Mix Examples-Work, Energy, Power and Collision Questions in English

(C) Step $1$: Elastic collision between $A$ and $B$.
Let $v_A = 9 \ m/s$ and $v_B = 0$. After the collision,let the velocities be $v_A'$ and $v_B'$.
By conservation of momentum: $m(9) + 2m(0) = m v_A' + 2m v_B' \Rightarrow 9 = v_A' + 2v_B'$.
By the coefficient of restitution for an elastic collision $(e=1)$: $v_B' - v_A' = e(v_A - v_B) = 1(9 - 0) = 9 \Rightarrow v_B' - v_A' = 9$.
Adding the two equations: $(v_A' + 2v_B') + (v_B' - v_A') = 9 + 9 \Rightarrow 3v_B' = 18 \Rightarrow v_B' = 6 \ m/s$.
Step $2$: Completely inelastic collision between $B$ and $C$.
Let $v_B' = 6 \ m/s$ and $v_C = 0$. After the collision,$B$ and $C$ move together with velocity $v_f$.
By conservation of momentum: $2m(v_B') + m(0) = (2m + m)v_f \Rightarrow 2m(6) = 3m v_f \Rightarrow 12 = 3v_f \Rightarrow v_f = 4 \ m/s$.
Thus,the final speed of object $C$ is $4 \ m/s$.

(C) Let $m_1 = 0.36 \text{ kg}$ and $m_2 = 0.72 \text{ kg}$.
For the block of mass $m_2$,the equation of motion is: $m_2 g - T = m_2 a$.
For the block of mass $m_1$,the equation of motion is: $T - m_1 g = m_1 a$.
Adding these two equations: $(m_2 - m_1) g = (m_1 + m_2) a$.
$a = \frac{(m_2 - m_1) g}{m_1 + m_2} = \frac{(0.72 - 0.36) \times 10}{0.72 + 0.36} = \frac{0.36 \times 10}{1.08} = \frac{3.6}{1.08} = \frac{10}{3} \text{ m/s}^2$.
The tension $T$ is given by: $T = m_1(g + a) = 0.36 \times (10 + \frac{10}{3}) = 0.36 \times \frac{40}{3} = 0.12 \times 40 = 4.8 \text{ N}$.
The displacement $s$ in $t = 1 \text{ s}$ starting from rest is: $s = \frac{1}{2} a t^2 = \frac{1}{2} \times \frac{10}{3} \times (1)^2 = \frac{5}{3} \text{ m}$.
The work done by the string on the block of mass $m_1$ is: $W = T \times s = 4.8 \times \frac{5}{3} = 1.6 \times 5 = 8 \text{ J}$.

(A) Let $u$ be the initial velocity of the $1 \, kg$ mass and $v$ be the final velocity of the $5 \, kg$ mass.
By conservation of linear momentum: $1 \cdot u + 5 \cdot 0 = 1 \cdot (-2) + 5 \cdot v \implies u = 5v - 2$ ... $(i)$
By Newton's experimental law of collision (for elastic collision, coefficient of restitution $e = 1$): $v - (-2) = 1 \cdot (u - 0) \implies u = v + 2$ ... $(ii)$
Solving $(i)$ and $(ii)$: $5v - 2 = v + 2 \implies 4v = 4 \implies v = 1 \, m/s$. Then $u = 3 \, m/s$.
$(A)$ Total momentum $P = 1 \cdot u = 1 \cdot 3 = 3 \, kg \cdot m/s$. (Correct)
$(B)$ Momentum of $5 \, kg$ mass $= 5 \cdot v = 5 \cdot 1 = 5 \, kg \cdot m/s$. (Incorrect)
$(C)$ Velocity of centre of mass $v_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{1 \cdot 3 + 5 \cdot 0}{1 + 5} = 0.5 \, m/s$. Kinetic energy of $CM$ $= \frac{1}{2} (m_1 + m_2) v_{cm}^2 = \frac{1}{2} \cdot 6 \cdot (0.5)^2 = 3 \cdot 0.25 = 0.75 \, J$. (Correct)
$(D)$ Total kinetic energy $= \frac{1}{2} \cdot 1 \cdot (-2)^2 + \frac{1}{2} \cdot 5 \cdot (1)^2 = 2 + 2.5 = 4.5 \, J$. (Incorrect)
Thus, statements $(A)$ and $(C)$ are correct.

(D) Let the mass of the bullet be $m = 0.01 \ kg$ and the mass of the ball be $M = 0.2 \ kg$. The ball is initially at rest at a height $h = 5 \ m$.
Since the motion is horizontal projectile motion,the time $t$ taken to reach the ground is given by $h = \frac{1}{2} g t^2$.
Taking $g = 10 \ m/s^2$,we have $5 = \frac{1}{2} \times 10 \times t^2$,which gives $t = 1 \ s$.
After the collision,the horizontal velocity of the ball is $v_b = \frac{\text{distance}}{\text{time}} = \frac{20 \ m}{1 \ s} = 20 \ m/s$.
The horizontal velocity of the bullet is $v_u = \frac{100 \ m}{1 \ s} = 100 \ m/s$.
By the principle of conservation of linear momentum in the horizontal direction:
$m V = m v_u + M v_b$
$0.01 \times V = (0.01 \times 100) + (0.2 \times 20)$
$0.01 \times V = 1 + 4 = 5$
$V = \frac{5}{0.01} = 500 \ m/s$.

(C) Let $m_1 = 1.0 \ kg$ and $m_2 = 2.0 \ kg$. Initial velocity of $m_1$ is $u_1 = 2.0 \ m \ s^{-1}$ and $m_2$ is $u_2 = 0$. Let $v_1$ and $v_2$ be their velocities after the elastic collision.
Using conservation of linear momentum: $m_1 u_1 = m_1 v_1 + m_2 v_2 \implies 1(2) = 1(v_1) + 2(v_2) \implies v_1 + 2v_2 = 2$ . . . . . $(1)$
Using the coefficient of restitution for an elastic collision $(e=1)$: $v_2 - v_1 = e(u_1 - u_2) = 1(2 - 0) = 2$ . . . . . $(2)$
Solving equations $(1)$ and $(2)$,we get $v_2 = 4/3 \ m \ s^{-1}$ and $v_1 = -2/3 \ m \ s^{-1}$.
The $2.0 \ kg$ block attached to the spring undergoes simple harmonic motion. The time period is $T = 2\pi \sqrt{m_2/k} = 2\pi \sqrt{2/2} = 2\pi \ s$.
The spring returns to its unstretched position after half the time period,i.e.,$\Delta t = T/2 = \pi \ s$.
During this time,the $1.0 \ kg$ block moves with constant velocity $v_1 = -2/3 \ m \ s^{-1}$ (away from the spring). Its displacement is $\Delta x_1 = v_1 \Delta t = (-2/3) \times \pi = -2\pi/3 \ m$.
The $2.0 \ kg$ block moves from $x=0$ to $x_{max}$ and back to $x=0$. Its displacement is $\Delta x_2 = 0$.
The distance between the blocks is $|\Delta x_1 - \Delta x_2| = |-2\pi/3 - 0| = 2\pi/3 \approx 2(3.14)/3 = 6.28/3 \approx 2.09 \ m$.

(B) Given $\frac{dK}{dt} = \gamma t$. Since $K = \frac{1}{2}mv^2$,we have $\frac{dK}{dt} = mv \frac{dv}{dt} = \gamma t$.
Thus,$v \frac{dv}{dt} = \frac{\gamma t}{m}$.
Integrating both sides,$\int v dv = \int \frac{\gamma}{m} t dt$,which gives $\frac{v^2}{2} = \frac{\gamma t^2}{2m}$.
Therefore,$v = \sqrt{\frac{\gamma}{m}} t$. This shows that speed is proportional to time,so $(B)$ is true.
The acceleration $a = \frac{dv}{dt} = \sqrt{\frac{\gamma}{m}}$,which is a constant. Since $F = ma$,the force $F = \sqrt{\gamma m}$ is also constant,so $(A)$ is true.
Since $v = \frac{dx}{dt} = \sqrt{\frac{\gamma}{m}} t$,integrating gives $x = \int \sqrt{\frac{\gamma}{m}} t dt = \frac{1}{2} \sqrt{\frac{\gamma}{m}} t^2$. The distance increases quadratically with time,not linearly,so $(C)$ is false.
Since the force is constant,it is a conservative force,so $(D)$ is true.
Thus,statements $(A), (B),$ and $(D)$ are true.

(A) For path $P$: $\vec{r} = \alpha t \hat{i} + \beta t \hat{j}$. Velocity $\vec{v} = \alpha \hat{i} + \beta \hat{j}$ (constant), so acceleration $\vec{a} = 0$ and force $\vec{F} = 0$. Linear momentum $\vec{p} = m\vec{v}$ is constant. Angular momentum $\vec{L} = \vec{r} \times \vec{p} = m(\alpha t \hat{i} + \beta t \hat{j}) \times (\alpha \hat{i} + \beta \hat{j}) = m(\alpha\beta t - \beta\alpha t)\hat{k} = 0$ (constant). Since $\vec{F}=0$, $K$, $U$, and $E$ are constant. Thus, $P \rightarrow 1, 2, 3, 4, 5$.
For path $Q$: $\vec{r} = \alpha \cos \omega t \hat{i} + \beta \sin \omega t \hat{j}$. This is an elliptical path. The force is central, so angular momentum $\vec{L}$ is conserved. Since the force is conservative, total energy $E$ is conserved. Thus, $Q \rightarrow 2, 5$.
For path $R$: $\vec{r} = \alpha(\cos \omega t \hat{i} + \sin \omega t \hat{j})$. This is a circular path with constant speed $v = \alpha\omega$. Thus, $K$ is constant. Since the force is central, $\vec{L}$ is conserved. For a circular path in a central potential, $U$ and $E$ are also constant. Thus, $R \rightarrow 2, 3, 4, 5$.
For path $S$: $\vec{r} = \alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}$. Velocity $\vec{v} = \alpha \hat{i} + \beta t \hat{j}$ (time-dependent), so $\vec{p}$ is not constant. Acceleration $\vec{a} = \beta \hat{j}$ (constant force). $\vec{L} = m(\alpha t \hat{i} + \frac{\beta}{2} t^2 \hat{j}) \times (\alpha \hat{i} + \beta t \hat{j}) = m(\alpha\beta t^2 - \frac{\alpha\beta}{2} t^2)\hat{k} = \frac{m\alpha\beta t^2}{2}\hat{k}$ (time-dependent). Only total energy $E$ is conserved. Thus, $S \rightarrow 5$.

(B) By the principle of conservation of mechanical energy between the bottom point and point $y$:
$\frac{1}{2} mv_0^2 = mgh + \frac{1}{2} mv_1^2$
$\therefore v_1^2 = v_0^2 - 2gh \quad \dots (i)$
At point $y$,the semicircular path lies on the inclined plane. The component of gravity acting towards the center of the circular path is $mg \sin 30^{\circ}$. This component provides the necessary centripetal force:
$mg \sin 30^{\circ} = \frac{mv_1^2}{R}$
$\frac{1}{2} mg = \frac{mv_1^2}{R} \implies v_1^2 = \frac{gR}{2}$
Substituting this into equation $(i)$:
$v_0^2 - 2gh = \frac{gR}{2}$. Thus,statement $(A)$ is correct.
At points $x$ and $z$,the skater is at a lower height than at point $y$. By energy conservation,the velocity $v$ at $x$ and $z$ is greater than $v_1$ at $y$. Since the centripetal force required is $F_c = \frac{mv^2}{R}$,and $v > v_1$,the required centripetal force at $x$ and $z$ is greater than that at $y$. Thus,statement $(D)$ is correct.

(A) Let the first particle be $P_1$ and the second be $P_2$. At the highest point,the velocity of $P_1$ is $v_{x1} = u_0 \cos \alpha$ and $v_{y1} = 0$.
The second particle $P_2$ is thrown vertically upward with speed $u_0$. At the highest point of $P_1$ (height $H = \frac{u_0^2 \sin^2 \alpha}{2g}$),the velocity of $P_2$ is $v_{y2} = \sqrt{u_0^2 - 2gH} = \sqrt{u_0^2 - u_0^2 \sin^2 \alpha} = u_0 \cos \alpha$.
Applying conservation of linear momentum in the horizontal direction:
$m(u_0 \cos \alpha) + m(0) = (2m)v_x \implies v_x = \frac{u_0 \cos \alpha}{2}$.
Applying conservation of linear momentum in the vertical direction:
$m(0) + m(u_0 \cos \alpha) = (2m)v_y \implies v_y = \frac{u_0 \cos \alpha}{2}$.
The angle $\theta$ with the horizontal is given by $\tan \theta = \frac{v_y}{v_x} = \frac{u_0 \cos \alpha / 2}{u_0 \cos \alpha / 2} = 1$.
Therefore,$\theta = 45^{\circ} = \frac{\pi}{4}$.

(B) For a bob to complete a vertical circle,the minimum velocity at the highest point is $v_{top} = \sqrt{g \ell}$.
Let the first bob have mass $m$ and string length $l_1$. Its velocity at the highest point before collision is $u_1 = \sqrt{g l_1}$.
The second bob has mass $m$ and string length $l_2$,and is initially at rest $(u_2 = 0)$.
Since the collision is elastic and the masses are equal,the two bobs exchange their velocities.
Therefore,the velocity of the second bob after collision is $v_2 = u_1 = \sqrt{g l_1}$.
For the second bob to complete a vertical circle,its minimum velocity at the lowest point must be $\sqrt{5 g l_2}$.
However,the collision occurs at the highest point of the second bob's path (which is the lowest point of its potential circular motion). Thus,the velocity acquired by the second bob after collision must be equal to the minimum velocity required at the lowest point to complete the circle.
So,$v_2 = \sqrt{5 g l_2}$.
Equating the two expressions for $v_2$: $\sqrt{g l_1} = \sqrt{5 g l_2}$.
Squaring both sides: $g l_1 = 5 g l_2$.
Therefore,$\frac{l_1}{l_2} = 5$.

(A) $1.$ Let the block be released from height $h = R = 40 \ m$. The vertical height of point $Q$ from the bottom is $h_Q = R - R \sin(30^{\circ}) = R - R/2 = R/2 = 20 \ m$.
Applying the work-energy theorem between the starting point and point $Q$:
$W_{gravity} + W_{friction} = \Delta K$
$Mg(R - h_Q) - 150 = \frac{1}{2} M v^2$
$1 \times 10 \times (40 - 20) - 150 = \frac{1}{2} \times 1 \times v^2$
$200 - 150 = 0.5 v^2 \Rightarrow 50 = 0.5 v^2 \Rightarrow v^2 = 100 \Rightarrow v = 10 \ m s^{-1}$.
Thus,the speed at $Q$ is $10 \ m s^{-1}$. (Option $B$)
$2.$ At point $Q$,the forces acting on the block are gravity $(Mg)$ and normal reaction $(N)$. The component of gravity perpendicular to the track is $Mg \cos(60^{\circ}) = Mg/2$.
The net centripetal force is $N - Mg \cos(60^{\circ}) = \frac{M v^2}{R}$.
$N - 1 \times 10 \times 0.5 = \frac{1 \times 100}{40}$
$N - 5 = 2.5 \Rightarrow N = 7.5 \ N$. (Option $A$)
Therefore,the correct pair is $(B, A)$.

(B) Before the collision,the ball falls under gravity. Its velocity $v$ at time $t$ is given by $v = gt$. The kinetic energy $K$ is given by $K = \frac{1}{2} mv^2 = \frac{1}{2} m(gt)^2 = \frac{1}{2} mg^2 t^2$. Since $K \propto t^2$,the graph of $K$ versus $t$ is parabolic,opening upwards.
During the collision,the ball is compressed. As the compression increases,the kinetic energy is converted into elastic potential energy. At the maximum compression,the velocity of the ball becomes zero,so the kinetic energy becomes zero. After reaching maximum compression,the ball starts to expand,and the elastic potential energy is converted back into kinetic energy until the ball leaves the surface.
Thus,the kinetic energy decreases to zero during the compression phase and increases back to its original value during the expansion phase. This behavior is represented by a parabolic increase followed by a sharp dip to zero and a subsequent parabolic increase. Among the given options,graph $B$ represents this variation most appropriately.

(A) The force is given by $F = -\frac{dU}{dx}$. Equilibrium points occur where $F = 0$, i.e., $\frac{dU}{dx} = 0$.
For $(A): U_1(x) = \frac{U_0}{2} \left[1 - (x/a)^2\right]^2$. 
$\frac{dU_1}{dx} = \frac{U_0}{2} \cdot 2 \left[1 - (x/a)^2\right] \cdot (-2x/a^2) = -\frac{2U_0 x}{a^2} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=0, a, -a$. For $|x| < a$, $F$ is attractive towards $x=0$. 
Thus $(A) \rightarrow (P, Q, R, S)$.
For $(B): U_2(x) = \frac{U_0}{2} (x/a)^2$. 
$\frac{dU_2}{dx} = \frac{U_0 x}{a^2}$. 
$F=0$ at $x=0$. For energy $U_0/4$, $U_2(x) = U_0/4 \Rightarrow (x/a)^2 = 1/2 \Rightarrow x = \pm a/\sqrt{2}$. 
It oscillates about $x=0$. 
Thus $(B) \rightarrow (Q, T)$.
For $(C): U_3(x) = \frac{U_0}{2} (x/a)^2 e^{-(x/a)^2}$. 
$\frac{dU_3}{dx} = \frac{U_0}{2} \left[\frac{2x}{a^2} e^{-(x/a)^2} + (x/a)^2 e^{-(x/a)^2} (-2x/a^2)\right] = \frac{U_0 x}{a^2} e^{-(x/a)^2} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=0, a, -a$. For energy $U_0/4$, it can oscillate about $x=a$ or $x=-a$ if $U_{min} < U_0/4 < U_{max}$. 
Thus $(C) \rightarrow (P, Q, R, T)$.
For $(D): U_4(x) = \frac{U_0}{2} \left[x/a - 1/3(x/a)^3\right]$. 
$\frac{dU_4}{dx} = \frac{U_0}{2a} \left[1 - (x/a)^2\right]$. 
$F=0$ at $x=a, -a$. For $|x| < a$, $F$ is negative for $x>0$ and positive for $x<0$, so it is attractive towards $x=0$. 
Thus $(D) \rightarrow (P, R, S)$.

(D) Let the top of the hoop be the reference level for gravitational potential energy $(U_g = 0)$.
At the top,the spring is stretched by $x_i = 2R - R = R$. The initial energy is $E_i = U_{g,i} + U_{s,i} + K_i = 0 + \frac{1}{2}kR^2 + 0 = \frac{1}{2}kR^2$.
When the spring length is $R$,the bead is at an angle $\theta = 60^\circ$ from the vertical,as the distance from the bottom is $R$ and the radius is $R$,forming an equilateral triangle.
The height of the bead from the top is $h = R + R \cos 60^\circ = R + \frac{R}{2} = \frac{3R}{2}$.
The gravitational potential energy at this point is $U_{g,f} = -mg(\frac{3R}{2})$.
The spring is at its natural length,so $U_{s,f} = 0$.
The final kinetic energy is $K_f = \frac{1}{2}mv^2$.
By the conservation of mechanical energy: $E_i = E_f \implies \frac{1}{2}kR^2 = -\frac{3mgR}{2} + \frac{1}{2}mv^2$.
Multiplying by $2/m$: $\frac{kR^2}{m} = -3gR + v^2$.
Thus,$v = \sqrt{3gR + \frac{kR^2}{m}}$.

(A) $1$. Velocity of bob $A$ just before hitting bob $B$:
Using the principle of conservation of mechanical energy, the change in height $h$ is $R - R \cos(60^{\circ}) = R - R/2 = R/2$.
$v = \sqrt{2gh} = \sqrt{2g(R/2)} = \sqrt{gR}$.
$2$. Let $u = \sqrt{gR}$ be the velocity of $A$ just before the collision.
Let $v_1$ and $v_2$ be the velocities of $A$ and $B$ respectively just after the elastic collision.
$3$. By the Law of Conservation of Momentum $(COM)$:
$mu = mv_1 + (m/2)v_2$
$u = v_1 + v_2/2$
$2u = 2v_1 + v_2$ --- $(i)$
$4$. For an elastic collision, the coefficient of restitution $e = 1$:
$e = (v_2 - v_1) / u = 1$
$v_2 - v_1 = u$ --- (ii)
$5$. Solving equations $(i)$ and (ii):
From (ii), $v_2 = u + v_1$.
Substitute into $(i)$: $2u = 2v_1 + (u + v_1)$
$2u = 3v_1 + u$
$u = 3v_1$
$v_1 = u/3 = \frac{1}{3} \sqrt{gR}$.

(C) By the law of conservation of linear momentum,the velocity of the center of mass $(v_{cm})$ when the blocks move together is given by:
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{cm}$
$10 \times 3 + 5 \times 0 = (10 + 5) v_{cm}$
$30 = 15 v_{cm} \Rightarrow v_{cm} = 2 \ m/s$
The energy stored in the spring is equal to the loss in kinetic energy during the collision:
$\frac{1}{2} kx^2 = K_i - K_f$
$\frac{1}{2} kx^2 = \frac{1}{2} m_1 v_1^2 - \frac{1}{2} (m_1 + m_2) v_{cm}^2$
$\frac{1}{2} (3000) x^2 = \frac{1}{2} (10) (3)^2 - \frac{1}{2} (15) (2)^2$
$1500 x^2 = 45 - 30$
$1500 x^2 = 15$
$x^2 = \frac{15}{1500} = \frac{1}{100}$
$x = 0.1 \ m$

(B) Given:
Mass $m = 5 \ kg$
Acceleration $a = 2 \ m/s^2$
Time $t = 3 \ s$
Initial velocity $u = 0$
Acceleration due to gravity $g = 10 \ m/s^2$
Step $1$: Calculate the tension $T$ in the thread.
Since the elevator is accelerating upward,the tension $T$ is given by:
$T = m(g + a)$
$T = 5(10 + 2) = 5(12) = 60 \ N$
Step $2$: Calculate the displacement $S$ of the block in $3 \ s$.
Using the equation of motion $S = ut + \frac{1}{2}at^2$:
$S = 0(3) + \frac{1}{2}(2)(3)^2$
$S = 0 + 1(9) = 9 \ m$
Step $3$: Calculate the work done by tension.
Work $W = T \cdot S \cdot \cos(\theta)$
Since the tension acts in the direction of displacement,$\theta = 0^{\circ}$,so $\cos(0^{\circ}) = 1$.
$W = (60 \ N)(9 \ m)(1) = 540 \ J$

(B) Given: Mass $m = 4 \ kg$,Applied force $F = 18 \ N$,Coefficient of kinetic friction $\mu_k = 0.2$,Time $t = 10 \ s$,Initial velocity $u = 0 \ m/s$.
$1$. Calculate the kinetic frictional force $f$:
$f = \mu_k \cdot N = \mu_k \cdot mg = 0.2 \times 4 \times 10 = 8 \ N$.
$2$. Calculate the net acceleration $a$ of the body:
$F_{net} = F - f = 18 - 8 = 10 \ N$.
$a = \frac{F_{net}}{m} = \frac{10}{4} = 2.5 \ m/s^2$.
$3$. Calculate the displacement $S$ in $10 \ s$:
$S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2.5 \times (10)^2 = 0.5 \times 2.5 \times 100 = 125 \ m$.
$4$. Calculate the work done by the applied force $W$:
$W = F \cdot S \cdot \cos(0^\circ) = 18 \times 125 \times 1 = 2250 \ J$.

(A) Given the velocity law $v = k\sqrt{S}$.
We know that acceleration $a = v \frac{dv}{dS}$.
Differentiating $v$ with respect to $S$: $\frac{dv}{dS} = k \cdot \frac{1}{2\sqrt{S}} = \frac{k^2}{2v}$.
Thus,$a = v \cdot \frac{k^2}{2v} = \frac{k^2}{2}$.
Since the acceleration is constant,we can use the equation of motion $v = u + at$. Starting from rest $(u=0)$,$v = \frac{k^2 t}{2}$.
According to the Work-Energy Theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta K.E. = \frac{1}{2}mv^2 - 0$.
Substituting the value of $v$: $W = \frac{1}{2}m \left(\frac{k^2 t}{2}\right)^2 = \frac{1}{2}m \left(\frac{k^4 t^2}{4}\right) = \frac{1}{8}mk^4 t^2$.

(B) In circular motion,the centripetal force acts towards the center of the circle,which is always perpendicular to the instantaneous velocity (displacement vector) of the body. Since the work done $W = \vec{F} \cdot \vec{d} = F d cos( \theta)$ and $ \theta = 90^{\circ}$,the work done by the centripetal force is $W_c = 0$.
The tangential force acts along the tangent to the circular path,which is in the direction of the instantaneous velocity (or displacement). Therefore,the angle between the tangential force and the displacement is $0^{\circ}$. Since $cos(0^{\circ}) = 1$,the work done by the tangential force is $W_t = F_t \cdot d $.

(C) Statement $A$ is incorrect because,by definition,an inelastic collision involves a loss of kinetic energy.
Statement $B$ is correct because,according to the law of conservation of linear momentum,the total linear momentum of a system remains constant if the net external force acting on the system is zero.
Therefore,$A$ is wrong and $B$ is correct.

(D) The momentum of the particle moving towards the east is $\vec{p_1} = mv \hat{i}$.
The momentum of the particle moving towards the north is $\vec{p_2} = mv \hat{j}$.
Let the velocity of the combined mass $2m$ after collision be $\vec{V'} = v_x \hat{i} + v_y \hat{j}$.
According to the law of conservation of linear momentum,$\vec{p_1} + \vec{p_2} = \vec{p_{final}}$.
$mv \hat{i} + mv \hat{j} = (2m) \vec{V'}$.
Dividing by $2m$,we get $\vec{V'} = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}$.
The magnitude of the resultant velocity is $V' = \sqrt{(\frac{v}{2})^2 + (\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.

(B) The velocity of the ball just before hitting the ground is $v_1 = \sqrt{2gh_1}$.
The velocity of the ball just after bouncing is $v_2 = \sqrt{2gh_2}$.
The loss in velocity is $\Delta v = v_1 - v_2$.
The factor by which the velocity is lost is given by $\frac{\Delta v}{v_1} = \frac{v_1 - v_2}{v_1} = 1 - \frac{v_2}{v_1}$.
Substituting the expressions for $v_1$ and $v_2$,we get $\frac{\Delta v}{v_1} = 1 - \sqrt{\frac{h_2}{h_1}}$.
Given $h_1 = 5 \ m$ and $h_2 = 1.8 \ m$,we have $\frac{\Delta v}{v_1} = 1 - \sqrt{\frac{1.8}{5}} = 1 - \sqrt{0.36} = 1 - 0.6 = 0.4$.
Converting $0.4$ to a fraction,we get $\frac{4}{10} = \frac{2}{5}$.

(B) The total vertical distance fallen by the ball is $(s+h)$.
Since the ball starts from rest,the loss in gravitational potential energy is $mg(s+h)$.
This energy is stored as elastic potential energy in the spring when it is compressed by a distance $h$.
The elastic potential energy stored in the spring is given by $\frac{1}{2} kh^2$.
According to the law of conservation of energy,the loss in potential energy of the ball equals the gain in potential energy of the spring:
$mg(s+h) = \frac{1}{2} kh^2$
Solving for the spring constant $k$:
$k = \frac{2mg(s+h)}{h^2}$

(D) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation:
$K = \frac{p^2}{2m}$
Rearranging this for momentum,we get:
$p = \sqrt{2mK}$
Since the kinetic energy $K$ is the same for both bodies,the momentum $p$ is directly proportional to the square root of the mass:
$p \propto \sqrt{m}$
Because the heavy body has a larger mass $m$ compared to the light body,it will have a greater momentum $p$.

(B) The work done $W$ by a constant force $\vec{F}$ is given by the dot product of force and displacement: $W = \vec{F} \cdot \vec{r}$.
Substituting the given values: $W = (3\hat{i} - 2\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} - 3\hat{k})$.
$W = (3 \times 2) + (-2 \times -3) + (-1 \times -3) = 6 + 6 + 3 = 15 \text{ J}$.
Power $P$ is defined as the rate of work done: $P = \frac{W}{t}$.
Given $t = 2 \text{ s}$,$P = \frac{15}{2} = 7.5 \text{ W}$.
Thus,the work done is $15 \text{ J}$ and the power is $7.5 \text{ W}$.

(C) The impulse applied to both bodies is given by $J = F \cdot t$. Since both bodies start from rest,the impulse is equal to the change in momentum for each body.
Thus,$p_{1} = p_{2} = F \cdot t$.
The kinetic energy $E$ of a body with mass $m$ and momentum $p$ is given by $E = \frac{p^{2}}{2m}$.
Therefore,$E_{1} = \frac{p_{1}^{2}}{2m_{1}}$ and $E_{2} = \frac{p_{2}^{2}}{2m_{2}}$.
Taking the ratio,we get $\frac{E_{1}}{E_{2}} = \frac{p_{1}^{2} / 2m_{1}}{p_{2}^{2} / 2m_{2}}$.
Since $p_{1} = p_{2}$,the ratio simplifies to $\frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}}$.

(C) As the block $A$ moves with velocity $u = 0.15 \,ms^{-1}$, it compresses the spring, which pushes block $B$ towards the right. The spring continues to compress until the velocity of both blocks becomes equal. Let this common velocity be $v$.
According to the law of conservation of linear momentum:
$m_A u = (m_A + m_B) v$
$v = \frac{m_A u}{m_A + m_B} = \frac{2 \times 0.15}{2 + 3} = \frac{0.3}{5} = 0.06 \,ms^{-1}$
According to the law of conservation of energy, the initial kinetic energy of block $A$ is converted into the final kinetic energy of the system and the potential energy stored in the spring at maximum compression $x$:
$\frac{1}{2} m_A u^2 = \frac{1}{2} (m_A + m_B) v^2 + \frac{1}{2} k x^2$
$\frac{1}{2} \times 2 \times (0.15)^2 = \frac{1}{2} \times (2 + 3) \times (0.06)^2 + \frac{1}{2} \times 10.8 \times x^2$
$0.0225 = 5 \times 0.0018 + 5.4 x^2$
$0.0225 = 0.009 + 5.4 x^2$
$5.4 x^2 = 0.0135$
$x^2 = \frac{0.0135}{5.4} = 0.0025$
$x = \sqrt{0.0025} = 0.05 \,m$

(B) Given: Mass of the ball $m = 0.2 \,kg$, height $h = 10 \,m$, acceleration due to gravity $g = 10 \,ms^{-2}$.
Total energy just before collision at the floor is $E_i = \frac{1}{2}mu^2 + mgh$.
After collision, the ball loses $50 \%$ of its energy, so the remaining energy is $E_f = \frac{E_i}{2}$.
The ball rises back to the same height $h$, meaning its potential energy at the peak is $mgh$.
Equating the remaining energy to the potential energy at the peak: $\frac{1}{2}(\frac{1}{2}mu^2 + mgh) = mgh$.
$\frac{1}{2}mu^2 + mgh = 2mgh$.
$\frac{1}{2}mu^2 = mgh$.
$u^2 = 2gh$.
$u = \sqrt{2 \times 10 \times 10} = \sqrt{200} \approx 14.14 \,ms^{-1}$.
Rounding to the nearest provided option, the initial velocity is $14 \,ms^{-1}$.

(C) Given: Initial velocity $u = 7 \text{ m/s}$, coefficient of restitution $e = 0.75$, acceleration due to gravity $g = 10 \text{ m/s}^2$.
Initial height $H = \frac{u^2}{2g} = \frac{7^2}{2 \times 10} = \frac{49}{20} = 2.45 \text{ m}$.
The ball bounces back with velocity $v_1 = eu = 0.75 \times 7 = 5.25 \text{ m/s}$.
The height reached after the first bounce is $H_1 = \frac{v_1^2}{2g} = e^2 H = (0.75)^2 \times 2.45 = 0.5625 \times 2.45 = 1.378125 \text{ m}$.
The total distance $S$ traveled by the ball is $S = H + 2H_1 + 2H_2 + \dots = H + 2(e^2H + e^4H + \dots) = H(1 + 2e^2(1 + e^2 + e^4 + \dots))$.
Using the sum of an infinite geometric series $S_{\infty} = \frac{a}{1-r}$, we get $S = H(1 + 2e^2(\frac{1}{1-e^2})) = H(\frac{1-e^2+2e^2}{1-e^2}) = H(\frac{1+e^2}{1-e^2})$.
Substituting the values: $S = 2.45 \times \frac{1 + (0.75)^2}{1 - (0.75)^2} = 2.45 \times \frac{1 + 0.5625}{1 - 0.5625} = 2.45 \times \frac{1.5625}{0.4375} = 2.45 \times 3.5714 = 8.75 \text{ m}$.

(B) Given: Mass of bullet $m_b = 0.03 \,kg$, initial velocity $v_b = 700 \,ms^{-1}$. Mass of block $m_B = 4 \,kg$, height $h = 0.2 \,m$.
Initial momentum of the system $= m_b v_b = 0.03 \times 700 = 21 \,kg \cdot ms^{-1}$.
Let $v_1$ be the velocity of the bullet and $v_2$ be the velocity of the block after the collision.
By conservation of linear momentum: $21 = 0.03 v_1 + 4 v_2$ ... $(i)$
For the block, by conservation of energy: $\frac{1}{2} m_B v_2^2 = m_B g h$.
$v_2 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.2} = \sqrt{3.92} \approx 1.98 \,ms^{-1}$.
Substituting $v_2$ into $(i)$: $21 = 0.03 v_1 + 4(1.98)$.
$21 = 0.03 v_1 + 7.92$.
$0.03 v_1 = 13.08$.
$v_1 = \frac{13.08}{0.03} = 436 \,ms^{-1}$.
Rounding to the nearest provided option, the velocity is approximately $433 \,ms^{-1}$.

(D) In an elastic collision,the total kinetic energy is conserved only before and after the collision. During the short interval of collision,the balls undergo deformation,and a part of the kinetic energy is converted into elastic potential energy. Thus,the kinetic energy is not conserved during the contact time.
Therefore,the Assertion $(A)$ is false.
The law of conservation of energy is a universal law that applies to all physical processes,including work done against friction (where energy is converted into heat). Therefore,the Reason $(R)$ is also false.
Since both statements are incorrect,the correct option is $(D)$.

(B) Given:
Mass of bullet,$m = 0.01 \,kg$
Initial speed of bullet,$u = 500 \,ms^{-1}$
Mass of block,$M = 2 \,kg$
Vertical rise of block,$h = 0.1 \,m$
Let $v_b$ be the speed of the bullet after emerging from the block and $V$ be the velocity of the block immediately after the collision.
Applying the law of conservation of energy to the block after the collision:
$\frac{1}{2} M V^2 = Mgh$
$V = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4 \,ms^{-1}$
Applying the law of conservation of linear momentum during the collision:
$m u = m v_b + M V$
$0.01 \times 500 = (0.01 \times v_b) + (2 \times 1.4)$
$5 = 0.01 v_b + 2.8$
$0.01 v_b = 5 - 2.8 = 2.2$
$v_b = \frac{2.2}{0.01} = 220 \,ms^{-1}$
Thus,the speed of the bullet after it emerges from the block is $220 \,ms^{-1}$.

(B) Let the masses be $m_P = 3m$,$m_R = 2m$,and $m_Q = 3m$. Particle $R$ moves with velocity $v$ towards $Q$.
$1$. First collision between $R$ and $Q$:
Using the elastic collision formula for velocities after collision: $v_1' = \frac{(m_1-m_2)v_1 + 2m_2v_2}{m_1+m_2}$ and $v_2' = \frac{(m_2-m_1)v_2 + 2m_1v_1}{m_1+m_2}$.
For $R$ and $Q$ $(m_R=2m, m_Q=3m)$: $v_R' = \frac{(2m-3m)v + 0}{2m+3m} = -v/5$ and $v_Q' = \frac{(3m-2m)0 + 2(2m)v}{2m+3m} = 4v/5$.
$2$. Now $R$ moves with velocity $-v/5$ towards $P$ $(m_P=3m)$:
For $R$ and $P$ $(m_R=2m, m_P=3m)$: $v_R'' = \frac{(2m-3m)(-v/5) + 0}{2m+3m} = v/25$ and $v_P'' = \frac{(3m-2m)0 + 2(2m)(-v/5)}{2m+3m} = -4v/25$.
Since $v_R''$ is positive and $v_Q'$ is positive $(4v/5 > v/25)$,$R$ will never catch $Q$ again. Also,$P$ moves in the negative direction,so no further collisions occur.
Total collisions = $3$.

(C) Let the mass of both bodies be $m$. Initially,body $A$ has momentum $P$,so its velocity is $v_A = P/m$. Body $B$ is stationary,so $v_B = 0$.
During the impact,body $A$ exerts an impulse $J$ on body $B$. By Newton's third law,body $B$ exerts an equal and opposite impulse $J$ on body $A$.
Statement $(b)$ is correct because the impulse exerted by $B$ on $A$ is equal in magnitude to the impulse exerted by $A$ on $B$.
After the impact,the momentum of $B$ is $P_B = J$,so its velocity is $v_B' = J/m$.
The momentum of $A$ becomes $P_A' = P - J$,so its velocity is $v_A' = (P - J)/m$.
The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_B' - v_A'}{v_A - v_B} = \frac{\frac{J}{m} - \frac{P-J}{m}}{\frac{P}{m} - 0} = \frac{2J - P}{P} = \frac{2J}{P} - 1$.
Thus,statement $(c)$ is also correct.
Therefore,statements $(b)$ and $(c)$ are correct.

(C) When a ball is dropped from height $h$, the time taken to reach the ground is $t_0 = \sqrt{\frac{2h}{g}}$ and the speed just before impact is $v_0 = \sqrt{2gh}$.
After the first collision, its speed becomes $v_1 = ev_0 = e\sqrt{2gh}$, where $e$ is the coefficient of restitution.
The ball rises and stops at time $t_1 = \frac{v_1}{g}$. It then falls back to the ground, taking the same time $t_1$. Thus, the time between the first and second collision is $2t_1 = \frac{2v_1}{g}$.
The total time $T$ before it ceases to rebound is:
$T = t_0 + 2t_1 + 2t_2 + \dots = t_0 + \frac{2v_1}{g} + \frac{2v_2}{g} + \dots = \sqrt{\frac{2h}{g}} + \frac{2e\sqrt{2gh}}{g} + \frac{2e^2\sqrt{2gh}}{g} + \dots$
$T = \sqrt{\frac{2h}{g}} [1 + 2e(1 + e + e^2 + \dots)] = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} \right)$.
Given $h = 180 \,m$, $g = 10 \,ms^{-2}$, and $e = 0.5$:
$T = \sqrt{\frac{2 \times 180}{10}} \left( \frac{1+0.5}{1-0.5} \right) = 6 \times 3 = 18 \,s$.
The total distance $H$ covered is:
$H = h + 2h_1 + 2h_2 + \dots = h + 2(e^2h) + 2(e^4h) + \dots = h \left( \frac{1+e^2}{1-e^2} \right)$.
$H = 180 \left( \frac{1 + 0.25}{1 - 0.25} \right) = 180 \times \frac{1.25}{0.75} = 180 \times \frac{5}{3} = 300 \,m$.
Average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{300}{18} = \frac{50}{3} \,ms^{-1}$.
Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{180}{18} = 10 \,ms^{-1}$.

(C) Let the mass of the bullet be $m = 10 \ g = 0.01 \ kg$,mass of plate $A$ be $M_A = 500 \ g = 0.5 \ kg$,and mass of plate $B$ be $M_B = 1.49 \ kg$.
Let the initial velocity of the bullet be $u$ and its velocity after passing through plate $A$ be $v_1$.
Let the velocity of plate $A$ after the bullet passes through it be $v_A$ and the velocity of plate $B$ after the bullet gets embedded in it be $v_B$.
According to the problem,$v_A = v_B = v$.
For plate $A$,by conservation of linear momentum:
$m u = m v_1 + M_A v \Rightarrow m(u - v_1) = M_A v \quad ... (1)$
For plate $B$,by conservation of linear momentum:
$m v_1 = (m + M_B) v \Rightarrow v = \frac{m v_1}{m + M_B} \quad ... (2)$
Substituting $v$ from $(2)$ into $(1)$:
$m(u - v_1) = M_A \left( \frac{m v_1}{m + M_B} \right)$
$u - v_1 = v_1 \left( \frac{M_A}{m + M_B} \right)$
$u = v_1 \left( 1 + \frac{M_A}{m + M_B} \right) = v_1 \left( \frac{m + M_B + M_A}{m + M_B} \right)$
$v_1 = u \left( \frac{m + M_B}{m + M_B + M_A} \right) = u \left( \frac{0.01 + 1.49}{0.01 + 1.49 + 0.5} \right) = u \left( \frac{1.5}{2.0} \right) = 0.75 u = \frac{3}{4} u$.
The initial kinetic energy of the bullet is $K_i = \frac{1}{2} m u^2$.
The kinetic energy of the bullet after passing through plate $A$ is $K_f = \frac{1}{2} m v_1^2 = \frac{1}{2} m \left( \frac{3}{4} u \right)^2 = \frac{9}{16} K_i$.
The percentage loss in kinetic energy is:
$\text{Percentage Loss} = \frac{K_i - K_f}{K_i} \times 100 = \left( 1 - \frac{9}{16} \right) \times 100 = \frac{7}{16} \times 100 = 43.75 \%$.

(D) Let $m$ be the mass of the ball, $h = 40 \,m$ be the height, and $g = 10 \,m/s^2$ be the acceleration due to gravity.
Total mechanical energy just before hitting the ground is $E_i = \frac{1}{2}mv^2 + mgh$.
After losing $\frac{1}{3}$ of its energy, the remaining energy is $E_f = \frac{2}{3}E_i$.
This remaining energy is used to rebound to the same height $h$, so the potential energy at the peak of the rebound is $mgh$.
Equating the energy: $\frac{2}{3}(\frac{1}{2}mv^2 + mgh) = mgh$.
Dividing by $m$: $\frac{2}{3}(\frac{1}{2}v^2 + gh) = gh$.
$\frac{1}{3}v^2 + \frac{2}{3}gh = gh$.
$\frac{1}{3}v^2 = gh - \frac{2}{3}gh = \frac{1}{3}gh$.
$v^2 = gh$.
$v = \sqrt{10 \times 40} = \sqrt{400} = 20 \,m/s$.

(D) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_{\text{sep}}}{v_{\text{app}}}$
For the first collision,we are given that the relative velocity of approach is twice the relative velocity of separation:
$v_{\text{app}, 1} = 2 v_{\text{sep}, 1} \implies e_1 = \frac{v_{\text{sep}, 1}}{v_{\text{app}, 1}} = \frac{1}{2}$
We are given the ratio of the coefficients of restitution as $e_1 : e_2 = 3 : 1$,which means:
$e_2 = \frac{e_1}{3} = \frac{1/2}{3} = \frac{1}{6}$
Since $e_2 = \frac{v_{\text{sep}, 2}}{v_{\text{app}, 2}} = \frac{1}{6}$,the ratio of the relative velocity of approach to the relative velocity of separation in the second collision is:
$\frac{v_{\text{app}, 2}}{v_{\text{sep}, 2}} = \frac{1}{e_2} = 6 = 6: 1$

(B) Initial momentum $p_i = 24 \ kg \ m \ s^{-1}$.
Mass $m = 8 \ kg$.
Initial velocity $v_i = \frac{p_i}{m} = \frac{24}{8} = 3 \ m \ s^{-1}$.
Initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 8 \times (3)^2 = 4 \times 9 = 36 \ J$.
Impulse $J = F \times \Delta t = 24 \times 3 = 72 \ N \ s$.
Change in momentum $\Delta p = J = p_f - p_i$.
Final momentum $p_f = p_i + J = 24 + 72 = 96 \ kg \ m \ s^{-1}$.
Final velocity $v_f = \frac{p_f}{m} = \frac{96}{8} = 12 \ m \ s^{-1}$.
Final kinetic energy $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 8 \times (12)^2 = 4 \times 144 = 576 \ J$.
Increase in kinetic energy $\Delta K = K_f - K_i = 576 - 36 = 540 \ J$.

(A) Step $1$: The bullet of mass $m$ strikes $M_1$ and gets embedded in it. This is a perfectly inelastic collision. By the law of conservation of linear momentum:
$m v = (m + M_1) v_1$
where $v_1$ is the velocity of the $(m + M_1)$ system after the first collision.
$v_1 = \frac{m v}{m + M_1}$
Step $2$: The system $(m + M_1)$ now moves with velocity $v_1$ and collides elastically with $M_2$,which is initially at rest $(u_2 = 0)$.
For an elastic collision between two masses $m_A$ and $m_B$ where $m_B$ is initially at rest,the final velocity $v_B$ of mass $m_B$ is given by:
$v_B = \frac{2 m_A}{m_A + m_B} u_A$
Here,$m_A = (m + M_1)$,$m_B = M_2$,and $u_A = v_1$.
Substituting these values:
$v_2 = \frac{2(m + M_1)}{(m + M_1) + M_2} v_1$
$v_2 = \frac{2(m + M_1)}{m + M_1 + M_2} \cdot \frac{m v}{m + M_1}$
$v_2 = \frac{2 m v}{m + M_1 + M_2}$
Thus,the velocity of $M_2$ after the collision is $\frac{2 m v}{M_1 + M_2 + m}$.

(B) Initial mass $M = 8 \ kg$,initial velocity $u = 2 \ m/s$. Initial momentum $P_i = M \cdot u = 8 \times 2 = 16 \ kg \cdot m/s$. Initial kinetic energy $K_i = \frac{1}{2} M u^2 = \frac{1}{2} \times 8 \times 2^2 = 16 \ J$. After the explosion,the body splits into two equal parts of mass $m = 4 \ kg$ each. Let their velocities be $v_1$ and $v_2$. By conservation of momentum: $m v_1 + m v_2 = P_i \implies 4(v_1 + v_2) = 16 \implies v_1 + v_2 = 4$. The final kinetic energy $K_f = K_i + \Delta E = 16 + 16 = 32 \ J$. Also,$K_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} \times 4 \times (v_1^2 + v_2^2) = 2(v_1^2 + v_2^2) = 32 \implies v_1^2 + v_2^2 = 16$. Solving $v_1 + v_2 = 4$ and $v_1^2 + v_2^2 = 16$: $(v_1 + v_2)^2 = v_1^2 + v_2^2 + 2 v_1 v_2 \implies 16 = 16 + 2 v_1 v_2 \implies v_1 v_2 = 0$. This means either $v_1 = 0$ or $v_2 = 0$. If $v_1 = 0$,then $v_2 = 4 \ m/s$. Thus,one part comes to rest and the other moves in the same direction as the original body.

(D) Let the masses be $m_1 = m$ and $m_2 = 4m$. Both are subjected to the same force $F$.
Acceleration of the first object: $a_1 = F/m$.
Acceleration of the second object: $a_2 = F/(4m) = a_1/4$.
Since they start from rest,their velocities after time $t$ are $v = at$.
Kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}m(at)^2 = \frac{1}{2}ma^2t^2$.
For the first object: $K_1 = \frac{1}{2}m(a_1)^2t_1^2$.
For the second object: $K_2 = \frac{1}{2}(4m)(a_2)^2t_2^2 = \frac{1}{2}(4m)(a_1/4)^2t_2^2 = \frac{1}{2}(4m)(a_1^2/16)t_2^2 = \frac{1}{2}m(a_1^2/4)t_2^2$.
Given $K_1 = K_2$,we have $\frac{1}{2}m a_1^2 t_1^2 = \frac{1}{2}m (a_1^2/4) t_2^2$.
Simplifying,$t_1^2 = t_2^2/4$,which implies $t_1^2/t_2^2 = 1/4$.
Therefore,$t_1/t_2 = 1/2$.

(D) For body $A$:
Maximum height $H_{\max} = \frac{u^2}{2g}$.
Given $m_A = 2m$, $u_A = u$.
The height of interest is $h = \frac{H_{\max}}{2} = \frac{u^2}{4g}$.
Using the work-energy theorem or conservation of energy for body $A$ at height $h$:
$(KE)_A = \text{Total Energy} - (PE)_A = \frac{1}{2} m_A u_A^2 - m_A gh = \frac{1}{2}(2m)u^2 - (2m)g\left(\frac{u^2}{4g}\right) = mu^2 - \frac{1}{2}mu^2 = \frac{1}{2}mu^2$.
For body $B$:
Given $m_B = m$, $u_B = 2u$.
The potential energy of body $B$ at height $h = \frac{u^2}{4g}$ is:
$(PE)_B = m_B gh = m g \left(\frac{u^2}{4g}\right) = \frac{1}{4}mu^2$.
The ratio of $(KE)_A$ to $(PE)_B$ is:
$\frac{(KE)_A}{(PE)_B} = \frac{\frac{1}{2}mu^2}{\frac{1}{4}mu^2} = 2:1$.

(A) Let the momentum of both vehicles be $p$. The braking force $F$ is the same for both.
Using the work-energy theorem,the work done by the braking force is equal to the change in kinetic energy:
$|W| = |\Delta K|$
$F \cdot S = \frac{p^2}{2m}$
Thus,the stopping distance $S$ is given by $S = \frac{p^2}{2mF}$.
Since $p$ and $F$ are constant for both vehicles,$S \propto \frac{1}{m}$.
Therefore,the ratio of the distance travelled by the truck $(S_T)$ to the distance travelled by the car $(S_C)$ is:
$\frac{S_T}{S_C} = \frac{m_C}{m_T} = \frac{M/10}{M} = \frac{1}{10}$.
Hence,the ratio is $1: 10$.

(C) According to the law of conservation of mechanical energy,the total mechanical energy at the point of projection is equal to the total mechanical energy at the ground level.
$E_{initial} = E_{final}$
$K_i + U_i = K_f + U_f$
Here,$K_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 25 \times (4)^2 = \frac{1}{2} \times 25 \times 16 = 200 \,J$.
$U_i = mgh = 25 \times 10 \times 2 = 500 \,J$.
At the ground,$U_f = 0$.
So,$200 + 500 = K_f + 0$.
$K_f = 700 \,J$.

(B) Given: Mass of body $A$ $(m_A = 20 \ kg)$,mass of body $B$ $(m_B = 5 \ kg)$,force $(F = 40 \ N)$.
Since both bodies start from rest,their kinetic energies are $K_A = K_B = K$.
Kinetic energy is given by $K = \frac{p^2}{2m}$,where $p$ is momentum.
Since $K_A = K_B$,we have $\frac{p_A^2}{2m_A} = \frac{p_B^2}{2m_B} \implies \frac{p_A}{p_B} = \sqrt{\frac{m_A}{m_B}} = \sqrt{\frac{20}{5}} = \sqrt{4} = 2$.
From the impulse-momentum theorem,$F \cdot t = \Delta p$. Since initial momentum is zero,$p = F \cdot t$.
Thus,$\frac{p_A}{p_B} = \frac{F \cdot t_A}{F \cdot t_B} = \frac{t_A}{t_B}$.
Therefore,$\frac{t_A}{t_B} = 2$,which means $t_A: t_B = 2: 1$.

(B) The radius of the well $r = 1 \ m$ and depth $h = 14 \ m$.
The volume of water $V = \pi r^2 h = \pi \times (1)^2 \times 14 = 14\pi \ m^3$.
The mass of water $m = \text{density} \times \text{volume} = 1000 \ kg/m^3 \times 14\pi \ m^3 = 14000\pi \ kg$.
The work done to empty the well is equal to the potential energy of the water,where the center of mass is at $h/2 = 7 \ m$.
$W = mgh_{cm} = (14000\pi) \times 10 \times 7 = 980000\pi \ J$.
Power $P = 1.4 \ kW = 1400 \ W$.
Time $t = W / P = (980000\pi) / 1400 = 700\pi \ s$.
Using $\pi \approx 3.14$,$t = 700 \times 3.14 = 2198 \ s \approx 2200 \ s$.

(B) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.
$W = \Delta K$
$F \cdot d = \frac{1}{2} m v^2$
Since the retarding force $F$ is the same for both bodies,the distance $d$ travelled before coming to rest is given by:
$d = \frac{m v^2}{2F}$
For body $A$:
$d_A = \frac{m_A v_A^2}{2F} = \frac{1.5 \times (20)^2}{2F} = \frac{1.5 \times 400}{2F} = \frac{600}{2F}$
For body $B$:
$d_B = \frac{m_B v_B^2}{2F} = \frac{3 \times (15)^2}{2F} = \frac{3 \times 225}{2F} = \frac{675}{2F}$
The ratio of the distances is:
$\frac{d_A}{d_B} = \frac{600}{675} = \frac{24}{27} = \frac{8}{9}$
Thus,the ratio is $8: 9$.