm{gH}} \pm \sqrt{\mathrm{gH}+8 \mathrm{gH}}}{2 \mathrm{~g}}$

$=\frac{\sqrt{\mathrm{gH}}+3 \sqrt{\mathrm{gH}}}{2 \mathrm{~g}}=2 \sqrt{\frac{\mathrm{H}}{\mathrm{g}}}$

Correct answer: (4)

Solution:

Let $x$-axis be taken along the horizontal and $\mathrm{y}$-axis along the vertical as show in Fig.

The actual velocity of the rain drops, is

$$ \begin{equation*} \mathbf{v} _{\mathbf{r}}=\mathrm{v} _{x} \hat{\mathrm{i}}+\mathrm{v} _{\mathrm{y}} \hat{\mathrm{j}} \tag{1} \end{equation*} $$

For the man, $\mathbf{v} _{\mathbf{m}}=3 \hat{i} \hspace{40mm} . . . . . .(2)$

The relative velocity of the rain drop with respect to the man is:

$$ \mathbf{v} _{\mathbf{m}}=\mathbf{v} _{\mathbf{r}}+\left(-\mathbf{v} _{\mathbf{m}}\right) $$

$$ =\mathrm{v} _{x} \hat{\mathrm{i}}+\mathrm{v} _{y} \hat{\mathrm{j}}+(-3 \hat{\mathrm{i}}) $$

$$ \begin{equation*} =\left(\mathrm{v} _{x}-3\right) \hat{\mathrm{i}}+\mathrm{v} _{\mathrm{y}} \hat{\mathrm{j}} \tag{3} \end{equation*} $$

The relative velocity $\left(\mathbf{v} _{\mathbf{r m}}\right)$ makes an angle $\theta$ with the velocity of the man. Therefore

$$ \begin{equation*} \tan \theta=\frac{\mathrm{v} _{\mathrm{y}}}{\mathrm{v} _{x}-3} \tag{4} \end{equation*} $$

In the first case, drops appear to fall vertically.

$\therefore \tan \theta=270^{\circ}=\propto=\frac{1}{0}$

or $\frac{1}{0}=\frac{\mathrm{v} _{\mathrm{y}}}{\mathrm{v} _{x}-3}$

$\therefore \mathrm{v} _{x}=3 \mathrm{kmh}^{-1} \hspace{40mm} . . . . . .(5)$

FromEqn. (1), $\mathbf{v} _{\mathbf{r}}=3 \hat{\mathrm{i}}+\mathrm{v} _{\mathrm{y}} \hat{\mathrm{j}} \hspace{40mm} . . . . . .(6)$

In the second case,

$$ \mathbf{v} _{m}=6 \hat{\dot{i}} $$

Here, $\mathbf{v} _{\mathbf{r m}}=\mathbf{v} _{\mathbf{r}}+\left(-\mathbf{v} _{\mathbf{m}}\right)$

$$ \begin{align*} & =\left(3 \hat{i}+v _{y} \hat{j}\right)+(-6 \hat{i}) \\ & =-3 \hat{i}+v _{y} \hat{j} \tag{7} \end{align*} $$

Since the rain drops now strike man at $45^{\circ}$ from the front, here $\theta=225^{\circ}$.

$\therefore \tan 225^{\circ}=\frac{\mathrm{v} _{\mathrm{y}}}{-3}$,

or $\quad 1=\frac{\mathrm{v} _{\mathrm{y}}}{-3} \quad \therefore \mathrm{v} _{\mathrm{y}}=-3 \mathrm{kmh}^{-1}$

Therefore, $\mathrm{v} _{\mathrm{r}}=3 \hat{\mathrm{i}}-3 \hat{\mathrm{j}} \mathrm{kmh}^{-1}\hspace{40mm} . . . . . .(8)$

or $\quad \mathrm{v} _{\mathrm{r}}=3 \sqrt{2} \mathrm{kmh}^{-1}$

Here, $\tan \theta=\frac{-3}{3}=-1$

or $\theta=270^{\circ}+45^{\circ}=315^{\circ}$, with $\hat{i}$

Hence, the speed of rain drops is $3 \sqrt{2} \mathrm{kmh}^{-1}$ and the direction is $45^{0}$ with the vertical and in the direction of man motion.

Correct answer: (2)

Solution:

Number of balls thrown is $1 \mathrm{~s}=\mathrm{n}$, Time interval between two balls $=\frac{1}{\mathrm{n}}$

Now, he throws one when the previous one thrown reaches the maximum height.

$\therefore$ Time taken by one ball to reach the maximum height is

$$ \begin{equation*} \mathrm{t}=\frac{1}{\mathrm{n}} \tag{1} \end{equation*} $$

Let $u$ be the velocity with which the ball is thrown. Then the velocity at the highest point is zero, and so,

$$ 0=\mathrm{u}-\mathrm{gt} $$

or $\quad \mathrm{u}=\frac{\mathrm{g}}{\mathrm{n}} \hspace{40mm} . . . . . .(2)$

Now, $\mathrm{v}^{2}=\mathrm{u}^{2}+2$ as

or $\quad 0=\mathrm{u}^{2}-2 \mathrm{gh}$

or $\quad 2 \mathrm{gh}=\frac{\mathrm{g}^{2}}{\mathrm{n}^{2}}$

$\therefore \mathrm{h}=\frac{\mathrm{g}}{2 \mathrm{n}^{2}}$

Correct answer: (1)

Solution:

Let $\mathrm{v} _{x}$ and $\mathrm{v} _{\mathrm{y}}$ be the instantaneous $x$ and $\mathrm{y}$-component of velocity. Then

$\mathrm{v} _{x}=\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{k} _{1} ; \quad \mathrm{v} _{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{k} _{1}\left(1-\mathrm{k} _{2} \mathrm{t}\right)+\mathrm{k} _{1} \mathrm{t}\left(-\mathrm{k} _{2}\right)=\mathrm{k} _{1}\left(1-2 \mathrm{k} _{2} \mathrm{t}\right)$ $\therefore \quad \mathrm{v}=\sqrt{\mathrm{v} _{x}^{2}+\mathrm{v} _{\mathrm{y}}^{2}}=\left[\mathrm{k} _{2}^{2}+\mathrm{k} _{1}^{2}\left(1-2 \mathrm{k} _{2} \mathrm{t}\right)^{2}\right]^{1 / 2}=\mathrm{k} _{1}\left[1+\left(1-2 \mathrm{k}^{2} \mathrm{t}\right)^{2}\right]^{1 / 2}$

The instantaneous $x$ and $y$ component of acceleration are

$$ \begin{aligned} & \mathrm{a} _{x}=\frac{\mathrm{dv} _{x}}{\mathrm{dt}}=0 ; \quad \mathrm{a} _{\mathrm{y}}=\frac{\mathrm{dv} _{\mathrm{y}}}{\mathrm{dt}}=-2 \mathrm{k} _{1} \mathrm{k} _{2} \\ & \therefore \quad \mathrm{a}=\sqrt{\mathrm{a} _{x}^{2}+\mathrm{a} _{\mathrm{y}}^{2}}=2 \mathrm{k} _{1} \mathrm{k} _{2} \end{aligned} $$