Thinking Process

To solve such type of questions, we have to use the formula for dot product or cross product.

Answer (b) Given, $\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$\mathbf{B}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

We know that

$ \begin{gathered} A \cdot B=|A||\mathbf{B}| \cos \theta \\ \Rightarrow \quad(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}})=(\sqrt{1+1})(\sqrt{1+1}) \cos \theta \end{gathered} $

where $\theta$ is the angle between $\mathbf{A}$ and $\mathbf{B}$

$ \begin{aligned} \Rightarrow & & \cos \theta & =\frac{1-0+0-1}{\sqrt{2} \sqrt{2}}=\frac{0}{2}=0 \\ \Rightarrow & & \theta & =90^{\circ} \end{aligned} $

• Option (a) $45^{\circ}$: This option is incorrect because the dot product of vectors $\mathbf{A}$ and $\mathbf{B}$ results in zero, which implies that the cosine of the angle between them is zero. The cosine of $45^{\circ}$ is $\frac{1}{\sqrt{2}}$, not zero.

• Option (c) $-45^{\circ}$: This option is incorrect for the same reason as option (a). The cosine of $-45^{\circ}$ is also $\frac{1}{\sqrt{2}}$, not zero. The dot product calculation shows that the angle must be $90^{\circ}$, where the cosine is zero.

• Option (d) $180^{\circ}$: This option is incorrect because the cosine of $180^{\circ}$ is $-1$. The dot product calculation shows that the cosine of the angle between $\mathbf{A}$ and $\mathbf{B}$ is zero, not $-1$. Therefore, the angle cannot be $180^{\circ}$.

Answer (d) A scalar quantity is independent of direction hence has the same value for observers with different orientations of the axes.

• (a) A scalar quantity is not necessarily conserved in a process. Conservation is a property that applies to certain physical quantities (like energy or momentum) but not to all scalar quantities.

• (b) A scalar quantity can take negative values. For example, temperature in Celsius or Fahrenheit can be negative, and electric potential can also be negative.

• (c) A scalar quantity can vary from one point to another in space. For instance, temperature and pressure are scalar quantities that can change from one location to another.

Thinking Process

In this question according to the diagram, we have to decide the components of a given vector.

Answer (b) Clearly from the diagram, $\mathbf{u}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}$

As $u$ is in the first quadrant, hence both components $a$ and $b$ will be positive.

For $\mathbf{v}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}$, as it is in positive $x$-direction and located downward hence $x$-component $p$ will be positive and $y$-component $q$ will be negative.

• Option (a): This option states that (a) and (p) are positive while (b) and (q) are negative. This is incorrect because (b) is positive as (\mathbf{u}) is in the first quadrant, where both components (a) and (b) are positive.

• Option (c): This option states that (a), (q), and (b) are positive while (p) is negative. This is incorrect because (q) is negative as (\mathbf{v}) is in the fourth quadrant, where the (y)-component (q) is negative.

• Option (d): This option states that (a), (b), (p), and (q) are all positive. This is incorrect because (q) is negative as (\mathbf{v}) is in the fourth quadrant, where the (y)-component (q) is negative.

Answer (b) Let $r$ makes an angle $\theta$ with positive $x$-axis component of $r$ along $X$-axis

As

$r_{x} =|\mathbf{r}| \cos \theta $

$(r_{x}) _{\text{maximum }} =|\mathbf{r}|(\cos \theta) _{\text{maximum }} $

$=|r| \cos 0^{\circ}=|r| \quad(\because \cos \theta \text{ is maximum of } \theta=0^{\circ})$

$ \theta = 0^o$ $r$ is along positive $x$-axis.

• (a) If ( r ) is along the positive ( Y )-axis, the angle ( \theta ) between ( r ) and the ( X )-axis is ( 90^\circ ). Since ( \cos 90^\circ = 0 ), the component of ( r ) along the ( X )-axis will be zero, which is not the maximum value.

• (c) If ( r ) makes an angle of ( 45^\circ ) with the ( X )-axis, the component of ( r ) along the ( X )-axis is ( |\mathbf{r}| \cos 45^\circ = |\mathbf{r}| \frac{1}{\sqrt{2}} ). This value is less than ( |\mathbf{r}| ), so it is not the maximum value.

• (d) If ( r ) is along the negative ( Y )-axis, the angle ( \theta ) between ( r ) and the ( X )-axis is ( 90^\circ ). Since ( \cos 90^\circ = 0 ), the component of ( r ) along the ( X )-axis will be zero, which is not the maximum value.

Answer (c) We know that

where $\theta$ is angle of projection

Given,

$ \theta=15^{\circ} \text { and } R=50 m $

$ \text { Range, } R=\frac{u^{2} \sin 2 \theta}{g} $

Putting all the given values in the formula, we get

$ \begin{matrix} \Rightarrow & & R & =50 m=\frac{u^{2} \sin (2 \times 15^{\circ})}{g} \\ \Rightarrow & & 50 \times g=u^{2} \sin 30^{\circ}=u^{2} \times \frac{1}{2} \end{matrix} $

$ \begin{matrix} \Rightarrow & 50 \times g \times 2 & =u^{2} \\ \Rightarrow & u^{2} & =50 \times 9.8 \times 2=100 \times 9.8=980 & \\ \Rightarrow & u & =\sqrt{980}=\sqrt{49 \times 20}=7 \times 2 \times \sqrt{5} m / s & \\ & & =14 \times 2.23 m / s=31.304 m / s & \\ \text { For } & \theta & =45^{\circ} ; R=\frac{u^{2} \sin 2 \times 45^{\circ}}{g}=\frac{u^{2}}{g} & (\because \sin 90^{\circ}=1) \\ \Rightarrow & R & =\frac{(14 \sqrt{5})^{2}}{g}=\frac{14 \times 14 \times 5}{9.8}=100 m \end{matrix} $

• Option (a) $60 m$: This option is incorrect because the range of a projectile fired at an angle of $45^{\circ}$ with the same initial speed is not $60 m$. The range at $45^{\circ}$ is maximized and should be greater than the range at $15^{\circ}$, which is $50 m$. The correct calculation shows the range to be $100 m.

• Option (b) $71 m$: This option is incorrect because the range of a projectile fired at an angle of $45^{\circ}$ with the same initial speed is not $71 m$. The range at $45^{\circ}$ is maximized and should be greater than the range at $15^{\circ}$, which is $50 m. The correct calculation shows the range to be $100 m.

• Option (d) $141 m$: This option is incorrect because the range of a projectile fired at an angle of $45^{\circ}$ with the same initial speed is not $141 m$. The range at $45^{\circ}$ is maximized and should be exactly double the range at $15^{\circ}$, which is $50 m. The correct calculation shows the range to be $100 m.

Answer (b) We know that impulse $J=F$. $\Delta t=\Delta p$, where $F$ is force, $\Delta t$ is time duration and $\Delta p$ is change in momentum. As $\Delta p$ is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector.

• Option (a) impulse, pressure and area: Pressure is a scalar quantity, not a vector. It is defined as force per unit area and does not have a direction associated with it.

• Option (c) area and gravitational potential: Gravitational potential is a scalar quantity, not a vector. It represents the potential energy per unit mass at a point in a gravitational field and does not have a direction.

• Option (d) impulse and pressure: As mentioned earlier, pressure is a scalar quantity and not a vector.

Thinking Process

As speed is a scalar quantity, hence it will be related with path length (scalar quantity) only.

Answer (d) We know that

$ \text { speed, } v_0=\frac{\text { total distance travelled }}{\text { time taken }} $

Hence, total distance travelled $=$ Path length

$ =(\text { speed }) \times \text { time taken } $

Note We should be very careful with the fact, that speed is related with total distance covered not with displacement.

• (a) The average velocity is not zero at any time:

  • This is incorrect because average velocity is defined as the total displacement divided by the total time taken. In a two-dimensional motion, if the object returns to its starting point, the displacement is zero, making the average velocity zero over that time interval.

• (b) Average acceleration must always vanish:

  • This is incorrect because average acceleration is defined as the change in velocity divided by the time taken. Even if the instantaneous speed is constant, the direction of the velocity vector can change, resulting in a non-zero average acceleration.

• (c) Displacements in equal time intervals are equal:

  • This is incorrect because displacement is a vector quantity that depends on both the magnitude and direction of motion. In two-dimensional motion, the direction of motion can change, leading to different displacements in equal time intervals even if the speed is constant.

Answer (c) As given motion is two dimensional motion and given that instantaneous speed $v_0$ is positive constant. Acceleration is rate of change of velocity (instantaneous speed) hence it will also be in the plane of motion.

• (a) The acceleration of the particle is zero: This is incorrect because a particle can have a constant speed but still be accelerating if its direction of motion is changing. For example, in uniform circular motion, the speed is constant but the direction of velocity changes continuously, resulting in a non-zero acceleration.

• (b) The acceleration of the particle is bounded: This is not necessarily true because the problem statement does not provide any information about the limits or bounds on the acceleration. The acceleration could be unbounded depending on the forces acting on the particle.

• (d) The particle must be undergoing a uniform circular motion: This is incorrect because constant speed in two-dimensional motion does not necessarily imply circular motion. The particle could be moving in any path where the speed remains constant, such as along a straight line or any other curved path, as long as the speed does not change.

Thinking Process

This question can solved by checking each options one by one.

Answer $(b, d)$

Given $A+B+C=0$

Hence, we can say that $A, B$ and $C$ are in one plane and are represented by the three sides of a triangle taken in one order. Now consider the options one by one.

(a) We can write

$B \times(A+B+C) =B \times 0=0 $

$\Rightarrow B \times A+B \times B+B \times C=0 $

$\Rightarrow B \times A+0 B \times C=0 $

$\Rightarrow B \times A=-B \times C $

$\Rightarrow A A B=B \times C $

$\therefore (A \times B) \times C =(B \times C) \times C$

It cannot be zero.

If $B | C$, then $B \times C=0$, than $(B \times C) \times C=0$.

(b) $(A \times B) \cdot C=(B \times AC) \cdot C=0$ whatever be the positions of $A, B$ and $C$. If $B | C$, then $B \times C=0$, then $(B \times C) \times C=0$.

(c) $(A \times B)=X=AB \sin \theta X$. The direction of $X$ is perpendicular to the plane containing $A$ and $B$. $(A \times B) \times C=X \times C$. Its direction is in the plane of $A, B$ and $C$.

(d) If $C^{2}=A^{2}+B^{2}$, then angle between $A$ and $B$ is $90^{\circ}$

$ \begin{aligned} (\mathbf{A} \times \mathbf{B}) \cdot C & =(A B \sin 90^{\circ} \mathbf{X}) \cdot \mathbf{C}=A B(X \cdot C) \\ & =A B C \cos 90^{\circ}=0 \end{aligned} $

• Option (a): The reasoning provided shows that ((\mathbf{A} \times \mathbf{B}) \times \mathbf{C}) is not zero unless (\mathbf{B}) and (\mathbf{C}) are parallel. This is because (\mathbf{A} + \mathbf{B} + \mathbf{C} = 0) implies (\mathbf{A}), (\mathbf{B}), and (\mathbf{C}) lie in the same plane, and the cross product (\mathbf{B} \times \mathbf{C}) is zero if (\mathbf{B}) and (\mathbf{C}) are parallel. Therefore, this option is correct and not false.

• Option (c): The reasoning provided shows that ((\mathbf{A} \times \mathbf{B}) \times \mathbf{C}) lies in the plane defined by (\mathbf{A}), (\mathbf{B}), and (\mathbf{C}). This is because (\mathbf{A} \times \mathbf{B}) is perpendicular to the plane containing (\mathbf{A}) and (\mathbf{B}), and the cross product of this result with (\mathbf{C}) will lie in the original plane. Therefore, this option is correct and not false.

• Option (d): The reasoning provided shows that if (C^2 = A^2 + B^2), then the angle between (\mathbf{A}) and (\mathbf{B}) is (90^\circ). This implies that ((\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}) should be zero because the dot product of a vector perpendicular to the plane (resulting from (\mathbf{A} \times \mathbf{B})) with a vector in the plane ((\mathbf{C})) is zero. Therefore, this option is false as stated.

Answer ( $a, b$ )

Given that

$ |\mathbf{A}+\mathbf{B}|=|\mathbf{A}| \text { or }|\mathbf{A}+\mathbf{B}|^{2}=|\mathbf{A}|^{2} $

$ \Rightarrow \quad|\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta=|\mathbf{A}|^{2} $

where $\theta$ is angle between $\mathbf{A}$ and $\mathbf{B}$.

$ \begin{aligned} \Rightarrow & & |\mathbf{B}|(|\mathbf{B}|+2|\mathbf{A}| \cos \theta) & =0 \\ \Rightarrow & & |\mathbf{B}| & =0 \text { or }|\mathbf{B}|+2|\mathbf{A}| \cos \theta=0 \\ \Rightarrow & & \cos \theta & =-\frac{|\mathbf{B}|}{2|\mathbf{A}|} \end{aligned} $

If $\mathbf{A}$ and $\mathbf{B}$ are antiparallel, then $\theta=180^{\circ}$

Hence, from Eq. (i)

$ -1=-\frac{|\mathbf{B}|}{2|\mathbf{A}|} \Rightarrow|\mathbf{B}|=2|\mathbf{A}| $

Hence, correct answer will be either $|\mathbf{B}|=0$ or $\mathbf{A}$ and $\mathbf{B}$ are antiparallel provided $|\mathbf{B}|=2|\mathbf{A}|$

• Option (c): A, B are perpendicular

  This option is incorrect because if vectors (\mathbf{A}) and (\mathbf{B}) are perpendicular, the angle (\theta) between them would be (90^\circ). In this case, (\cos \theta = 0). Substituting (\cos \theta = 0) into the equation (|\mathbf{A}+\mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos \theta) would yield (|\mathbf{A}+\mathbf{B}|^2 = |\mathbf{A}|^2 + |\mathbf{B}|^2), which does not satisfy the given condition (|\mathbf{A}+\mathbf{B}| = |\mathbf{A}|).

• Option (d): (\mathbf{A} . \boldsymbol{B} \leq 0)

  This option is incorrect because the condition (\mathbf{A} . \mathbf{B} \leq 0) implies that the dot product of (\mathbf{A}) and (\mathbf{B}) is non-positive, which means the angle (\theta) between them is either (90^\circ) (perpendicular) or greater than (90^\circ) (obtuse). However, the given condition (|\mathbf{A}+\mathbf{B}| = |\mathbf{A}|) specifically leads to either (\mathbf{B} = 0) or (\mathbf{A}) and (\mathbf{B}) being antiparallel with (|\mathbf{B}| = 2|\mathbf{A}|). The condition (\mathbf{A} . \mathbf{B} \leq 0) is not sufficient to guarantee these specific cases.

Thinking Process

In this problem, we have to apply equation for maximum height reached $H=$ $\frac{u^{2} \sin ^{2} \theta}{2 g}$, where $\theta$ is angle of projection and $u$ is speed of projection of a projectile

Answer

$(a, b, c)$

We know that maximum height reached by a projectile,

$ \begin{aligned} & H=\frac{u^{2} \sin ^{2} \theta}{2 g} \\ & H_1=\frac{v_0^{2} \sin ^{2} \theta_1}{2 g} \quad \text { (for first particle) } \\ & H_2=\frac{v_0^{2} \sin ^{2} \theta_2}{2 g} \quad \text { (for second particle) } \end{aligned} $

According to question, we know that

$ \begin{aligned} & H_1>H_2 \\ & \Rightarrow \quad \frac{v_0^{2} \sin ^{2} \theta_1}{2 g}>\frac{v_0^{2} \sin ^{2} \theta_2}{2 g} \\ & \Rightarrow \quad \sin ^{2} \theta_1>\sin ^{2} \theta_2 \\ & \Rightarrow \quad \sin ^{2} \theta_1-\sin ^{2} \theta_2>0 \\ & \Rightarrow \quad(\sin \theta_1-\sin \theta_2)(\sin \theta_1+\sin \theta_2)>0 \\ & \text { Thus, either } \quad \sin \theta_1+\sin \theta_2>0 \\ & \Rightarrow \quad \sin \theta_1-\sin \theta_2>0 \\ & \Rightarrow \quad \sin \theta_1>\sin \theta_2 \text { or } \theta_1>\theta_2 \\ & T=\frac{2 u \sin \theta}{g}=\frac{2 v_0 \sin \theta}{g} \\ & T_1=\frac{2 v_0 \sin \theta_1}{g} \\ & T_2=\frac{2 v_0 \sin \theta_2}{g} \end{aligned} $

As,

Hence, (Here, $T_1=$ Time of flight of first particle and $T_2=$ Time of flight of second particle).

$ \sin \theta_1>\sin \theta_2 $

$ T_1>T_2 $

We know that

Given,

$\Rightarrow$

$\Rightarrow \quad \frac{R_1}{R_2}=\frac{\sin 2 \theta_1}{\sin 2 \theta_2}>1$

$\Rightarrow \quad R_1>R_2$

Total energy for the first particle,

$ \begin{aligned} & U_1=K E+P E=\frac{1}{2} m_1 v_0^{2} \\ & \text { (This value will be constant throughout the journey) } \\ & U_2=K E+P E=\frac{1}{2} m_2 v_0^{2} \quad \text { (Total energy for the second particle) } \end{aligned} $

Total energy for the second particle

If

$ \begin{aligned} & m_1=m_2 \text { then } U_1=U_2 \\ & m_1>m_2 \text { then } U_1>U_2 \\ & m_1<m_2, \text { then } U_1 < U_2 \end{aligned} $

• Option (d) Total energy ( U_1 > U_2 ): The total mechanical energy of a projectile is given by the sum of its kinetic energy (KE) and potential energy (PE). For both particles, the total energy is given by ( U = \frac{1}{2} m v_0^2 ). Since the speed ( v_0 ) is the same for both particles and assuming the masses ( m_1 ) and ( m_2 ) are equal, the total energy ( U_1 ) and ( U_2 ) will be the same. Therefore, ( U_1 ) cannot be greater than ( U_2 ) unless the masses are different, which is not specified in the problem. Hence, option (d) is incorrect.

Thinking Process

In this type of question, nature of track is very important of consider, as friction is not in this track, total energy of the particle will remain constant throughout the journey.

Answer (c)

As the given track $y=x^{2}$ is a frictionless track thus, total energy (KE + PE) will be same throughout the journey.

Hence, total energy at $A=$ Total energy at $P$. At $B$, the particle is having only KE but at $P$ some KE is converted to $P$.

Hence, $(KE)_B >(KE)_P$

Total energy at $A=PE=$ Total energy at $B=KE$

$ \begin{aligned} & =\text { Total energy at } P \\ & =PE+KE \end{aligned} $

The potential energy at $A$, is converted to $KE$ and $PE$ at $P$, hence

$ (PE) P<(PE) A $

Hence,

$ \text { (Height) } P<\text { (Height) A } $

As, height of $P<$ Height of $A$

Hence, path length $A B>$ path length $B P$

Hence, time of travel from $A$ to $B \neq$ Time of travel from $B$ to $P$.

• (a) $KE$ at $P=KE$ at $B$

  This option is incorrect because at point $B$, the particle has only kinetic energy (KE) as it is at the lowest point of the parabola. However, at point $P$, the particle is at the highest point of its projectile motion, meaning some of its kinetic energy has been converted into potential energy (PE). Therefore, the kinetic energy at $P$ is less than the kinetic energy at $B$.

• (b) height at $P=$ height at $A$

  This option is incorrect because the height at point $P$ is less than the height at point $A$. As the particle slides down the frictionless track, it loses potential energy and gains kinetic energy. When it reaches point $C$ and moves as a projectile, it cannot reach a height greater than or equal to the initial height at $A$ due to energy conservation. Therefore, the height at $P$ is less than the height at $A$.

• (d) time of travel from $A$ to $B=$ time of travel from $B$ to $P$

  This option is incorrect because the time of travel from $A$ to $B$ is not equal to the time of travel from $B$ to $P$. The path from $A$ to $B$ is along the parabolic track, which is longer and involves a continuous change in direction and speed. In contrast, the path from $B$ to $P$ is a projectile motion, which is generally shorter and follows a different trajectory. Therefore, the times of travel for these two segments are not equal.

Answer (a, c)

If an object undergoes a displacement $\Delta r$ in time $\Delta t$, its average velocity is given by $v=\frac{\Delta r}{\Delta t}=\frac{r_2-r_1}{t_2-t_1}$; where $r_1$ and $r_2$ are position vectors corresponding to time $t_1$ and $t_2$.

It the velocity of an object changes from $v_1$ to $v_2$ in time $\Delta t$. Average acceleration is given by

$ a_{av}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_2} $

But, when acceleration is non-uniform

We can write

$ v_{av} \neq \frac{v_1+v_2}{2} $

$ \text { Hence, } \quad \Delta r=r_2-r_1=(v_2-v_1)(t_2-t_1) $

• Option (a): The average velocity ( v_{av} ) is not necessarily equal to the arithmetic mean of the initial and final velocities, ( \frac{1}{2}[v(t_1) + v(t_2)] ), unless the acceleration is constant. For non-uniform acceleration, this relation does not hold.

• Option (c): The expression ( r = \frac{1}{2}(v(t_2) - v(t_1))(t_2 - t_1) ) is incorrect because it does not correctly represent the displacement. The correct expression for displacement when acceleration is constant is ( r = v_1(t_2 - t_1) + \frac{1}{2}a(t_2 - t_1)^2 ).

• Option (d): This option is correct. The average acceleration ( a_{av} ) is given by ( \frac{v(t_2) - v(t_1)}{t_2 - t_1} ), which is a standard definition.

Answer $(a, b, c)$

For a particle performing uniform circular motion

(i) speed will be constant throughout.

(ii) velocity will be tangential in the direction of motion at a particular point.

(iii) acceleration $a=\frac{v^{2}}{r}$, will always be towards centre of the circular path.

(iv) angular momentum ( $m v r$ ) is constant in magnitude and direction/out of the plane perpendicularly, as well.

Note In uniform circular motion, magnitude of velocity and acceleration is constant but direction changes continuously.

• Option (d) is incorrect because, in uniform circular motion, the direction of the angular momentum vector changes continuously as the particle moves around the circle. While the magnitude of the angular momentum remains constant, its direction is always perpendicular to the plane of motion and changes as the particle’s position changes.

Answer $(b, d)$

Given, $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$

$ \begin{matrix} \Rightarrow & \sqrt{|\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta}=\sqrt{|\mathbf{A}|^{2}+|\mathbf{B}|^{2}-2|\mathbf{A}||\mathbf{B}| \cos \theta} \\ \Rightarrow & |\mathbf{A}|^{2}+|\mathbf{B}|^{2}+2|\mathbf{A}||\mathbf{B}| \cos \theta=|\mathbf{A}|^{2}+|\mathbf{B}|^{2}-2|\mathbf{A}||\mathbf{B}| \cos \theta \\ \Rightarrow & 4|\mathbf{A}||\mathbf{B}| \cos \theta=0 \\ \Rightarrow & |\mathbf{A}||\mathbf{B}| \cos \theta=0 \\ \Rightarrow & |\mathbf{A}|=0 \text { or }|\mathbf{B}|=0 \text { or } \cos \theta=0 \\ \Rightarrow & \theta=90^{\circ} \end{matrix} $

When $\theta=90^{\circ}$, we can say that $\mathbf{A} \perp \mathbf{B}$

• Option (a): This option states that ( |\mathbf{A}| = |\mathbf{B}| \neq 0 ). However, this condition alone does not guarantee that ( |\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}| ). For example, if (\mathbf{A}) and (\mathbf{B}) are parallel, the magnitudes of (\mathbf{A} + \mathbf{B}) and (\mathbf{A} - \mathbf{B}) will not be equal.

• Option (c): This option states that ( |\mathbf{A}| = |\mathbf{B}| \neq 0 ) and (\mathbf{A}) and (\mathbf{B}) are parallel or anti-parallel. If (\mathbf{A}) and (\mathbf{B}) are parallel, then (|\mathbf{A} + \mathbf{B}|) will be twice the magnitude of (\mathbf{A}) or (\mathbf{B}), and (|\mathbf{A} - \mathbf{B}|) will be zero, which are not equal. If (\mathbf{A}) and (\mathbf{B}) are anti-parallel, (|\mathbf{A} + \mathbf{B}|) will be zero, and (|\mathbf{A} - \mathbf{B}|) will be twice the magnitude of (\mathbf{A}) or (\mathbf{B}), which are also not equal. Therefore, this condition does not satisfy the given equation.

Answer As shown in the adjacent figure. The cyclist covers the path $O P R Q O$. As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

Hence, acceleration at $R$ is a $=\frac{v^{2}}{r}$

$\Rightarrow \quad a=\frac{(10)^{2}}{1 km}=\frac{100}{10^{3}}=0.1 m / s^{2}$ along $R O$.

Thinking Process

When a particle is under projectile motion, horizontal component of velocity will always be constant and acceleration is always vertically downward and is equal to $g$.

Answer Consider the adjacent diagram in which a particle is projected at an angle $\theta$.

$v_{x}=$ Horizontal component of velocity $=v \cos \theta=$ constant.

$v_{y}=$ Vertical component of velocity $=v \sin \theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to $g$ (acceleration due to gravity).

Answer Consider the adjacent diagram in which a ball is projected from point $O$, and covering the path $O A B C$.

(a) At point $B$, it will gain the same speed $u$ and after that speed increases and will be maximum just before reaching $C$.

(b) During upward journey from $O$ to $A$ speed decreases and will be minimum at point $A$.

(c) Acceleration is always constant throughout the journey and is vertically downward equal to $g$.

Answer (a) Consider the adjacent diagram in which a football is kicked into the air vertically upwards. Acceleration of the football will always be vertical downward and is equal to $g$.

(b) When the football reaches the highest point velocity will be zero as it is continuously retarded by acceleration due to gravity $g$.

Answer The path of the ball observed by a boy standing on the footpath is parabolic. The horizontal speed of the ball is same as that of the car, therefore, ball as well car travells equal horizontal distance. Due to its vertical speed, the ball follows a parabolic path.

Note We must be very clear that we are working with respect to ground. When we observe with respect to the car motion will be along vertical direction only.

Answer Consider the diagram below

(b)

The boy throws the ball at an angle of $60^{\circ}$.

$\therefore$ Horizontal component of velocity $=4 \cos \theta$

$ \begin{aligned} & =(10 m / s) \cos 60^{\circ}=10 \times \frac{1}{2}=5 m / s . \\ \text { Speed of the car } & =18 km / h=5 m / s . \end{aligned} $

As horizontal speed of ball and car is same, hence relative velocity of car and ball in the horizontal direction will be zero.

Only vertical motion of the ball will be seen by the boy in the car, as shown in fig. (b)

Thinking Process

When air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.

Answer Due to air resistance, particle energy as well as horizontal component of velocity keep on decreasing making the fall steeper than rise as shown in the figure.

When we are neglecting air resistance path was symmetric parabola $(O A B)$. When air resistance is considered path is asymmetric parabola $(O A C)$.

Thinking Process

When the bomb is dropped from the plane, the bomb will have same velocity as that of plane.

Answer Consider the adjacent diagram. Let a fighter plane, when it be at position $P$, drops a bomb to hit a target $T$.

Let

$ <P^{\prime} P T=\theta $

Speed of the plane $=720 km / h=720 \times \frac{5}{18} m / s=200 m / s$

Altitude of the plane $(P^{\prime} T)=1.5 km=1500 m$

If bomb hits the target after time $t$, then horizontal distance travelled by the bomb,

$ P P^{\prime}=u \times t=200 t $

Vertical distance travelled by the bomb,

$ \begin{aligned} & P^{\prime} T=\frac{1}{2} g t^{2} \Rightarrow 1500=\frac{1}{2} \times 9.8 t^{2} \\ \Rightarrow \quad t^{2} & =\frac{1500}{4.9} \Rightarrow t=\sqrt{\frac{1500}{49}}=17.49 s \end{aligned} $

Using value of $t$ in Eq. (i),

Now,

$ \begin{aligned} P P^{\prime} & =200 \times 17.49 m \\ \tan \theta & =\frac{P^{\prime} T}{P^{\prime} P}=\frac{1500}{200 \times 17.49}=.49287=\tan 23^{\circ} 12^{\prime} \\ \theta & =23^{\circ} 12^{\prime} \end{aligned} $

Note Angle is with respect to target. As seen by observer in the plane motion of the bomb will be vertically downward below the plane.

Answer (a) Radius of the earth $(R)=6400 km=6.4 \times 10^{6} m$

Time period $(T)=1$ day $=24 \times 60 \times 60 s=86400 s$

Centripetal acceleration $(a_{c})=\omega^{2} R=R(\frac{2 \pi}{T})^{2}=\frac{4 \pi^{2} R}{T}$

$ \begin{aligned} & =\frac{4 \times(22 / 7)^{2} \times 6.4 \times 10^{6}}{(24 \times 60 \times 60)^{2}} \\ & =\frac{4 \times 484 \times 64 \times 10^{6}}{49 \times(24 \times 3600)^{2}} \\ & =0.034 m / s^{2} \end{aligned} $

At equator,

latitude $\theta=0^{\circ}$

$\therefore$

$ \frac{a_{c}}{g}=\frac{0.034}{9.8}=\frac{1}{288} $

(b) Orbital radius of the earth around the sun $(R)=1.5 \times 10^{11} m$

Time period $=1 yr=365$ day

$ =365 \times 24 \times 60 \times 60 s=3.15 \times 10^{7} s $

Centripetal acceleration $(a_{c})=R \omega^{2}=\frac{4 \pi^{2} R}{T^{2}}$

$ \begin{aligned} & =\frac{4 \times(22 / 7)^{2} \times 1.5 \times 10^{11}}{(3.15 \times 10^{7})^{2}} \\ & =5.97 \times 10^{-3} m / s^{2} \\ \frac{a_{C}}{g} & =\frac{5.97 \times 10^{-3}}{9.8}=\frac{1}{1642} \end{aligned} $

Thinking Process

In this problem, triangular law of vector addition will be applied.

Answer Consider the adjacent diagram in which vectors $\mathbf{A}$ and $\mathbf{B}$ are corrected by head and tail.

Resultant vector $\mathbf{C}=\mathbf{A}+\mathbf{B}$

(a) from (iv) it is clear that $\mathbf{c}=\mathbf{a}+\mathbf{b}$

(b) from (iii) $\mathbf{c}+\mathbf{b}=\mathbf{a} \Rightarrow \mathbf{a}-\mathbf{c}=\mathbf{b}$

(c) from (i) $\mathbf{b}=\mathbf{a}+\mathbf{c} \Rightarrow \mathbf{b}-\mathbf{a}=\mathbf{c}$

(d) from (ii) $-\mathbf{c}=\mathbf{a}+\mathbf{b} \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$

Answer Given $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$

(a) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=0 \quad \Rightarrow \quad 2 \times 4 \cos \theta=0$ $\Rightarrow \cos \theta=0=\cos 90^{\circ} \Rightarrow \theta=90^{\circ}$

$\therefore$ Option (a) matches with option (ii).

(b) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=8 \quad \Rightarrow \quad 2 \times 4 \cos \theta=8$

$\Rightarrow \cos \theta=1=\cos 0^{\circ} \Rightarrow \quad \theta=0^{\circ}$

$\therefore$ Option (b) matches with option (i).

(c) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=4 \quad \Rightarrow 2 \times 4 \cos \theta=4$

$\Rightarrow \cos \theta=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=60^{\circ}$

$\therefore$ Option (c) matches with option (iv).

(d) $\mathbf{A} \cdot \mathbf{B}=A B \cos \theta=-8 \quad \Rightarrow \quad 2 \times 4 \cos \theta=-8$

$\Rightarrow \cos \theta=-1=\cos 180^{\circ} \Rightarrow \quad \theta=180^{\circ}$

$\therefore$ Option (d) matches with option (iii).

Answer Given $|\mathbf{A}|=2$ and $|\mathbf{B}|=4$

(a) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=0 \quad \Rightarrow 2 \times 4 \times \sin \theta=0$

$\Rightarrow \sin \theta=0=\sin 0^{\circ} \Rightarrow \quad \theta=0^{\circ}$

$\therefore$ Option (a) matches with option (iv).

(b) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=8 \quad \Rightarrow \quad 2 \times 4 \sin \theta=8$

$\Rightarrow \sin \theta=1=\sin 90^{\circ} \Rightarrow \theta=90^{\circ}$

$\therefore$ Option (b) matches with option (iii).

(c) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=4 \quad \Rightarrow \quad 2 \times 4 \sin \theta=4$

$\Rightarrow \sin \theta=\frac{1}{2}=\sin 30^{\circ} \Rightarrow \quad \theta=30^{\circ}$

$\therefore$ Option (c) matches with option (i).

(d) $|\mathbf{A} \times \mathbf{B}|=A B \sin \theta=4 \sqrt{2} \quad \Rightarrow \quad 2 \times 4 \sin \theta=4 \sqrt{2}$

$\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \Rightarrow \theta=45^{\circ}$

$\therefore$ Option (d) matches with option (ii).

Answer Given, speed of packets $=125 m / s$

Height of the hill $=500 m$.

To cross the hill, the vertical component of the velocity should be sufficient to cross such height.

But

$ \begin{aligned} u_{y} & \geq \sqrt{2 g h} \\ & \geq \sqrt{2 \times 10 \times 500} \\ & \geq 100 m / s \\ u^{2} & =u_x^{2}+u_y^{2} \end{aligned} $

$\therefore$ Horizontal component of initial velocity,

$ u_{x}=\sqrt{u^{2}-u_y^{2}}=\sqrt{(125)^{2}-(100)^{2}}=75 m / s $

Time taken to reach the top of the hill,

$ t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 s $

Time taken to reach the ground from the top of the hill $t^{\prime}=t=10 s$. Horizontal distance travelled in $10 s$

$ x=u_{x} \times t=75 \times 10=750 m $

$\therefore$ Distance through which canon has to be moved $=800-750=50 m$

Speed with which canon can move $=2 m / s$

$\therefore \quad$ Time taken by canon $=\frac{50}{2} \Rightarrow t^{\prime \prime}=25 s$

$\therefore$ Total time taken by a packet to reach on the ground $=t^{\prime \prime}+t+t^{\prime}=25+10+10=45 s$

Thinking Process

Horizontal range of a projectile is maximum when it is thrown at an angle $45^{\circ}$ from the horizontal and $R_{\max }=\frac{u^{2}}{g}$, where $u$ is speed of projection of the projectile.

Answer This problem can be approached in two different ways

(i) Refer to the diagram, target $T$ is at horizontal distance $x=R+\Delta x$ and between point of projection $y=-h$.

(ii) From point $P$ in the diagram projection at speed $v_0$ at an angle $\theta$ below horizontal with height $h$ and horizontal range $\Delta x A$ )

Applying method (i)

Maximum horizontal range

$ R=\frac{v_0^{2}}{g}, \text { for } \theta=45^{\circ} $

Let the gun be raised through a height $h$ from the ground so that it can hit the target. Let vertically downward direction is taken as positive

Horizontal component of initial velocity $=v_0 \cos \theta$

Vertical component of initial velocity $=-v_0 \sin \theta$

Taking motion in vertical direction, $h=(-v_0 \sin \theta) t+\frac{1}{2} g t^{2}$

Taking motion in horizontal direction

$(R+\Delta x) =v_0 \cos \theta \times t $

$\Rightarrow t =\frac{(R+\Delta x)}{v_0 \cos \theta}$

Substituting value of $t$ in Eq. (ii), we get

$ \begin{aligned} & h=(-v_0 \sin \theta) \times(\frac{R+\Delta x}{v_0 \cos \theta})+\frac{1}{2} g(\frac{R+\Delta x}{v_0 \cos \theta})^{2} \\ & h=-(R+\Delta x) \tan \theta+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2} \cos ^{2} \theta} \end{aligned} $

As angle of projection is $\theta=45^{\circ}$, therefore

$ \begin{aligned} h & =-(R+\Delta x)+\tan 45^{\circ}+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2} \cos ^{2} 45^{\circ}} \\ h & =-(R+\Delta x) \times 1+\frac{1}{2} g \frac{(R+\Delta x)^{2}}{v_0^{2}(1 / 2)} \\ h & .=-(R+\Delta x)+\frac{(R+\Delta x)^{2}}{R} \quad \quad(\because \tan 45^{\circ}=1 \text { and } \cos 45^{\circ}=\frac{1}{\sqrt{2}}) \quad \text { [Using Eq. (i), } R=v_0^{2} / g] \\ & =-(R+\Delta x)+\frac{1}{R}(R^{2}+\Delta x^{2}+2 R \Delta x) \\ & =-R-\Delta x+(R+\frac{\Delta x^{2}}{R}+2 \Delta x) \\ & =\Delta x+\frac{\Delta x^{2}}{R} \quad \text { Hence proved. } \\ h & =\Delta x \biggl (1+\frac{\Delta x}{R} \biggl) \quad \end{aligned} $

Note We should not confuse with the positive direction of motion. May be vertically upward direction or vertically downward direction is taken as positive according to convenience.

Thinking Process

To solve problems involving projectile motion on an inclined plane, we have to choose two mutually perpendicular axes, one along inclined plane and other perpendicular to the inclined plane.

Answer Consider the adjacent diagram.

Mutually perpendicular $x$ and $y$-axes are shown in the diagram.

Particle is projected from the point $O$.

Let time taken in reaching from point $O$ to point $P$ is $T$.

(b) Considering motion along vertical upward direction perpendicular to $O X$.

For the journey $O$ to $P$.

$ y=0, u_{y}=v_0 \sin \beta, a_{y}=-g \cos \alpha, t=T $

Applying equation,

y = $u_{y} t+\frac{1}{2} a_{y} t^{2} $

$\rightarrow 0 =v_0 \sin \beta T+\frac{1}{2}(-g \cos \alpha) T^{2} $

$\rightarrow \quad T[v _0 \sin \beta-\frac{g \cos \alpha}{2} T] =0 $

$rightarrow T =0, T=\frac{2 v _0 \sin \beta}{g \cos \alpha}$

As $T=0$, corresponding to point $O$

Hence,

$ T=\text { Time of flight }=\frac{2 v_0 \sin \beta}{g \cos \alpha} $

(a) Considering motion along $O X$.

$ \begin{aligned} x & =L, u_{x}=v_0 \cos \beta, a_{x}=-g \sin \alpha \\ t & =T=\frac{2 v_0 \sin \beta}{g \cos \alpha} \\ x & =u_{x} t+\frac{1}{2} a_{x} t^{2} \end{aligned} $

$ \begin{aligned} \Rightarrow \quad L & =v_0 \cos \beta T+\frac{1}{2}(-g \sin \alpha) T^{2} \\ \quad L & =v_0 \cos \beta T-\frac{1}{2} g \sin \alpha T^{2} \\ & =T[v_0 \cos \beta-\frac{1}{2} g \sin \alpha T] \\ & =T[v_0 \cos \beta-\frac{1}{2} g \sin \alpha \times \frac{2 v_0 \sin \beta}{g \cos \alpha}] \\ & =\frac{2 v_0 \sin \beta}{g \cos \alpha}[v_0 \cos \beta-\frac{v_0 \sin \alpha \sin \beta}{\cos \alpha}] \\ & =\frac{2 v_0^{2} \sin \beta}{g \cos ^{2} \alpha}[\cos \beta \cdot \cos \alpha-\sin \alpha \cdot \sin \beta] \\ \Rightarrow \quad L & =\frac{2 v_0^{2} \sin \beta}{g \cos ^{2} \alpha} \cos (\alpha+\beta) \end{aligned} $

(c) For range ( $L$ ) to be maximum,

$\sin \beta \cdot \cos (\alpha+\beta)$ should be maximum.

$ \begin{aligned} & \text { Let, } \quad \begin{aligned} Z & =\sin \beta \cdot \cos (\alpha+\beta) \\ & =\sin \beta[\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta) \\ & =\frac{1}{2}[\cos \alpha \cdot \sin 2 \beta-2 \sin \alpha \cdot \sin ^{2} \beta] \\ & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha(1-\cos 2 \beta)] \\ \Rightarrow \quad Z & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha+\sin \alpha \cdot \cos 2 \beta] \\ & =\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha+\cos 2 \beta \cdot \sin \alpha-\sin \alpha] \\ & =\frac{1}{2}[\sin (2 \beta+\alpha)-\sin \alpha] \end{aligned} \end{aligned} $

For $z$ to be maximum,

$ \begin{aligned} & \sin (2=\beta+\alpha)=\text { maximum }=1 \\ \Rightarrow \quad 2 \beta+\alpha & =\frac{\pi}{2} \text { or, } \beta=\frac{\pi}{4}-\frac{\alpha}{2} \end{aligned} $

Thinking Process

When particle rebounded elastically speed will remain same.

Answer Considering $x$ and $y$-axes as shown in the diagram. For the motion of the projectile from $O$ to $A$.

$ \begin{aligned} y & =0, u_{y}=v_0 \cos \theta \\ a_{y} & =-g \cos \theta, t=T \end{aligned} $

Applying equation of kinematics,

$ \begin{aligned} y & =u_{y} t+\frac{1}{2} a_{y} t^{2} \\ \Rightarrow \quad 0 & =v_0 \cos \theta T+\frac{1}{2}(-g \cos \theta) T^{2} \\ \Rightarrow \quad T,[v_0 \cos \theta-\frac{g \cos \theta T}{2}] & =0 \\ T & =\frac{2 v_0 \cos \theta}{g \cos \theta} \end{aligned} $

As $T=0$, corresponds to point $O$

Hence, $\quad T=\frac{2 v_0}{g}$

Now considering motion along $O X$.

$ x=L, u_{x}=v_0 \sin \theta, a_{x}=g \sin \theta, t=T=\frac{2 v_0}{g} $

Applying equation of kinematics,

$ \begin{aligned} x & =u_{x} t+\frac{1}{2} a_{x} t^{2} \\ L \quad L & =v_0 \sin \theta t+\frac{1}{2} g \sin \theta t^{2}=(v_0 \sin \theta)(T)+\frac{1}{2} g \sin \theta T^{2} \\ & =(v_0 \sin \theta)(\frac{2 v_0}{g})+\frac{1}{2} g \sin \theta \times(\frac{2 v_0}{g})^{2} \\ & =\frac{2 v_0^{2}}{g} \sin \theta+\frac{1}{2} g \sin \theta \times \frac{4 v_0^{2}}{g^{2}}=\frac{2 v_0^{2}}{g}[\sin \theta+\sin \theta] \\ \quad L & =\frac{4 v_0^{2}}{g} \sin \theta \end{aligned} $

Thinking Process

Draw the vector diagram is for the information given and find $a$ and $b$. We may draw all vectors in the reference frame of ground based observer.

Answer Assume north to be $\hat{i}$ direction and vertically downward to be $-\hat{j}$. Let the rain velocity $v_{r}$ be $a \hat{i}+b \hat{j}$.

$v_r$ = $ a \hat{i} + b \hat{j} $

Case I Given velocity of girl =$ v_{g}=(5 m / s) \hat{i}$

Let $v_{rg}$ = Velocity of rain w.r.t girl

$ =v_{r}-v_{g} = (a \hat{i} + b \hat{j})-5 \hat{i} $

$ =(a-5) \hat{i} + b \hat{j}$

According to question rain, appears to fall vertically downward.

Hence,

$ a-5=0 \Rightarrow a=5 $

Case II Given velocity of the girl, $\mathbf{v} _{g}=(10 m / s) \hat{\mathbf{i}}$

$\therefore \quad v_{rg}=v_{r}-v_{g}$

$ =(a \hat{i}+b \hat{j})-10 \hat{i}=(a-10) \hat{i}+b \hat{j} $

According to question rain appears to fall at $45^{\circ}$ to the vertical hence $\tan 45^{\circ}=\frac{b}{a-10}=1$

$\Rightarrow \quad b=a-10=5-10=-5$

Hence, velocity of rain $=a \hat{i}+b \hat{j}$

$\Rightarrow v_r = 5 \hat{i}-5 \hat{j}$

Speed of rain =$|v_{r}|=\sqrt{(5)^{2}+(-5)^{2}}=\sqrt{50}=5 \sqrt{2} m / s$

Answer Given, Speed of the river $(v_{r})=3 m / s$ (east)

Speed of swimmer $(v_{s})=4 m / s$ (east)

(a) When swimmer starts swimming due north then his resultant velocity

$ \begin{aligned} v & =\sqrt{v_r^{2}+v_s^{2}}=\sqrt{(3)^{2}+(4)^{2}} \\ & =\sqrt{9+16}=\sqrt{25}=5 m / s \\ \tan \theta & =\frac{v_{r}}{v_{s}}=\frac{3}{4} \\ & =0.75=\tan 36^{\circ} 54^{\prime} \\ \theta & =36^{\circ} 54^{\prime} N \end{aligned} $

Hence,

(b) To reach opposite points $B$, the swimmer should swim at an angle $\theta$ of north.

Resultant speed of the swimmer

$ \begin{aligned} v & =\sqrt{v_s^{2}-v_r^{2}}=\sqrt{(4)^{2}-(3)^{2}} \\ & =\sqrt{16-9}=\sqrt{7} m / s \\ \tan \theta & =\frac{v_{r}}{v}=\frac{3}{\sqrt{7}} \\ \theta \quad \theta & =\tan ^{-1}(\frac{3}{\sqrt{7}}) \text { of north } \end{aligned} $

(c) In case (a),

Time taken by the swimmer to cross the river, $t_1=\frac{d}{v_{s}}=\frac{d}{4} s$

In case (b),

Time taken by the swimmer to cross the river

$ \begin{aligned} & t_1=\frac{d}{v}=\frac{d}{\sqrt{7}} \\ & \frac{d}{4}<\frac{d}{\sqrt{7}}, \text { therefore } t_1<t_2 \end{aligned} $

As

Hence, the swimmer will cross the river in shorter time in case (a).

Answer Consider the adjacent diagram.

(a) Initial velocity in

$x -direction, u_x = u+v_0 \cos \theta $

$u_{y} =\text { Initial velocity in } y \text {-direction } $

$ =v_0 \sin \theta$

where angle of projection is $\theta$.

Now, we can write

$ \begin{aligned} \tan \theta & =\frac{u_{y}}{u_{x}}=\frac{u_0 \sin \theta}{u+u_0 \cos \theta} \\ \Rightarrow \quad \theta & =\tan ^{-1}(\frac{v_0 \sin \theta}{u+v_0 \cos \theta}) \end{aligned} $

(b) Let $T$ be the time of flight.

As net displacement is zero over time period $T$.

$ y=0, u_{y}=v_0 \sin \theta, a_{y}=-g, t=T $

$\text { We know that } y =u_{y} t+\frac{1}{2} a_{y} t^{2} $

$\Rightarrow \quad 0 =v_0 \sin \theta T+\frac{1}{2}(-g) T^{2} $

$\Rightarrow \quad T[v_0 \sin \theta-\frac{g}{2} T] =0$

$\Rightarrow T=0, \frac{2 v_0 \sin \theta}{g} $

$T =0_1, \text { corresponds to point } O . $

${ Hence, } T =\frac{2 u_0 \sin \theta}{g}$

(c) Horizontal range, $R=(u+v_0 \cos \theta) T=(u+v_0 \cos \theta) \frac{2 v_0 \sin \theta}{g}$

$ =\frac{v_0}{g}[2 u \sin \theta+v_0 \sin 2 \theta] $

(d) For horizontal range to be maximum, $\frac{d R}{d \theta}=0$

$ \begin{aligned} & \Rightarrow \quad \frac{v_0}{g}[2 u \cos \theta+v_0 \cos 2 \theta \times 2]=0 \\ & \Rightarrow \quad 2 u \cos \theta+2 v_0[2 \cos ^{2} \theta-1]=0 \\ & \Rightarrow \quad 4 v_0 \cos ^{2} \theta+2 u \cos \theta-2 v_0=0 \\ & \Rightarrow \quad 2 v_0 \cos ^{2} \theta+u \cos \theta-v_0=0 \\ & \Rightarrow \quad \cos \theta=\frac{-u \pm \sqrt{u^{2}+8 v_0^{2}}}{4 v_0} \\ & \Rightarrow \quad \theta_{\max }=\cos ^{-1}[\frac{-u \pm \sqrt{u^{2}+8 v_0^{2}}}{4 v_0}] \\ & =\cos ^{-1}[\frac{-u+\sqrt{u^{2}+8 v_0^{2}}}{4 v_0}] \end{aligned} $

(e) If $u=v_0$,

$ \cos \theta=\frac{-v_0 \pm \sqrt{v_0^{2}+8 v_0^{2}}}{4 v_0}=\frac{-1+3}{4}=\frac{1}{2} $

$ \Rightarrow \quad \theta=60^{\circ} $

If $u«v_0$, then $8 v_0{ }^{2}+u^{2} \approx 8 v_0{ }^{2}$

$ \theta_{\max }=\cos ^{-1}[\frac{-u \pm 2 \sqrt{2} v_0}{4 v_0}]=\cos ^{-1}[\frac{1}{\sqrt{2}}-\frac{u}{4 v_0}] $

If $u«v_0$, then $\quad \theta_{\max }=\cos ^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$

If $u>u_0$ and

$ u»v_0 $

$ \theta_{\max }=\cos ^{-1}[\frac{-u \pm u}{4 v_0}]=0 \Rightarrow \theta_{\max }=\frac{\pi}{2} $

(f) If $u=0, \theta_{\max }=\cos ^{-1}[\frac{0 \pm \sqrt{8 v^{2} 0}}{4 v_0}]=\cos ^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$

Answer (a) Given, unit vector

$ \begin{aligned} & \hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j} \\ & \hat{\theta}=-\sin \theta \hat{i}+\cos \theta \hat{j} \end{aligned} $

Multiplying Eq. (i) by $\sin \theta$ and Eq. (ii) with $\cos \theta$ and adding

$\hat{r} \sin \theta+\hat{\theta} \cos \theta=\sin \theta \cdot \cos \theta \hat{i}+\sin ^{2} \theta \hat{j}+\cos ^{2} \theta \hat{j}-\sin \theta \cdot \cos \theta \hat{i}$

$=\hat{j}(\cos ^{2} \theta+\sin ^{2} \theta)=\hat{j}$

$ \Rightarrow \quad \hat{r} \sin \theta+\hat{\theta} \cos \theta=\hat{j} $

By Eq. (i) $\times \cos \theta-$ Eq. (ii) $\times \sin \theta$

$ n(\hat{r} \cos \theta-\hat{\theta} \sin \theta)=\hat{i} $

(b) $\hat{r} \cdot \hat{\theta}=(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}) \cdot(-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}})=-\cos \theta \cdot \sin \theta+\sin \theta \cdot \cos \theta=0$

$ \Rightarrow \quad \theta=90^{\circ} \text { Angle between } \hat{\mathbf{r}} \text { and } \hat{\theta} \text {. } $

(c) Given, $\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$

$ \begin{aligned} \frac{d \hat{\mathbf{r}}}{d t} & =\frac{d}{d t}(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})=-\sin \theta \cdot \frac{d \theta}{d t} \hat{\mathbf{i}}+\cos \theta \cdot \frac{d \theta}{d t} \hat{\mathbf{j}} \\ & =\omega[-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}}] \quad[\because \theta=\frac{d \theta}{d t}] \end{aligned} $

(d) Given, $r=a \theta \hat{r}$, here, writing dimensions $[r]=[a][\theta][\hat{r}]$

$ \Rightarrow \quad L=[a] % 1 \Rightarrow[a]=L=[M^{0} L^{1} T^{0}] $

(e) Given, $a=1$ unit

$ \begin{aligned} \mathbf{r} & =\theta \hat{\mathbf{r}}=\theta[\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}}] \\ V & =\frac{d r}{d t}=\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \frac{d}{d t} \hat{\mathbf{r}}=\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \frac{d}{d t}[(\cos \theta \hat{\mathbf{i}}+\sin \theta \hat{\mathbf{j}})] \\ & =\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta[(-\sin \theta \hat{\mathbf{i}}+\cos \theta \hat{\mathbf{j}}) \frac{d \theta}{d t}] \\ & =\frac{d \theta}{d t} \hat{\mathbf{r}}+\theta \hat{\theta} \omega=\omega \hat{\mathbf{r}}+\omega \theta \hat{\theta} \end{aligned} $

Velocity,

Acceleration,

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

Answer Consider adjacent diagram.

Time taken to go from $A$ to $C$ via straight line path $A P Q C$ through the $S$ and

$ \begin{aligned} T_{\text {sand }} & =\frac{A P+Q C}{1}+\frac{P Q}{v}=\frac{25 \sqrt{2}+25 \sqrt{2}}{1}+\frac{50 \sqrt{2}}{v} \\ & =50 \sqrt{2}+\frac{50 \sqrt{2}}{v}=50 \sqrt{2}(\frac{1}{v}+1) \end{aligned} $

Clearly from figure the shortest path outside the sand will be $A R C$.

Time taken to go from $A$ to $C$ via this path

Clearly,

$ T_{\text {outside }}=\frac{A R+R C}{1} s $

Clearly

$ A R=\sqrt{75^{2}+25^{2}}=\sqrt{75 \times 75+25 \times 25} $

$ \begin{aligned} & =5 \times 5 \sqrt{9+1}=25 \sqrt{10} m \\ R C & =A R=\sqrt{75^{2}+25^{2}}=25 \sqrt{10} m \end{aligned} $

$\Rightarrow$

$T_{\text {outside }}=2 A R=2 \times 25 \sqrt{10} s=50 \sqrt{10} s$

For

$ T_{\text {sand }}<T_{\text {outside }} $

$\Rightarrow \quad 50 \sqrt{2}(\frac{1}{v}+1)<2 \times 25 \sqrt{10}$

$\Rightarrow \quad \frac{2 \sqrt{2}}{2}(\frac{1}{v}+1)<\sqrt{10}$

$\Rightarrow \quad \frac{1}{v}+1<\frac{2 \sqrt{10}}{2 \sqrt{2}}=\frac{\sqrt{5}}{2} \times 2=\sqrt{5}$

$\Rightarrow \quad \frac{1}{v}<\frac{\sqrt{5}}{2} \times 2-1 \Rightarrow \frac{1}{v}<\sqrt{5}-1$

$\Rightarrow \quad v>\frac{1}{\sqrt{5}-1} \approx 0.81 m / s$

$\Rightarrow \quad v>0.81 m / s$