Answer (c) In a uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant.

The situation is shown in adjacent diagram where a body

$A$ is in uniform translatory motion.

• Option (a) is incorrect: If the ball were at rest, it would not be in motion at all, let alone in uniform translatory motion. Uniform translatory motion implies movement with a constant velocity, not a state of rest.

• Option (b) is incorrect: While the path can indeed be straight or circular, uniform translatory motion specifically requires that all parts of the ball have the same velocity in both magnitude and direction. This option does not address the requirement for all parts of the ball to have the same velocity.

• Option (d) is incorrect: This option describes a scenario where the center of the ball moves with constant velocity while the ball spins about its center. This is not uniform translatory motion but rather a combination of translatory and rotational motion. In uniform translatory motion, there is no spinning; all parts of the ball move with the same velocity.

Answer (b) To solve this question we have to apply Newton’s second law of motion, in terms of force and change in momentum.

We known that

$ F=\frac{d p}{d t} $

given that meter scale is moving with uniform velocity, hence, $d p=0$

$ \text { Force }=F=0 \text {. } $

As all part of the scale is moving with uniform velocity and total force is zero, hence, torque will also be zero.

• Option (a): The force acting on the scale is zero, but a torque about the centre of mass can act on the scale.

  • Incorrect because if a torque were acting on the scale, it would cause rotational acceleration, contradicting the condition of uniform velocity.

• Option (c): The total force acting on it need not be zero but the torque on it is zero.

  • Incorrect because if the total force were not zero, it would result in a change in momentum, contradicting the condition of uniform velocity.

• Option (d): Neither the force nor the torque need to be zero.

  • Incorrect because if neither the force nor the torque were zero, the scale would experience both linear and rotational acceleration, contradicting the condition of uniform velocity.

Answer (c) Given,

$ \mathbf{u}=(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) m / s $

and

$ \mathbf{v}=-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) m / s $

Mass of the ball $=150 g=0.15 kg$

$\Delta p=$ Change in momentum

$ \begin{aligned} & =\text { Final momentum }- \text { Initial momentum } \\ & =m \mathbf{v}-m \mathbf{u} \\ & =m(\mathbf{v}-\mathbf{u})=(0.15)[-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})-(3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})] \\ & =(0.15)[-6 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}] \\ & =-[0.15 \times 6 \hat{\mathbf{i}}+0.15 \times 8 \hat{\mathbf{j}}] \\ & =-[0.9 \hat{\mathbf{i}}+1.20 \hat{\mathbf{j}}] \\ \Delta \mathbf{p} & =-[0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}}] \end{aligned} $

Hence,

• Option (a) zero: This option is incorrect because the change in momentum is not zero. The initial and final velocities are not the same, which means there is a change in momentum. The initial momentum is in the positive direction, while the final momentum is in the negative direction, resulting in a non-zero change in momentum.

• Option (b) $-(0.45 \hat{\mathbf{i}}+0.6 \hat{\mathbf{j}})$: This option is incorrect because the calculated change in momentum is $-(0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}})$. The values given in this option are half of the correct values, indicating an error in the calculation.

• Option (d) $-5(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \hat{\mathbf{i}}$: This option is incorrect because it does not correctly represent the change in momentum. The expression $-5(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \hat{\mathbf{i}}$ is not dimensionally consistent and does not match the correct calculated change in momentum, which is $-(0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}})$.

Answer (c) By previous solution $\Delta p=-(0.9 \hat{\mathbf{i}}+1.2 \hat{\mathbf{j}})$

$ \begin{aligned} \text { Magnitude } & =|\Delta \mathbf{p}|=\sqrt{(0.9)^{2}+(1.2)^{2}} \\ & =\sqrt{0.81+1.44}=1.5 kg-m s^{-1} \end{aligned} $

• Option (a) zero: This is incorrect because the momentum transferred during the hit is not zero. The calculation shows a non-zero value for the change in momentum.

• Option (b) $0.75 , \text{kg-m s}^{-1}$: This is incorrect because the magnitude of the momentum transferred, as calculated, is $1.5 , \text{kg-m s}^{-1}$, not $0.75 , \text{kg-m s}^{-1}$.

• Option (d) $14 , \text{kg-m s}^{-1}$: This is incorrect because the magnitude of the momentum transferred is much smaller, specifically $1.5 , \text{kg-m s}^{-1}$, not $14 , \text{kg-m s}^{-1}$.

Thinking Process

For conservation of momentum we have to see whether net external force is acting on a system or not.

Answer (d) We know that for a system $\quad F_{\text {ext }}=\frac{d p}{d t} \quad$ (from Newton’s second law) If $F_{\text {ext }}=0, d p=0 \Rightarrow p=$ constant

Hence, momentum of a system will remain conserve if external force on the system is zero.

In case of collision’ between particles equal and opposite forces will act on individuel particles by Newtons third law,

Hence total force on the system will be zero.

Note We should not confuse with system and individual particles. As total force on the system of both particles is zero but force acts on individual particles.

• (a) Conservation of energy: Conservation of energy does not directly imply conservation of momentum. Energy can be conserved in forms other than kinetic energy, such as potential energy, and does not necessarily dictate the momentum of the system.

• (b) Newton’s first law only: Newton’s first law states that an object will remain at rest or in uniform motion unless acted upon by an external force. It does not provide a direct relationship between forces and changes in momentum during collisions.

• (c) Newton’s second law only: While Newton’s second law relates force to the rate of change of momentum, it alone does not account for the interaction forces between colliding particles. Newton’s third law, which states that every action has an equal and opposite reaction, is also necessary to fully understand the conservation of momentum in collisions.

Answer (c) Consider the adjacent diagram

Let $\quad O A=p_1$

$=$ Initial momentum of player northward

Clearly

$A B=p_2=$ Final momentum of player towards west.

Change in momentum $=p_2-p_1$

$=AB-OA=AB+(-OA)$

$=$ Clearly resultant $\mathbf{A R}$ will be along south-west.

• (a) Frictional force along westward: This option is incorrect because frictional force acts in the direction opposite to the motion. Since the player is turning westward, the frictional force would act eastward, not westward.

• (b) Muscle force along southward: This option is incorrect because the muscle force required to change direction from northward to westward would not be purely southward. It would need to have a component in the westward direction as well to achieve the turn.

• (d) Muscle force along south-West: This option is incorrect because the muscle force is not the primary force responsible for the change in momentum. The frictional force between the player’s skates and the ice is what allows the player to change direction, not the muscle force alone.

Thinking Process

We have to apply differentiations to calculate acceleration and then Newton’s second law will be applied.

Answer (a) Given, mass $=2 kg$

$ \begin{aligned} x(t) & =p t+q t^{2}+r t^{3} \\ v & =\frac{d x}{d t}=p+2 q t+3 r t^{2} \\ a & =\frac{d v}{d t}=0+2 q+6 r t \\ \text { at } t & =2 s ; a=2 q+6 \times 2 \times r \\ & =2 q+12 r \\ & =2 \times 4+12 \times 5 \\ & =8+60=68 m / s \\ \text { Force } & =F=m a \\ & =2 \times 68=136 N \end{aligned} $

• Option (b) $134 N$ is incorrect because the calculated force using the given values and the correct formula results in $136 N$, not $134 N$.
• Option (c) $158 N$ is incorrect because the calculated force using the given values and the correct formula results in $136 N$, not $158 N$.
• Option (d) $68 N$ is incorrect because this value represents the acceleration ($a$) at $t=2 s$, not the force. The force is calculated by multiplying the mass ($2 kg$) by the acceleration ($68 m/s^2$), resulting in $136 N$.

Answer (b) Given, mass $=m=5 kg$

Acting force $=\mathbf{F}=(-3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) N$

Initial velocity at $t=0, \mathbf{u}=(6 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}) m / s$

Retardation, $\hat{\mathbf{a}}=\frac{\mathbf{F}}{m}=(-\frac{3 \hat{\mathbf{i}}}{\mathbf{5}}+\frac{4 \hat{\mathbf{j}}}{5}) m / s^{2}$

As final velocity is along $Y$-axis only, its $x$-component must be zero.

From $v=u+a t$, for $X$-component only, $0=6 \hat{\mathbf{i}}-\frac{3 \hat{\mathbf{i}}}{5} t$

$ t=\frac{5 \times 6}{3}=10 s $

• Option (a) never: This option is incorrect because the body will eventually have a velocity along the Y-axis when the X-component of its velocity becomes zero. The calculation shows that this happens at ( t = 10 ) seconds, so it is not “never.”

• Option (c) 2 s: This option is incorrect because, based on the calculation, the time required for the X-component of the velocity to become zero is ( t = 10 ) seconds, not 2 seconds. At ( t = 2 ) seconds, the X-component of the velocity would still be non-zero.

• Option (d) 15 s: This option is incorrect because the calculation shows that the X-component of the velocity becomes zero at ( t = 10 ) seconds. At ( t = 15 ) seconds, the X-component would have already been zero for 5 seconds, so this is not the correct time.

Answer (b) Given, mass of the car $=m$

As car starts from rest, $u=0$

Velocity acquired along east $=v \hat{\mathbf{i}}$

Duration $=t=2 s$.

We know that

$ \begin{aligned} v & =u+a t \\ \Rightarrow v \hat{\mathbf{i}} & =0+a \times 2 \\ \Rightarrow \mathbf{a} & =\frac{\mathbf{v}}{2} \hat{\mathbf{i}} \end{aligned} $

Force,

$ \mathbf{F}=m \mathbf{a}=\frac{\mathbf{m} \mathbf{v}}{2} \hat{\mathbf{i}} $

Hence, force acting on the car is $\frac{m v}{2}$ towards east. As external force on the system is only friction hence, the force $\frac{m v}{2}$ is by friction. Hence, force by engine is internal force.

• Option (a): This option is incorrect because it attributes the force to the car engine. However, the force calculated ($\frac{m v}{2}$) is due to the friction exerted by the road on the tyres, not directly by the car engine.

• Option (c): This option is incorrect because it suggests that the force exerted by the engine is more than $\frac{m v}{2}$ to overcome friction. The problem states that the car moves with uniform acceleration, implying that the net force is $\frac{m v}{2}$, which is the force due to friction. There is no indication that the engine exerts a force greater than this value.

• Option (d): This option is incorrect because it states that the force $\frac{m v}{2}$ is exerted by the engine. However, the force $\frac{m v}{2}$ is actually due to the friction between the tyres and the road, not directly by the engine.

Thinking Process

Here, position of the particle is given for different time intervals. Hence, we have to find velocity and acceleration corresponding to the intervals.

Answer $(a, b, d)$

Given,

$ \begin{aligned} x & =0 \text { for } t<0 s . \\ x(t) & =A \sin 4 \pi t ; \text { for } 0<t<\frac{1}{4} s \\ x & =0 ; \text { for } t>\frac{1}{4} s \\ v(t) & =\frac{d x}{d t}=4 \pi \text { Acos } 4 \pi t \\ a(t) & =\text { acceleration } \\ & =\frac{d v(t)}{d t}=-16 \pi^{2} A \sin 4 \pi t \\ \text { At } t & =\frac{1}{8} s, a(t)=-16 \pi^{2} A \sin 4 \pi \times \frac{1}{8}=-16 \pi^{2} A \\ F & =m a(t)=-16 \pi^{2} A \times m=-16 \pi^{2} mA \\ \text { Impulse } & =\text { Change in linear momentum } \\ & =F \times t=(-16 \pi^{2} Am) \times \frac{1}{4} \\ & =-4 \pi^{2} Am \end{aligned} $

$ \text { For, } 0<t<\frac{1}{4} s $

The impulse (Change in linear momentum)

$ \text { at } t=0 \text { is same as, } t=\frac{1}{4} s \text {. } $

Clearly, force depends upon $A$ which is not constant. Hence, force is also not constant.

Note We have to keep in mind that the force is varying for different time intervals. Hence, we should apply differential formulae for each interval separately.

• Option (c): The particle is not acted upon by any force.

  • This is incorrect because the particle experiences a force during the interval (0 < t < \frac{1}{4} \text{ s}). The force is given by (F = ma(t) = -16 \pi^2 m A \sin 4 \pi t), which is not zero during this interval.

• Option (e): There is no impulse acting on the particle.

  • This is incorrect because the particle experiences impulses at (t = 0 \text{ s}) and (t = \frac{1}{4} \text{ s}). The magnitude of the impulse is (4 \pi^2 A m) at these times, indicating that impulses do act on the particle.

Thinking Process

In this problem we have to find frictional forces on each surface and accordingly we will decide maximum force.

Answer $(a, b, d, e)$

Consider the adjacent diagram. Frictional force on $B(f_1)$ and frictional force on $A(f_2)$ will be as shown.

Let $A$ and $B$ are moving together $a_{\text {common }}=\frac{F-f_1}{m_{A}+m_{B}}=\frac{F-f_1}{(m / 2)+m}=\frac{2(F-f_1)}{3 m}$

Pseudo force on $A=(m_{A}) \times a_{\text {common }}$

$=m_{A} \times \frac{2(F-f_1)}{3 m}=\frac{m}{2} \times \frac{2(F-f_1)}{3 m}=\frac{(F-f_1)}{3}$

The force $(F)$ will be maximum when

Pseudo force on $A=$ Frictional force on $A$

$ \begin{aligned} & \Rightarrow \quad \frac{F_{\max }-f_1}{3}=\mu m_{A} g \\ & =0.2 \times \frac{m}{2} \times g=0.1 mg \\ & \Rightarrow \quad F_{\max }=0.3 mg+f_1 \\ & =0.3 mg+(0.1) \frac{3}{2} mg=0.45 mg \end{aligned} $

$\Rightarrow$ Hence, maximum force upto which bodies will move together is $F_{\max }=0.45 mg$

(a) Hence, for $F=0.25 mg<F_{\max }$ bodies will move together.

(b) For $F=0.5 mg>F_{\max }$, body $A$ will slip with respect to $B$.

(c) For $F=0.5 mg>F_{\text {max }}$, bodies slip.

$ (f_1) _ {\max } $ = $\mu m_{B} g=(0.1) \times \frac{3}{2} m \times g=0.15 mg $

$ (f_2) _ {\max} $= $\mu m_A g=(0.2)(\frac{m}{2})(g)=0.1 mg $

Hence, minimum force required for movement of the system $(A+B)$

$F_{\min} = (f_1) _ {\max } + (f_2)_{\max } $

$=0.15 mg+0.1 mg=0.25 mg$

(d) Given, force $F=0.1 mg<F_{\text {min }}$.

Hence, the bodies will be at rest.

(e) Maximum force for combined movement $F_{\max }=0.45 mg$.

• Option (c) is incorrect: For ( F = 0.5 , mg ), the force exceeds the maximum force ( F_{\text{max}} = 0.45 , mg ) for which the two bodies will move together. Therefore, the bodies will not move together; instead, body ( A ) will slip with respect to body ( B ).

Thinking Process

The friction force always have tendency to oppose the motion. Consider the adjacent diagram.

Answer $(b, d)$

Let $m_1$, moves up the plane. Different forces involved are shown in the diagram.

$ \begin{aligned} N & =\text { Normal reaction } \\ f & =\text { Frictional force } \\ T & =\text { Tension in the string } \\ f & =\mu N=\mu m_1 g \cos \theta \end{aligned} $

For the system $(m_1+m_2)$ to move up

$ \begin{matrix} \Rightarrow & m_2 g-(m_1 g \sin \theta+\mu m_1 g \cos \theta)>0 \\ \Rightarrow & m_2>m_1(\sin \theta+\mu \cos \theta) \end{matrix} $

Hence, option (b) is correct.

Let the body moves down the plane, in this case $f$ acts up the plane.

Hence,

$ \begin{aligned} & m_1 g \sin \theta-f>m_2 g \\ & m_1 g \sin \theta-\mu m_1 g \cos \theta>m_2 g \\ & m_1(\sin \theta-\mu \cos \theta)>m_2 \\ & m_2<m_1(\sin \theta-\mu \cos \theta) \end{aligned} $

Hence, option (d) is correct.

• Option (a): This option states that if ( m_2 > m_1 \sin \theta ), the body will move up the plane. However, this condition does not account for the frictional force acting on ( m_1 ). The correct condition for the body to move up the plane must include the frictional force, which is why the correct condition is ( m_2 > m_1 (\sin \theta + \mu \cos \theta) ). Therefore, option (a) is incorrect because it neglects the frictional force.

• Option (c): This option states that if ( m_2 < m_1 (\sin \theta + \mu \cos \theta) ), the body will move up the plane. However, this condition actually implies that the force due to ( m_2 ) is not sufficient to overcome the combined effect of the gravitational component along the plane and the frictional force acting on ( m_1 ). Therefore, the body will not move up the plane under this condition. Hence, option (c) is incorrect because it misinterprets the condition for the body to move up the plane.

Answer $(b, c)$

Let $A$ moves up the plane frictional force on $A$ will be downward as shown.

When $A$ just starts moving up

$m g \sin \theta_1+f =m g \sin \theta_2 $

$\Rightarrow m g \sin \theta_1+\mu m g \cos \theta_1 =m g \sin \theta_2 $

$\Rightarrow \mu =\frac{\sin \theta_2-\sin \theta_1}{\cos \theta_1}$

When A moves upwards

$ \Rightarrow \quad \sin \theta_2>\sin \theta_1 \Rightarrow \theta_2>\theta_1 $

• (a) A will never move up the plane: This statement is incorrect because it is possible for body A to move up the plane if the conditions specified in option (b) are met, i.e., when the coefficient of friction $\mu$ is equal to $\frac{\sin \theta_2 - \sin \theta_1}{\cos \theta_1}$.

• (d) B will always slide down with constant speed: This statement is incorrect because the speed of body B depends on the net force acting on it. If the forces are balanced, B will move with constant speed, but if there is a net force (e.g., due to the tension in the string or the gravitational component along the plane), B will accelerate or decelerate accordingly.

Answer (c, $d)$

Given,

Initial velocity $(u)=u_1=u_2=5 m / s$

Final velocity $(v)=v_1=v_2=-5 m / s$

Time duration of collision $=10^{-3} s$.

Change in linear momentum $=m(v-u)$

$ \begin{aligned} & =\frac{1}{20}[-5-5]=-0.5 N-s . \\ \text { Force } & =\frac{\text { Impulse }}{\text { Time }}=\frac{\text { Change in momentum }}{10^{-3} s} \\ & =\frac{0.5}{10^{-3}}=500 N \end{aligned} $

Impulse and force are opposite in directions.

• Option (a): The impulse imparted to each ball is indeed $0.5 , \text{N-s}$, not $0.25 , \text{kg-m/s}$. Additionally, the force on each ball is $500 , \text{N}$, not $250 , \text{N}$.

• Option (b): The impulse imparted to each ball is $0.5 , \text{N-s}$, not $0.25 , \text{kg-m/s}$. Furthermore, the force exerted on each ball is $500 , \text{N}$, not $25 \times 10^{-5} , \text{N}$.

• Option (c): This option is correct and thus not incorrect.

• Option (d): This option is correct and thus not incorrect.

Thinking Process

In this problem, we have to use the concept of resultant of two vectors, when they are perpendicular.

Answer (a, c)

Consider the adjacent diagram

Given,

$ \begin{aligned} \text { mass } & =m=10 kg . \\ F_1 & =6 N, F_2=8 N \\ \text { Resultant force } & =F=\sqrt{F_1^{2}+F_2^{2}}=\sqrt{36+64} \\ & =10 N \\ a & =\frac{F}{m}=\frac{10}{10}=1 m / s^{2} ; \text { along } R . \end{aligned} $

Let $\theta_1$ be angle between $R$ and $F_1$

$ \begin{aligned} \tan \theta_1 & =\frac{8}{6}=\frac{4}{3} \\ \theta_1 & =\tan ^{-1}(4 / 3) \text { w.r.t. } F_1=6 N \end{aligned} $

Let $\theta_2$ be angle between $F$ and $F_2$

$ \begin{aligned} \tan \theta_2 & =\frac{6}{8}=\frac{3}{4} \\ \theta_2 & =\tan ^{-1}(\frac{3}{4}) \text { w.r.t. } F_2=8 N \end{aligned} $

• Option (b) is incorrect because the calculated acceleration is (1 , \text{m/s}^2), not (0.2 , \text{m/s}^2). The resultant force is (10 , \text{N}) and the mass is (10 , \text{kg}), leading to an acceleration of (1 , \text{m/s}^2).

• Option (d) is incorrect because the calculated acceleration is (1 , \text{m/s}^2), not (0.2 , \text{m/s}^2). The resultant force is (10 , \text{N}) and the mass is (10 , \text{kg}), leading to an acceleration of (1 , \text{m/s}^2).

Thinking Process

In this problem, we have to apply conservation of linear momentum.

Answer Given, total mass of girl, bicycle and stone $=m_1=(50+0.5) kg=50.5 kg$.

Velocity of bicycle $u_1=5 m / s$, Mass of stone $m_2=0.5 kg$

Velocity of stone $u_2=15 m / s$, Mass of girl and bicycle $m=50 kg$

Yes, the speed of the bicycle changes after the stone is thrown.

Let after throwing the stone the speed of bicycle be $v m / s$.

According to law of conservation of linear momentum,

$ \begin{aligned} m_1 u_1 & =m_2 u_2+m v \\ 50.5 \times 5 & =0.5 \times 15+50 \times v \\ 252.5-7.5 & =50 v \\ \text { or } \quad v & =\frac{245.0}{50} \\ v & =4.9 m / s \\ \text { Change in speed } & =5-4.9=0.1 m / s . \end{aligned} $

Answer When a lift descends with a downward acceleration a the apparent weight of a body of mass $m$ is given by

$ w^{\prime}=R=m(g-a) $

Mass of the person $m=50 kg$

Descending acceleration $a=9 m / s^{2}$

Acceleration due to gravity $g=10 m / s^{2}$

Apparent weight of the person,

$ \begin{aligned} R & =m(g-a) \\ & =50(10-9) \\ & =50 N \\ \therefore \quad \text { Reading of the weighing scale } & =\frac{R}{g}=\frac{50}{10}=5 kg . \end{aligned} $

Answer Given, mass of the body $(m)=2 kg$

From the position-time graph, the body is at $x=0$ when $t=0$, i.e., body is at rest.

$\therefore$ Impulse at $t=0, s=0$, is zero

From $t=0 s$ to $t=4 s$, the position-time graph is a straight line, which shows that body moves with uniform velocity.

Beyond $t=4 s$, the graph is a straight line parallel to time axis, i.e., body is at rest $(v=0)$.

Velocity of the body $=$ slope of position-time graph

$ =\tan \theta=\frac{3}{4} m / s $

Impulse (at $t=4 s)=$ change in momentum

$ \begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =2(0-\frac{3}{4}) \\ & =-\frac{3}{2} kg-m / s=-1.5 kg-m / s \end{aligned} $

Answer When a person driving a car suddenly applies the brakes, the lower part of the body slower down with the car while upper part of the body continues to move forward due to inertia of motion.

If driver is not wearing seat belt, then he falls forward and his head hit against the steering wheel.

Answer Given, mass of the body $m=2 kg$.

Velocity of the body $\mathbf{v}(t)=2 t \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}$

$\therefore$ Velocity of the body at $t=2 s$

$ \mathbf{v}=2 \times 2 \hat{\mathbf{i}}+(2)^{2} \hat{\mathbf{j}}=(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) $

Momentum of the body $(p)=m \mathbf{v}$

$ =2(4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})=(8 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) kg-m / s $

Acceleration of the body $(a)=\frac{d v}{d t}$

$ \begin{aligned} & =\frac{d}{d t}(2 t \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}}) \\ & =(2 \hat{\mathbf{i}}+2 t \hat{\mathbf{j}}) \\ \text { At } t & =2 s \\ \mathbf{a} & =(2 \hat{\mathbf{i}}+2 \times 2 \hat{\mathbf{j}}) \\ & =(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) \end{aligned} $

Force acting on the body $(\mathbf{F})=$ ma

$ =2(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}) $

$ =(4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}) N $

Answer The approximate graph is shown in the diagram

The frictional force $f$ is shown on vertical axis and the applied force $F$ is shown on the horizontal axis. The portion $O A$ of graph represents static friction which is self adjusting. In this portion, $f=F$.

The point $B$ corresponds to force of limiting friction which is the maximum value of static friction. $C D | O X$ represents kinetic friction, when the body actually starts moving. The force of kinetic friction does not increase with applied force, and is slightly less than limiting friction.

Answer When a child falls on a cement floor, her body comes to rest instantly. But $F \times \Delta t=$ change in momentum $=$ constant. As time of stopping $\Delta t$ decreases, therefore $F$ increases and hence, child feel more pain.

When she falls on a soft muddy ground in the garden the time of stopping increases and hence, $F$ decreases and she feels lesser pain.

Answer Mass of the object $(m)=500 g=0.5 kg$

Speed of the object $(v)=25 m / s$

(a) Impulse imparted to the object = change in momentum

$ \begin{aligned} & =m v-m u \\ & =m(v-u) \\ & =0.5(25-0)=12.5 N-s \end{aligned} $

(b)

$ \begin{aligned} & \text { ounding } \\ & =-\frac{25}{2} m / s \\ & v^{\prime}=-12.5 m / s \\ & \therefore \quad \text { Change in momentum }=m(v^{\prime}-v) \\ & =0.5(-12.5-25)=-18.75 N-s \end{aligned} $

Answer While going up a mountain, the force of friction acting on a vehicle of mass $m$ is $f=\mu R=\mu m g \cos \theta$, where $\theta$ is the angle of slope of the road with the horizontal. To avoid skidding force of friction (f) should be large and therefore, $\cos \theta$ should be large and hence, $\theta$ should be small.

That’s why mountain roads are generally made winding upwards rather than going straight upto avoid skidding.

Thinking Process

As the whole system is going upward with an acceleration we have to apply Newton’s laws.

Answer Given, $m_1=5 kg, m_2=3 kg$

$g=9.8 m / s^{2}$ and $a=2 m / s^{2}$

For the upper block

$ \begin{aligned} & \Rightarrow \\ & \text { For the lower block } \\ & \Rightarrow \\ & \text { From Eq. (i) } \end{aligned} $

$ \begin{aligned} T_1-T_2-5 g & =5 a \\ T_1-T_2 & =5(g+a) \\ T_2-3 g & =3 a \\ T_2 & =3(g+a)=3(9.8+2)=35.4 N \\ T_1 & =T_2+5(g+a) \\ & =35.4+5(9.8+2)=94.4 N \end{aligned} $

Answer In equilibrium, the force $m g \sin \theta$ acting on block $A$ parallel to the plane should be balanced by the tension in the string, i.e.,

$ \begin{aligned} & m g \sin \theta=T=F \quad[\because T=\text { Fgiven }] \\ & W=T=F \end{aligned} $

where, $w$ is the weight of block $B$.

From Eqs. (i) and (ii), we get,

$\therefore$

$ \begin{aligned} W & =m g \sin \theta \\ & =100 \times \sin 30^{\circ} \\ & =100 \times \frac{1}{2} N=50 N \end{aligned} $

$ (\because m g=100 N) $

Note While finding normal reaction in such cases, we should be careful it will be $N=m g \cos \theta$, where $\theta$ is angle of inclination.

Answer Given, mass of the block $=M$

Coefficient of friction between the block and the wall $=\mu$

Let a force $F$ be applied on the block to hold the block against the wall. The normal reaction of mass be $N$ and force of friction acting upward be $f$. In equilibrium, vertical and horizontal forces should be balanced separately.

$\therefore$

$f=M g$

and

$F=N$

But force of friction $(f)=\mu N$

$$ \mu F \quad \quad \quad [\text{using Eq.} (ii)]…(iii)$$

From Eqs. (i) and (iii), we get $$ \mu F = Mg$$ $$F= \frac{Mg}{\mu}$$

Answer Given, mass of the gun $(m_1)=100 kg$

Mass of the ball $(m_2)=1 kg$

Height of the cliff $(h)=500 m$

Horizontal distance travelled by the ball $(x)=400 m$

From

$ h=\frac{1}{2} g t^{2} \quad( \because \text {Initial velocity in downward direction is zero) }$

From

$ \begin{aligned} 500 & =\frac{1}{2} \times 10 t^{2} \\ t & =\sqrt{100}=10 s \\ x=u t, u & =\frac{x}{t}=\frac{400}{10}=40 m / s \end{aligned} $

If $v$ is recoil velocity of gun, then according to principle of conservation of linear momentum,

$ \begin{aligned} m_1 v & =m_2 u \\ v & =\frac{m_2 u}{m_1}=\frac{1}{100} \times 40=0.4 m / s \end{aligned} $

Thinking Process

To solve this question, we have to find the relation for $x$ and time $(t), y$ and time (t) from the given diagram.

Answer Clearly from diagram (a), the variation can be related as

$ \begin{aligned} & x=t \Rightarrow \frac{d x}{d t}=1 m / s \\ & a_{x}=0 \end{aligned} $

From diagram (b)

$\Rightarrow$

Hence,

Hence, net force,

$ \begin{matrix} y & =t^{2} \\ \frac{d y}{d t} & =2 t \quad \text { or } a_{y}=\frac{d^{2} y}{d t^{2}}=2 m / s^{2} \\ I_{y} & =m a_{y}=500 \times 10^{-3} \times 2=1 N & \\ F_{x} & =m a_{x}=0 \\ F & =\sqrt{F_x^{2}+F_y^{2}}=F_{y}=1 N & \quad(\because m=500 g) \\ & \text { (along } y \text {-axis) } \end{matrix} $

Answer Here, initial speed of the coin $(u)=20 m / s$

Acceleration of the elevator $(a)=2 m / s^{2}$

(upwards)

Acceleration due to gravity $(g)=10 m / s^{2}$

$\therefore$ Effective acceleration $a^{\prime}=g+a=10+2=12 m / s^{2} \quad$ (here, acceleration is w.r.t. the lift)

If the time of ascent of the coin is $t$, then

or

$ \begin{aligned} & v=u+a t \\ & 0=20+(-12) \times t \\ & t=\frac{20}{12}=\frac{5}{3} s \end{aligned} $

Time of ascent $=$ Time of desent

$\therefore$ Total time after which the coin fall back into hand $=(\frac{5}{3}+\frac{5}{3}) s=\frac{10}{3} s=3.33 s$

Note While calculating net acceleration we should be aware that if lift is going upward net acceleration is $(g+a)$ and for downward net acceleration is $(g-a)$.

Thinking Process

As the body is found to move with uniform velocity hence, we can say that total force acting will be zero.

Answer As the body is moving with uniform speed (velocity) its acceleration $a=0$.

$\therefore$ The sum of the forces is zero, $F_1+F_2+F_3=0$

(a) Let $F_1, F_2, F_3$ be the three forces passing through a point. Let $F_1$ and $F_2$ be in the plane $A$ (one can always draw a plane having two intersecting lines such that the two lines lie on the plane). Then $F_1+F_2$ must be in the plane $A$.

Since, $F_3=-(F_1+F_2), F_3$ is also in the plane $A$.

(b) Consider the torque of the forces about $P$. Since, all the forces pass through $P$, the torque is zero. Now, consider torque about another point $O$. Then torque about $O$ is

Torque $=OP \times(F_1+F_2+F_3)$

Since, $F_1+F_2+F_3=0$, torque $=0$

Answer Consider the diagram where a body slides down from along an inclined plane of inclination $\theta(=45^{\circ})$.

On smooth inclined plane Acceleration of a body sliding down a smooth inclined plane Here,

$ \begin{aligned} & a=g \sin \theta \\ & \theta=45^{\circ} \\ & a=g \sin 45^{\circ}=\frac{g}{\sqrt{2}} \end{aligned} $

$ \therefore $

Let the travelled distance be $s$

Using equation of motion, $s=u t+\frac{1}{2} a t^{2}$, we get

$ \begin{aligned} s & =0 . t+\frac{1}{2} \frac{g}{\sqrt{2}} T^{2} \\ \text { or } \quad s & =\frac{g T^{2}}{2 \sqrt{2}} \end{aligned} $

On rough inclined plane Acceleration of the body $a=g(\sin \theta-\mu \cos \theta)$

$ \begin{aligned} & =g(\sin 45^{\circ}-\mu \cos 45^{\circ}) \\ & =\frac{g(1-\mu)}{\sqrt{2}} \end{aligned} \quad(\text { As, } \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}) $

Again using equation of motion, $s=u t+\frac{1}{2} a t^{2}$, we get

or

$ \begin{aligned} & s=0(p T)+\frac{1}{2} \frac{g(1-\mu)}{\sqrt{2}}(p T)^{2} \\ & s=\frac{g(1-\mu) p^{2} T^{2}}{2 \sqrt{2}} \end{aligned} $

From Eqs. (i) and (ii), we get

or

$ \begin{aligned} \frac{g T^{2}}{2 \sqrt{2}} & =\frac{g(1-\mu) p^{2} T^{2}}{2 \sqrt{2}} \\ (1-\mu) p^{2} & =1 \\ 1-\mu & =\frac{1}{p^{2}} \\ \mu & =(1-\frac{1}{p^{2}}) \end{aligned} $

Answer Consider figure (a)

$ \begin{aligned} v_{x} & =2 t \text { for } 0<t<1 s \\ & =2(2-t) \text { for } 1<t<2 s \\ & =0 \text { for } t>2 s . \\ v_{y} & =t \text { for } 0<t<1 s \\ & =1 \text { for } t>1 s \\ F_{x} & =m a_{x}=m \frac{d v_{x}}{d t} \\ & =1 \times 2 \\ & =1(-2) \end{aligned} $

From figure (b)

for $0<t<1$ s

for $1<t<2 s$ for $2<t$

(a)

(b)

Hence,

$ \begin{aligned} F_{y} & =m a_{y}=m \frac{d v_{y}}{d t} \\ & =1 \times 1 \text { for } 0<t<1 s=0 \text { for } 1<t \\ F & =F_{x} \hat{\mathbf{i}}+F_{y} \hat{\mathbf{j}} \\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & =-2 \hat{\mathbf{i}} \\ & =0 \end{aligned} $

$ \text { for } 0<t<1 s $

for $1<t<2 s$

for $t>2 s$.

Thinking Process

The necessary centripetal force required for the circular motion will be provided by the frictional force.

Answer Balancing frictional force for centripetal force $\frac{m v^{2}}{r}=f=\mu N=\mu m g$ where, $N$ is normal reaction.

$\therefore \quad V=\sqrt{\mu r g} \quad$ (where, $r$ is radius of the circular track)

For path $A B C$

$ \text { Path length }=\frac{3}{4}(2 \pi 2 R)=3 \pi R=3 \pi \times 100 $

$ \begin{aligned} & =300 \pi m \\ v_1=\sqrt{\mu 2 R g} & =\sqrt{0.1 \times 2 \times 100 \times 10} \\ & =14.14 m / s \end{aligned} $

$\therefore \quad t_1=\frac{300 \pi}{14.14}=66.6 s$

For path $D E F$

$ \text { Path length }=\frac{1}{4}(2 \pi R)=\frac{\pi \times 100}{2}=50 \pi $

$ \begin{aligned} & v_2=\sqrt{\mu R g}=\sqrt{0.1 \times 100 \times 10}=10 m / s \\ & t_2=\frac{50 \pi}{10}=5 \pi s=15.7 s \end{aligned} $

For paths, $C D$ and $F A$

$ \text { Path length }=R+R=2 R=200 m $

$ t_3=\frac{200}{50}=4.0 s . $

$\therefore$ Total time for completing one round

$ t=t_1+t_2+t_3=66.6+15.7+4.0=86.3 s $

Thinking Process

To find trajectory, we will relate $x$ and $y$ in terms of constants $A$ and $B$.

Answer (a) Displacement vector of the particle of mass $m$ is given by

$ \mathbf{r}(t)=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t $

$\therefore$ Displacement along $x$-axis is,

$ \begin{aligned} x & =A \cos \omega t \\ \frac{x}{A} & =\cos \omega t \end{aligned} $

$ \text { or } $

Displacement along $y$-axis is,

and

$ \begin{aligned} y & =B \sin \omega t \\ \frac{y}{B} & =\sin \omega t \end{aligned} $

Squaring and then adding Eqs. (i) and (ii), we get

$ \frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=\cos ^{2} \omega t+\sin ^{2} \omega t=1 $

This is an equation of ellipse.

Therefore, trajectory of the particle is an ellipse. (b) Velocity of the particle

$ \begin{aligned} & \mathbf{v}=\frac{d \mathbf{r}}{d t}=\hat{\mathbf{i}} \frac{d}{d t}(A \cos \omega t)+\hat{\mathbf{j}} \frac{d}{d t}(B \sin \omega t) \\ & =\hat{\mathbf{i}}[A(-\sin \omega t) \cdot \omega]+\hat{\mathbf{j}}[B(\cos \omega t) \cdot \omega] \\ & =-\hat{\boldsymbol{i}} A \omega \sin \omega t+\hat{\boldsymbol{j}} B \omega \cos \omega t \\ & \mathbf{a}=-\hat{\mathbf{i}} A \omega \frac{d}{d}(\sin \omega t)+\hat{\mathbf{j}} B \omega \frac{d}{d t}(\cos \omega t) \\ & =-\hat{\mathbf{i}} A \omega[\cos \omega t] \cdot \omega+\hat{\mathbf{j}} B \omega[-\sin \omega t] \cdot \omega \\ & =-\hat{\mathbf{i}} A \omega^{2} \cos \omega t-\hat{\mathbf{j}} B \omega^{2} \sin \omega t \\ & =-\omega^{2}[\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} B \sin \omega t] \\ & =-\omega^{2} \mathbf{r} \end{aligned} $

or

$\therefore$ Force acting on the particle,

$ F=m \mathbf{a}=-\mathbf{m} \omega^{2} \mathbf{r} $

Hence proved.

Thinking Process

The horizontal component of velocity will remain unaffected by gravity.

Answer (a) When ball is given only horizontal velocity Horizontal velocity at the time of release $(u_{x})=v_{s}$

During projectile motion, horizontal velocity remains unchanged,

Therefore,

In vertical direction,

$ \begin{aligned} & v_{x}=u_{x}=v_{s} \\ & v_y^{2}=u_y^{2}+2 g H \end{aligned} $

$ v_{y}=\sqrt{2 g H} $

$\therefore$ Resultant speed of the ball at bottom,

$ \begin{aligned} v & =\sqrt{v_x^{2}+v_y^{2}} \\ & =\sqrt{v_s^{2}+2 g H} \end{aligned} $

(b) When ball is given horizontal velocity and a small downward velocity

Let the ball be given a small downward velocity $u$.

In horizontal direction

In vertical direction

$ \begin{aligned} & v_x^{\prime}=u_{x}=v_{s} \\ & v_y^{\prime 2}=u^{2}+2 g H \\ & v_y^{\prime}=\sqrt{u^{2}+2 g H} \end{aligned} $

or

$\therefore$ Resultant speed of the ball at the bottom

$ v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^{2}+u^{2}+2 g H} $

From Eqs. (i) and (ii), we get $\quad v^{\prime}>v$

Thinking Process

To balance the forces, we have to resolve them along two mutually perpendicular directions.

Answer Consider the adjacent diagram, in which forces are resolved.

On resolving forces into rectangular components, in equilibrium forces $(F_1+\frac{1}{\sqrt{2}}) N$ are equal to $\sqrt{2} N$ and $F_2$ is equal to $(\sqrt{2}+\frac{1}{\sqrt{2}}) N$.

$\therefore \quad F_1+\frac{1}{\sqrt{2}}=\sqrt{2}$

and

$ F _1=\sqrt{2}-\frac{1}{\sqrt{2}}=\frac{2-1}{\sqrt{2}}=\frac{1}{\sqrt{2}}=0.707 N $

and $\quad F_2=\sqrt{2}+\frac{1}{\sqrt{2}}=\frac{2+1}{\sqrt{2}} N=\frac{3}{\sqrt{2}} N=2.121 N$

Answer (a) Consider the adjacent diagram, force of friction on the box will act up the plane. For the box to just starts sliding down $m g$

or $\quad \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu)$

(b) When angle of inclination is increased to $\alpha>\theta$, then net force acting on the box, down the plane is

$ \begin{aligned} F_1 & =m g \sin \alpha-f=m g \sin \alpha-\mu N \\ & =m g(\sin \alpha-\mu \cos \alpha) \end{aligned} $

(c) To keep the box either stationary or just move it up with uniform speed, upward force needed, $F_2=m g \sin \alpha+f=m g(\sin \alpha+\mu \cos \alpha)$ (In this case, friction would act down the plane).

(d) If the box is to be moved with an upward acceleration a, then upward force needed, $F_3=m g(\sin \alpha+\mu \cos \alpha)+m a$.

Answer Given, mass of helicopter $(m_1)=2000 kg$

Mass of the crew and passengeres $m_2=500 kg$

Acceleration in vertical direction $a=15 m / s^{2}(\uparrow)$ and $g=10 m / s^{2}(\downarrow)$

(a) Force on the floor of the helicopter by the crew and passengers

$ \begin{aligned} & m_2(g+a)=500(10+15) N \\ & 500 \times 25 N=12500 N \end{aligned} $

(b) Action of the rotor of the helicopter on the surrounding air $=(m_1+m_2)(g+a)$

$ \begin{aligned} & =(2000+500) \times(10+15)=2500 \times 25 \\ & =62500 N(\text { downward }) \end{aligned} $

(c) Force on the helicopter due to the surrounding air

$ \begin{aligned} & =\text { reaction of force applied by helicopter. } \\ & =62500 N \text { (upward) } \end{aligned} $

Note We should be very clear when we are balancing action and reaction forces. We must know that which part is action and which part is reaction due to the action.