Elastic Collision Questions in English

(B) During the short time of collision,the billiard balls undergo deformation,which leads to the storage of potential energy $(PE)$.
Since some kinetic energy $(KE)$ is converted into potential energy during the deformation phase,the total kinetic energy is not conserved during the contact time.
However,for the system of two balls,the resultant external force acting on the system is zero.
According to the law of conservation of linear momentum,if the net external force is zero,the total linear momentum of the system remains conserved throughout the collision process,including the time of contact.

(N/A) When ball $A$ reaches the bottom point,its velocity is horizontal. Since the collision is elastic and the masses are equal,the velocities are exchanged.
$(a)$ In an elastic collision between two bodies of equal mass where one is at rest,the moving body comes to rest and the stationary body acquires the velocity of the moving body. Therefore,bob $A$ comes to rest at the bottom point after the collision. The height to which it will rise is $0\,m$.
$(b)$ The speed with which bob $B$ starts moving is equal to the speed with which bob $A$ hits bob $B$. Using the law of conservation of energy for bob $A$ as it falls from height $h = 1\,m$:
$v = \sqrt{2gh}$
$v = \sqrt{2 \times 9.8 \times 1}$
$v = \sqrt{19.6} \approx 4.43\,m/s$.

(C) In an elastic collision,the total kinetic energy is conserved. However,the question asks for the loss in kinetic energy of the colliding body. Let the mass of the first body be $m_1 = m$ and its initial velocity be $u_1 = v$. Let the mass of the second body be $m_2 = 2m$ and its initial velocity be $u_2 = 0$.
After the collision,the final velocity $v_1$ of the first body is given by the formula: $v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2$.
Substituting the values: $v_1 = \frac{m - 2m}{m + 2m} v + 0 = \frac{-m}{3m} v = -v/3$.
The initial kinetic energy of the first body is $K_i = 1/2 \ mv^2$.
The final kinetic energy of the first body is $K_f = 1/2 \ m(v_1)^2 = 1/2 \ m(-v/3)^2 = 1/2 \ m(v^2/9) = 1/18 \ mv^2$.
The loss in kinetic energy of the colliding body is $\Delta K = K_i - K_f = 1/2 \ mv^2 - 1/18 \ mv^2$.
$\Delta K = (9/18 - 1/18) \ mv^2 = 8/18 \ mv^2 = 4/9 \ mv^2$.

(D) From the law of conservation of linear momentum:
$\vec{P}_i = \vec{P}_f$
$m(u\hat{i}) + 3m(0) = m(v\hat{j}) + 3m\vec{v}_1$
$m(u\hat{i} - v\hat{j}) = 3m\vec{v}_1$
$\vec{v}_1 = \frac{u\hat{i} - v\hat{j}}{3}$
Taking the magnitude squared:
$v_1^2 = \frac{u^2 + v^2}{9} \quad \dots(1)$
Since the collision is perfectly elastic, kinetic energy is conserved:
$K_i = K_f$
$\frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)v_1^2$
$u^2 = v^2 + 3v_1^2$
Substituting equation $(1)$ into the energy equation:
$u^2 = v^2 + 3\left(\frac{u^2 + v^2}{9}\right)$
$u^2 = v^2 + \frac{u^2 + v^2}{3}$
$3u^2 = 3v^2 + u^2 + v^2$
$2u^2 = 4v^2$
$v^2 = \frac{u^2}{2}$
$v = \frac{u}{\sqrt{2}}$

(A) From the law of conservation of momentum:
$Mv + m \times 0 = Mv_1 + mv_2$
$\Rightarrow M(v - v_1) = mv_2 \dots (i)$
From the law of conservation of kinetic energy (since the collision is elastic):
$\frac{1}{2}Mv^2 + 0 = \frac{1}{2}Mv_1^2 + \frac{1}{2}mv_2^2$
$\Rightarrow M(v^2 - v_1^2) = mv_2^2 \dots (ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{M(v - v_1)(v + v_1)}{M(v - v_1)} = \frac{mv_2^2}{mv_2}$
$v + v_1 = v_2 \dots (iii)$
Solving equations $(i)$ and $(iii)$ for the final velocity of the light object $(v_2)$:
$v_2 = \frac{2Mv}{M + m}$
Since $M \gg m$,we can approximate $M + m \approx M$:
$v_2 = \frac{2Mv}{M} = 2v$.

(C) For a one-dimensional elastic collision, the final velocity of body $A$ $(v_1)$ is given by:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$
Given $m_1 = 4m$, $u_1 = u$, $m_2 = 2m$, and $u_2 = 0$:
$v_1 = \left( \frac{4m - 2m}{4m + 2m} \right) u + 0 = \left( \frac{2m}{6m} \right) u = \frac{1}{3} u$
Initial kinetic energy of body $A$ is $K_i = \frac{1}{2} (4m) u^2 = 2mu^2$.
Final kinetic energy of body $A$ is $K_f = \frac{1}{2} (4m) (\frac{1}{3} u)^2 = 2m (\frac{1}{9} u^2) = \frac{2}{9} mu^2$.
Energy lost by body $A$ is $\Delta K = K_i - K_f = 2mu^2 - \frac{2}{9} mu^2 = \frac{16}{9} mu^2$.
The fraction of energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{16}{9} mu^2}{2mu^2} = \frac{8}{9}$.

(B) According to the law of conservation of linear momentum,the total momentum of the system before the collision is equal to the total momentum of the system after the collision.
Let $v_{cm}$ be the velocity of the centre of mass of the rod after the collision.
Before the collision,the momentum of the particle is $p_i = mv$ and the rod is at rest $(p_{rod} = 0)$.
After the collision,the particle comes to rest,so its momentum is $0$. The rod moves with velocity $v_{cm}$.
Applying conservation of linear momentum:
$mv + 0 = 0 + Mv_{cm}$
Solving for $v_{cm}$:
$v_{cm} = \frac{mv}{M}$

(A) Let the initial velocity of mass $m_{1}$ be $u$ and the final velocities of $m_{1}$ and $m_{2}$ be $v$ in opposite directions. Since the collision is elastic (implied by the context of such problems),we use the conservation of linear momentum and the coefficient of restitution.
Conservation of linear momentum:
$m_{1}u = m_{2}v - m_{1}v$
$m_{1}u = v(m_{2} - m_{1})$ --- $(1)$
Coefficient of restitution $e = 1$ for elastic collision:
$e = \frac{v_{sep}}{v_{app}} = \frac{v - (-v)}{u - 0} = 1$
$2v = u$ --- $(2)$
Substitute $u = 2v$ into equation $(1)$:
$m_{1}(2v) = v(m_{2} - m_{1})$
$2m_{1} = m_{2} - m_{1}$
$m_{2} = 3m_{1}$
$\frac{m_{2}}{m_{1}} = 3$
Thus,the ratio $m_{2} : m_{1}$ is $3:1$.

(C) For a one-dimensional elastic collision where the second body is initially at rest,the final velocity $V_2$ of the second body is given by the formula:
$V_2 = \frac{2m_1 u_1}{m_1 + m_2}$
Here,$m_1 = M$,$m_2 = m$,and $u_1 = u$. Substituting these values:
$V_2 = \frac{2Mu}{M + m}$
Given the condition $m << M$,we can approximate $M + m \approx M$.
Therefore,$V_2 \approx \frac{2Mu}{M} = 2u$.
Thus,Assertion $A$ is correct.
Reason $R$ states that momentum and kinetic energy are conserved in an elastic collision,which is the fundamental definition and physical principle used to derive the velocity equations. Therefore,$R$ is the correct explanation for $A$.

(C) Given $\theta_{1} = \theta_{2} = \theta$.
From the conservation of linear momentum:
In the $x$-direction: $M V_{0} = M V_{1} \cos \theta + m V_{2} \cos \theta$
In the $y$-direction: $0 = M V_{1} \sin \theta - m V_{2} \sin \theta$
From the $y$-direction equation,we get $M V_{1} = m V_{2}$,so $V_{2} = \frac{M V_{1}}{m}$.
Substituting this into the $x$-direction equation: $M V_{0} = (M V_{1} + m \cdot \frac{M V_{1}}{m}) \cos \theta = 2 M V_{1} \cos \theta$,which gives $V_{0} = 2 V_{1} \cos \theta$.
From the conservation of kinetic energy in an elastic collision:
$\frac{1}{2} M V_{0}^{2} = \frac{1}{2} M V_{1}^{2} + \frac{1}{2} m V_{2}^{2}$
Substituting $V_{0} = 2 V_{1} \cos \theta$ and $V_{2} = \frac{M V_{1}}{m}$:
$M (4 V_{1}^{2} \cos^{2} \theta) = M V_{1}^{2} + m (\frac{M V_{1}}{m})^{2}$
$4 M \cos^{2} \theta = M + \frac{M^{2}}{m}$
Dividing by $M$: $4 \cos^{2} \theta = 1 + \frac{M}{m}$
Since $\cos^{2} \theta \leq 1$,we have $1 + \frac{M}{m} \leq 4$,which implies $\frac{M}{m} \leq 3$.
Thus,the largest possible value of the ratio $M/m$ is $3$.

(C) Let the mass of the moving particle be $m_1 = m$ and its initial velocity be $u_1 = u_0$.
Let the mass of the stationary particle be $m_2 = 5m$ and its initial velocity be $u_2 = 0$.
For a head-on elastic collision,the final velocity $V_2$ of the second particle is given by:
$V_2 = \frac{2m_1 u_1}{m_1 + m_2} = \frac{2m u_0}{m + 5m} = \frac{2m u_0}{6m} = \frac{u_0}{3}$.
The kinetic energy transferred to the stationary particle is its final kinetic energy $K_2 = \frac{1}{2} m_2 V_2^2$.
$K_2 = \frac{1}{2} (5m) \left(\frac{u_0}{3}\right)^2 = \frac{5}{2} m \frac{u_0^2}{9} = \frac{5}{18} m u_0^2$.
The initial kinetic energy of the moving particle is $K_1 = \frac{1}{2} m u_0^2$.
The percentage of kinetic energy transferred is $\frac{K_2}{K_1} \times 100$.
$\frac{\frac{5}{18} m u_0^2}{\frac{1}{2} m u_0^2} \times 100 = \frac{5}{18} \times 2 \times 100 = \frac{5}{9} \times 100 = 55.55\% \approx 55.5\%$.

(B) Mass of each ball,$m = 0.05\,kg$.
Initial velocity of one ball,$u = 10\,m/s$.
Final velocity of the same ball after collision,$v = -10\,m/s$ (since it rebounds in the opposite direction).
Change in momentum of one ball,$\Delta P = m(v - u) = 0.05 \times (-10 - 10) = 0.05 \times (-20) = -1\,kg\cdot m/s$.
The magnitude of change in momentum is $|\Delta P| = 1\,kg\cdot m/s$.
The average force exerted is given by $F = \frac{|\Delta P|}{\Delta t}$.
Given $\Delta t = 0.005\,s$,we have $F = \frac{1}{0.005} = \frac{1000}{5} = 200\,N$.

(D) Since the collision is elastic,both linear momentum and kinetic energy are conserved.
Let $u_1$ be the initial velocity of mass $M$,and $v_1, v_2$ be the final velocities of masses $M$ and $M/2$ respectively.
Conservation of linear momentum:
$M u_1 = M v_1 + (M/2) v_2$
$u_1 = v_1 + v_2/2$
$2 u_1 = 2 v_1 + v_2$ ... $(i)$
Conservation of kinetic energy:
$(1/2) M u_1^2 = (1/2) M v_1^2 + (1/2) (M/2) v_2^2$
$u_1^2 = v_1^2 + v_2^2/2$
$2 u_1^2 = 2 v_1^2 + v_2^2$ ... (ii)
From $(i)$,$u_1 = (2 v_1 + v_2) / 2$. Substituting this into (ii):
$2 ((2 v_1 + v_2) / 2)^2 = 2 v_1^2 + v_2^2$
$2 (4 v_1^2 + v_2^2 + 4 v_1 v_2) / 4 = 2 v_1^2 + v_2^2$
$(4 v_1^2 + v_2^2 + 4 v_1 v_2) / 2 = 2 v_1^2 + v_2^2$
$4 v_1^2 + v_2^2 + 4 v_1 v_2 = 4 v_1^2 + 2 v_2^2$
$4 v_1 v_2 = v_2^2$
Since $v_2 \neq 0$,we have $v_2 = 4 v_1$.
The ratio of final velocities of $M$ and $M/2$ is $x = v_1 / v_2 = 1 / 4$.

(D) Let the velocity of the ball of mass $m$ just before the collision be $u$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$, so $u = \sqrt{2gh}$.
In an elastic collision, both kinetic energy and linear momentum are conserved.
Let $v_1$ be the velocity of the ball (in the opposite direction) and $v_2$ be the velocity of the block after the collision.
Conservation of momentum: $mu = 2mv_2 - mv_1 \Rightarrow u = 2v_2 - v_1 \quad \dots (i)$
Conservation of kinetic energy: $\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(2m)v_2^2 \Rightarrow u^2 = v_1^2 + 2v_2^2 \quad \dots (ii)$
From $(i)$, $2v_2 = u + v_1$. Substituting this into $(ii)$:
$u^2 = v_1^2 + 2\left(\frac{u+v_1}{2}\right)^2 = v_1^2 + \frac{(u+v_1)^2}{2}$
$2u^2 = 2v_1^2 + u^2 + v_1^2 + 2uv_1$
$u^2 - 2uv_1 - 3v_1^2 = 0$
$(u - 3v_1)(u + v_1) = 0$
Since $v_1$ is the speed in the opposite direction, $v_1 = \frac{u}{3}$.
The new height $h'$ reached by the ball is given by $mgh' = \frac{1}{2}mv_1^2$.
$h' = \frac{v_1^2}{2g} = \frac{(u/3)^2}{2g} = \frac{u^2}{18g} = \frac{2gh}{18g} = \frac{h}{9}$.

(D) The collision is elastic,so both linear momentum and kinetic energy are conserved.
According to the conservation of linear momentum:
$M V = M V^{\prime} + m v \implies M(V - V^{\prime}) = m v \dots (i)$
According to the conservation of kinetic energy:
$\frac{1}{2} M V^2 = \frac{1}{2} M V^{\prime 2} + \frac{1}{2} m v^2 \implies M(V^2 - V^{\prime 2}) = m v^2 \dots (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{M(V^2 - V^{\prime 2})}{M(V - V^{\prime})} = \frac{m v^2}{m v}$
$\frac{(V - V^{\prime})(V + V^{\prime})}{(V - V^{\prime})} = v$
$V + V^{\prime} = v$
Thus,option $(d)$ is correct.

(B) Let the velocity of the ball of mass $2m$ be $u_0$ and the system of two balls be at rest. After the elastic collision, let the velocity of the ball of mass $2m$ be $v_1$ and the velocity of the first ball of mass $m$ be $v_2$.
By conservation of linear momentum: $2m u_0 = 2m v_1 + m v_2 \implies 2u_0 = 2v_1 + v_2$.
Since the collision is perfectly elastic, the coefficient of restitution $e = 1 = \frac{v_2 - v_1}{u_0} \implies v_2 - v_1 = u_0$.
Solving these two equations: $v_2 = u_0 + v_1$. Substituting this into the momentum equation: $2u_0 = 2v_1 + (u_0 + v_1) = 3v_1 + u_0 \implies 3v_1 = u_0 \implies v_1 = \frac{u_0}{3}$.
Then $v_2 = u_0 + \frac{u_0}{3} = \frac{4u_0}{3}$.
The vibrational energy is the kinetic energy of the two-ball system in the center-of-mass frame. The velocity of the center of mass of the two-ball system is $v_{cm} = \frac{m(v_2) + m(0)}{2m} = \frac{v_2}{2} = \frac{2u_0}{3}$.
The vibrational energy $E_v = \frac{1}{2} \mu v_{rel}^2$, where $\mu = \frac{m \cdot m}{m+m} = \frac{m}{2}$ and $v_{rel} = v_2 - 0 = \frac{4u_0}{3}$.
$E_v = \frac{1}{2} \left( \frac{m}{2} \right) \left( \frac{4u_0}{3} \right)^2 = \frac{m}{4} \cdot \frac{16u_0^2}{9} = \frac{4mu_0^2}{9}$.
The initial kinetic energy of the ball of mass $2m$ is $K_i = \frac{1}{2} (2m) u_0^2 = mu_0^2$.
The ratio is $\frac{E_v}{K_i} = \frac{4mu_0^2 / 9}{mu_0^2} = \frac{4}{9}$.

(D) Let the initial velocity of particle $A$ be $u_A$ and particle $B$ be $0$. Let the final velocities be $v_A$ and $v_B$. Given that they move in opposite directions with equal speeds,let $v_A = -v$ and $v_B = v$.
From the conservation of linear momentum:
$m_A u_A = m_A (-v) + m_B v$
$m_A u_A = (m_B - m_A) v$ --- $(1)$
From the conservation of kinetic energy (perfectly elastic collision):
$\frac{1}{2} m_A u_A^2 = \frac{1}{2} m_A (-v)^2 + \frac{1}{2} m_B v^2$
$m_A u_A^2 = (m_A + m_B) v^2$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{m_A u_A^2}{m_A u_A} = \frac{(m_A + m_B) v^2}{(m_B - m_A) v}$
$u_A = \frac{(m_A + m_B) v}{(m_B - m_A)}$
Substitute $u_A$ back into equation $(1)$:
$m_A \left[ \frac{(m_A + m_B) v}{(m_B - m_A)} \right] = (m_B - m_A) v$
$m_A (m_A + m_B) = (m_B - m_A)^2$
$m_A^2 + m_A m_B = m_B^2 - 2 m_A m_B + m_A^2$
$3 m_A m_B = m_B^2$
$3 m_A = m_B$

(B) The force exerted on the surface is equal to the rate of change of momentum of the balls.
Since the collision is elastic and normal, the velocity of each ball changes from $u$ to $-u$.
The change in momentum for a single ball is $\Delta p = m(-u) - mu = -2mu$.
The magnitude of the change in momentum for one ball is $|\Delta p| = 2mu$.
Since $n$ balls strike the surface per unit time, the total rate of change of momentum is $F = n \times |\Delta p|$.
Therefore, the force on the surface is $F = 2mun$.

(B) During an elastic collision between two spheres,the impulsive force acts along the line joining their centers (the line $SR$).
This force causes a change in the momentum of both objects along the line $SR$.
However,there is no impulsive force acting perpendicular to the line $SR$.
Therefore,the component of momentum of each object perpendicular to the line $SR$ remains unchanged (conserved).
Thus,the momentum of $M_1$ is conserved in the direction perpendicular to $SR$.

(D) The velocity of the ball before collision can be resolved into two components: horizontal component $u_x = u \cos \theta$ and vertical component $u_y = -u \sin \theta$ (taking downward as negative).
Since the surface is smooth,there is no impulse in the horizontal direction,so the horizontal velocity remains unchanged: $v_x = u \cos \theta$.
In the vertical direction,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $e = \frac{v_y}{u_y}$.
Thus,the vertical velocity after collision is $v_y = e u \sin \theta$ (upward).
The impulse $I$ imparted to the ball by the surface is equal to the change in momentum of the ball: $\vec{I} = m(\vec{v} - \vec{u})$.
Since there is no change in horizontal momentum,the impulse is only in the vertical direction: $I = m(v_y - u_y) = m(e u \sin \theta - (-u \sin \theta)) = m u(1+e) \sin \theta$.
The magnitude of the impulse imparted to the surface by the ball is equal to the magnitude of the impulse imparted to the ball by the surface,which is $m u(1+e) \sin \theta$.

(D) In an elastic collision between two bodies of equal mass $(m_1 = m_2)$ where one is initially at rest,if the collision is oblique,the two bodies move at an angle of $90^{\circ}$ to each other,not in opposite directions. Thus,statement $(d)$ is incorrect.
For $(a)$: When $m_1 \ll m_2$,the momentum transfer is indeed maximum.
For $(b)$: If $m_1 \gg m_2$,the velocity of the second ball becomes $v_2 \approx 2u_1$,which is a standard result,but the specific claim in $(b)$ is generally considered a distractor or incorrect depending on the mass ratio context,however,$(d)$ is physically impossible for oblique collisions.
For $(c)$: When $m_1 = m_2$ and $m_2$ is at rest,the first ball comes to rest and the second moves with the initial velocity of the first,resulting in $100\%$ transfer of $K.E$.

(B) In a one-dimensional elastic head-on collision,the final velocity $v_1$ of the first mass $m_1$ is given by the formula:
$v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2}$
where $u_1$ is the initial velocity of $m_1$ and $m_2$ is initially at rest $(u_2 = 0)$.
According to the problem,the mass $m_1$ rebounds,meaning its final velocity $v_1$ is in the opposite direction to its initial velocity $u_1$. Thus,$v_1$ must be negative.
For $v_1$ to be negative,the numerator $(m_1 - m_2)$ must be negative,which implies $m_1 - m_2 < 0$ or $m_1 < m_2$.
Therefore,the correct option is $B$.

(A) Since there is no external force acting on the system,the linear momentum of the system is conserved.
Let $v_1$ and $v_2$ be the velocities of the blocks after the collision.
Initial momentum = $mv + (2m)(0) = mv$.
Final momentum = $m(0) + (2m)v_2 = 2mv_2$.
By conservation of linear momentum: $mv = 2mv_2$,which gives $v_2 = v/2$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.
Velocity of approach = $v - 0 = v$.
Velocity of separation = $v_2 - 0 = v_2 = v/2$.
Therefore,$e = \frac{v/2}{v} = 0.5$.

(B) The Assertion $(A)$ is true. By the definition of an elastic collision,both kinetic energy and linear momentum are conserved. For a one-dimensional elastic collision between two bodies of masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$ and final velocities $v_1$ and $v_2$,the coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$. For an elastic collision,$e = 1$,which implies $v_2 - v_1 = u_1 - u_2$. This confirms that the relative speed of separation after collision is equal to the relative speed of approach before collision.
The Reason $(R)$ is also true. In any collision (elastic or inelastic),the total linear momentum of the system is conserved provided no external force acts on the system.
However,the conservation of linear momentum alone does not explain why the relative speed remains constant; that property is a consequence of the conservation of kinetic energy in an elastic collision. Therefore,the Reason is not the correct explanation of the Assertion.

(B) Let $m_1 = 1\,kg$,$m_2 = 3\,kg$,$u_1$ be the initial velocity of the $1\,kg$ mass,and $u_2 = 0$ be the initial velocity of the $3\,kg$ mass.
After the collision,the velocity of the $1\,kg$ mass is $v_1 = -2\,m/s$ (as it reverses direction).
By the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$1(u_1) + 3(0) = 1(-2) + 3(v_2)$
$u_1 = -2 + 3v_2 \quad \dots(1)$
For an elastic collision,the coefficient of restitution $e = 1$:
$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$
$1 = \frac{v_2 - (-2)}{u_1 - 0} \Rightarrow u_1 = v_2 + 2 \quad \dots(2)$
Substituting $v_2 = u_1 - 2$ into equation $(1)$:
$u_1 = -2 + 3(u_1 - 2)$
$u_1 = -2 + 3u_1 - 6$
$2u_1 = 8$
$u_1 = 4\,m/s$.

(C) $1$. The ball $P$ slides down a height $h = 20\,cm = 0.2\,m$.
$2$. By the law of conservation of energy,the velocity $v_P$ of ball $P$ just before the collision is given by $v_P = \sqrt{2gh}$.
$3$. Substituting the values: $v_P = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2\,m/s$.
$4$. Since the collision is elastic and the masses of the two balls are equal,the velocities are interchanged after the collision.
$5$. Before the collision,ball $P$ has velocity $v_P = 2\,m/s$ and ball $Q$ is at rest $(v_Q = 0)$.
$6$. After the collision,ball $P$ comes to rest $(v_P' = 0)$ and ball $Q$ moves with the velocity that ball $P$ had before the collision $(v_Q' = v_P = 2\,m/s)$.

(B) In an elastic collision, both linear momentum and kinetic energy are conserved.
Let two bodies of masses $m_1$ and $m_2$ have initial velocities $u_1$ and $u_2$, and final velocities $v_1$ and $v_2$.
Conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ => $m_1(u_1 - v_1) = m_2(v_2 - u_2)$ (Equation $1$).
Conservation of kinetic energy: $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$ => $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$ (Equation $2$).
Dividing Equation $2$ by Equation $1$: $u_1 + v_1 = v_2 + u_2$ => $u_1 - u_2 = v_2 - v_1$.
This shows that the relative velocity of approach $(u_1 - u_2)$ equals the relative velocity of separation $(v_2 - v_1)$. Thus, $Statement-1$ is true.
$Statement-2$ is also true because linear momentum is always conserved in any collision (elastic or inelastic) in the absence of external forces. However, the conservation of momentum alone is not sufficient to derive the relative speed property; the conservation of kinetic energy is required. Therefore, $Statement-2$ is not the correct explanation for $Statement-1$.

(C) Since the collision is elastic and the masses are equal,the velocities of the particles will be interchanged after each collision.
Let the particles collide at time $t$ at an angle $\theta$ from point $A$.
The distance traveled by the first particle is $R\theta = vt$.
The distance traveled by the second particle is $R(2\pi - \theta) = 2vt$.
Dividing the two equations: $\frac{\theta}{2\pi - \theta} = \frac{vt}{2vt} = \frac{1}{2}$.
This gives $2\theta = 2\pi - \theta$,so $3\theta = 2\pi$,which means $\theta = \frac{2\pi}{3} = 120^{\circ}$.
After the first collision at $120^{\circ}$,the velocities are interchanged. The particle that had speed $v$ now has speed $2v$,and the particle that had speed $2v$ now has speed $v$.
They will collide again after traveling another $120^{\circ}$ relative to each other,which occurs at an angle of $240^{\circ}$ from point $A$.
After this second collision,the velocities are interchanged again. The particles will then travel the remaining $120^{\circ}$ to reach point $A$ simultaneously.
Thus,they reach point $A$ after $2$ collisions.

(A) When two objects of identical mass undergo a perfectly elastic one-dimensional collision,they exchange their velocities.
Initially,we have $v_{A} = 5 \ m/s$,$v_{B} = 2 \ m/s$,and $v_{C} = 4 \ m/s$.
First,sphere $A$ collides with sphere $B$. Since they have identical masses,they exchange velocities: $v_{A}' = 2 \ m/s$ and $v_{B}' = 5 \ m/s$.
Now,sphere $B$ (moving with $5 \ m/s$) collides with sphere $C$ (moving with $4 \ m/s$). They exchange velocities: $v_{B}'' = 4 \ m/s$ and $v_{C}' = 5 \ m/s$.
Thus,the final velocities are $v_{A} = 2 \ m/s$,$v_{B} = 4 \ m/s$,and $v_{C} = 5 \ m/s$.
The Assertion $(A)$ states the final velocities are $4 \ m/s, 2 \ m/s, 5 \ m/s$,which is incorrect.
The Reason $(R)$ is a correct physical principle.
Therefore,$(A)$ is false but $(R)$ is true.

(B) Let the velocity of the wall be $v_w = 20 \ m/s$ (to the right,positive direction).
The velocity of the ball before collision is $u = -10 \ m/s$ (to the left,negative direction).
The mass of the ball is $m = 500 \ g = 0.5 \ kg$.
In an elastic collision with a massive wall moving with velocity $v_w$,the velocity of the ball after collision $v$ is given by the formula $v = 2v_w - u$.
Substituting the values: $v = 2(20) - (-10) = 40 + 10 = 50 \ m/s$.
The change in kinetic energy of the ball is $\Delta KE = KE_f - KE_i = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$.
$\Delta KE = \frac{1}{2} \times 0.5 \times (50^2 - (-10)^2) = 0.25 \times (2500 - 100) = 0.25 \times 2400 = 600 \ J$.

(C) In a two-dimensional perfectly elastic collision between two identical masses $(m_1 = m_2 = m)$,where one mass is initially at rest,the conservation of linear momentum and kinetic energy leads to the condition that the angle between the final velocities of the two particles is $90^{\circ}$.
Given that $\theta_1 = 30^{\circ}$,we have the relation $\theta_1 + \theta_2 = 90^{\circ}$.
Therefore,$\theta_2 = 90^{\circ} - 30^{\circ} = 60^{\circ}$.

(B) Let the initial velocity of the particle of mass $2m$ be $v_1$ and the final velocities of particles $2m$ and $m$ be $v_{1f}$ and $v_{2f}$ at angles $\theta$ and $\phi$ respectively.
Conservation of linear momentum along the $x$-axis:
$2mv_1 = 2mv_{1f} \cos \theta + mv_{2f} \cos \phi$ ---$(i)$
Conservation of linear momentum along the $y$-axis:
$0 = 2mv_{1f} \sin \theta - mv_{2f} \sin \phi \implies 2mv_{1f} \sin \theta = mv_{2f} \sin \phi$ ---(ii)
Conservation of kinetic energy:
$\frac{1}{2}(2m)v_1^2 = \frac{1}{2}(2m)v_{1f}^2 + \frac{1}{2}mv_{2f}^2 \implies 2v_1^2 = 2v_{1f}^2 + v_{2f}^2$ ---(iii)
From $(i)$ and (ii),$mv_{2f} \cos \phi = 2m(v_1 - v_{1f} \cos \theta)$ and $mv_{2f} \sin \phi = 2mv_{1f} \sin \theta$.
Squaring and adding these:
$(mv_{2f})^2 = 4m^2(v_1^2 + v_{1f}^2 - 2v_1v_{1f} \cos \theta)$.
Substitute $v_{2f}^2 = 2v_1^2 - 2v_{1f}^2$ from (iii):
$m^2(2v_1^2 - 2v_{1f}^2) = 4m^2(v_1^2 + v_{1f}^2 - 2v_1v_{1f} \cos \theta)$.
$v_1^2 - v_{1f}^2 = 2v_1^2 + 2v_{1f}^2 - 4v_1v_{1f} \cos \theta$.
$3v_{1f}^2 - (4v_1 \cos \theta)v_{1f} + v_1^2 = 0$.
For $v_{1f}$ to be real,the discriminant $D \geq 0$:
$(4v_1 \cos \theta)^2 - 4(3)(v_1^2) \geq 0 \implies 16v_1^2 \cos^2 \theta - 12v_1^2 \geq 0$.
$\cos^2 \theta \geq \frac{12}{16} = \frac{3}{4} \implies \cos \theta \geq \frac{\sqrt{3}}{2}$.
Thus,the maximum value of $\theta$ is $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

$(A)$ In a head-on elastic collision,the relative velocity of approach is equal to the relative velocity of separation.
Let the masses of the two particles be $m_1$ and $m_2$.
Initial velocities are $u_1 = 5 \,m/s$ and $u_2 = 3 \,m/s$.
Final velocities are $v_1 = 4 \,m/s$ and $v_2 = ?$.
The relative velocity of approach is $u_1 - u_2 = 5 - 3 = 2 \,m/s$.
The relative velocity of separation is $v_2 - v_1 = v_2 - 4$.
Since the collision is elastic,$u_1 - u_2 = v_2 - v_1$.
Substituting the values: $2 = v_2 - 4$.
Therefore,$v_2 = 2 + 4 = 6 \,m/s$.
Since the result is positive,the second particle moves in the same direction.

(B) Let the mass of both spheres be $m$. The initial velocity of the first sphere is $u_1 = 3u$ and the second sphere is $u_2 = 0$.
Let $v_1$ and $v_2$ be the velocities of the first and second spheres after the collision,respectively.
By the law of conservation of linear momentum: $m(3u) + m(0) = mv_1 + mv_2$,which simplifies to $v_1 + v_2 = 3u$ (Equation $1$).
By the definition of the coefficient of restitution $e$: $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{3u - 0}$,which gives $v_2 - v_1 = 3ue$ (Equation $2$).
Adding Equation $1$ and Equation $2$: $2v_2 = 3u(1 + e) \implies v_2 = \frac{3u(1 + e)}{2}$.
Subtracting Equation $2$ from Equation $1$: $2v_1 = 3u(1 - e) \implies v_1 = \frac{3u(1 - e)}{2}$.
The ratio of the velocity of the second sphere to the first sphere is $\frac{v_2}{v_1} = \frac{3u(1 + e) / 2}{3u(1 - e) / 2} = \frac{1 + e}{1 - e}$.

(A) Let $u$ be the initial velocity of mass $m$ and $0$ be the initial velocity of mass $M$.
Let $v_1$ be the final velocity of mass $m$ and $v_2$ be the final velocity of mass $M$.
Given that the particle $m$ stops after the collision,$v_1 = 0$.
By the law of conservation of linear momentum: $mu + M(0) = m(0) + Mv_2$.
This simplifies to $mu = Mv_2$,so $v_2 = \frac{mu}{M}$.
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Substituting the known values: $e = \frac{\frac{mu}{M} - 0}{u - 0} = \frac{mu/M}{u} = \frac{m}{M}$.

(B) In an elastic collision,both linear momentum and kinetic energy are conserved.
Let the mass of both balls be $m$.
Initial kinetic energy $K_i = \frac{1}{2} m v^2 + 0 = \frac{1}{2} m v^2$.
Final kinetic energy $K_f = \frac{1}{2} m (v/3)^2 + \frac{1}{2} m v_2'^2$,where $v_2'$ is the speed of the second ball.
Since the collision is elastic,$K_i = K_f$.
$\frac{1}{2} m v^2 = \frac{1}{2} m (v^2/9) + \frac{1}{2} m v_2'^2$.
Dividing by $\frac{1}{2} m$,we get $v^2 = \frac{v^2}{9} + v_2'^2$.
$v_2'^2 = v^2 - \frac{v^2}{9} = \frac{8v^2}{9}$.
Therefore,$v_2' = \sqrt{\frac{8}{9}} v = \frac{2\sqrt{2}}{3} v$.

(B) In an elastic collision between two bodies of masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$,the final velocities $v_1$ and $v_2$ are given by the conservation of momentum and kinetic energy.
Given that after the collision,the masses exchange their velocities such that $m_1$ moves with $v_2$ and $m_2$ moves with $v_1$,this is a characteristic property of an elastic collision between two bodies of equal mass.
If $m_1 = m_2$,then the bodies exchange their velocities after an elastic collision.
Therefore,the ratio $\frac{m_2}{m_1} = 1$.

(C) In an elastic collision,both linear momentum and kinetic energy are conserved.
Let the initial velocities be $v_{a}$ and $-v_{b}$ (since they move in opposite directions).
After the collision,the velocities are $-v_{b}$ and $v_{a}$ respectively.
According to the law of conservation of linear momentum:
$m_{a}v_{a} - m_{b}v_{b} = m_{a}(-v_{b}) + m_{b}v_{a}$
Rearranging the terms:
$m_{a}v_{a} + m_{a}v_{b} = m_{b}v_{a} + m_{b}v_{b}$
$m_{a}(v_{a} + v_{b}) = m_{b}(v_{a} + v_{b})$
Since $(v_{a} + v_{b}) \neq 0$,we can divide both sides by $(v_{a} + v_{b})$:
$m_{a} = m_{b}$
Therefore,the ratio $\frac{m_{a}}{m_{b}} = 1$.

(A) In an elastic collision with a stationary wall,the molecule rebounds with the same speed $v$ in the opposite direction.
Change in momentum of the molecule in one collision = $mv - (-mv) = 2mv$.
By the law of conservation of momentum,the change in momentum of the wall in one collision is equal and opposite to the change in momentum of the molecule,which is $2mv$.
Since there are $5$ collisions per second,the total change in momentum of the wall per second is $5 \times 2mv = 10mv$.

(B) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
For a perfectly elastic collision,the kinetic energy and momentum are conserved.
In such a collision,the relative velocity of separation is equal to the relative velocity of approach.
Therefore,$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$.

(B) In an elastic collision,both linear momentum and kinetic energy are conserved.
Since the collision is elastic,the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Let $v_2$ be the speed of the second block after the collision.
Initial kinetic energy $KE_i = \frac{1}{2}mv^2 + \frac{1}{2}m(0)^2 = \frac{1}{2}mv^2$.
Final kinetic energy $KE_f = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$.
Equating $KE_i = KE_f$:
$\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2$
Dividing by $\frac{1}{2}m$:
$v^2 = v_1^2 + v_2^2$
$v_2^2 = v^2 - v_1^2$
$v_2 = \sqrt{v^2 - v_1^2}$.

(C) Let the initial velocity of mass $m_1$ be $v_1$ and the final velocity of mass $m_2$ be $v_2$.
According to the given condition,the final velocity of mass $m_1$ is $v_1' = \frac{v_1}{1.5} = \frac{2}{3} v_1$.
Assuming an elastic collision,the coefficient of restitution $e = 1$.
The formula for the coefficient of restitution is $e = \frac{v_2 - v_1'}{v_1 - 0}$.
Substituting the values: $1 = \frac{v_2 - \frac{2}{3}v_1}{v_1} \implies v_2 = v_1 + \frac{2}{3}v_1 = \frac{5}{3}v_1$.
By the law of conservation of linear momentum: $m_1 v_1 = m_1 v_1' + m_2 v_2$.
Substituting the velocities: $m_1 v_1 = m_1 (\frac{2}{3} v_1) + m_2 (\frac{5}{3} v_1)$.
Rearranging the terms: $m_1 v_1 - \frac{2}{3} m_1 v_1 = m_2 \frac{5}{3} v_1$.
$\frac{1}{3} m_1 v_1 = \frac{5}{3} m_2 v_1$.
Therefore,$\frac{m_1}{m_2} = \frac{5}{1}$.

(C) $I$. In an elastic collision,the total kinetic energy of the system is conserved,meaning the initial kinetic energy equals the final kinetic energy.
$II$. Linear momentum is conserved in all types of collisions (elastic or inelastic) provided no external force acts on the system.
$III$. During the collision interval $\Delta t$,the bodies deform,and part of the kinetic energy is temporarily converted into elastic potential energy. Therefore,the kinetic energy is not conserved during the collision process itself.
$\therefore$ All three statements are correct.

(D) Let the mass of the first sphere be $m_1$ and the second be $m_2$. Since both are steel spheres,their density $\rho$ is the same. Thus,$m = \rho V = \rho (\frac{4}{3} \pi r^3)$.
So,$m_1 \propto r_1^3$ and $m_2 \propto r_2^3$.
For an elastic collision in one dimension where the second body is at rest,the final velocity of the first body $v_1'$ is given by $v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1$.
Given $v_1' = \frac{7}{9} v_1$,we have $\frac{m_1 - m_2}{m_1 + m_2} = \frac{7}{9}$.
Cross-multiplying: $9m_1 - 9m_2 = 7m_1 + 7m_2$.
$2m_1 = 16m_2 \implies m_1 = 8m_2$.
Since $m \propto r^3$,we have $r_1^3 = 8r_2^3$.
Taking the cube root: $r_1 = 2r_2$.
Given $r_1 = 1.2 \ cm$,we get $1.2 = 2r_2 \implies r_2 = 0.6 \ cm$.

(A) For a one-dimensional elastic collision between two bodies of equal mass $(m_1 = m_2 = m)$,the bodies simply exchange their velocities.
Given:
$u_1 = 20 \ ms^{-1}$
$u_2 = -30 \ ms^{-1}$ (moving in the opposite direction)
According to the property of elastic collision of equal masses:
$v_1 = u_2 = -30 \ ms^{-1}$
$v_2 = u_1 = 20 \ ms^{-1}$
Thus,the velocity of the first body after the collision is $-30 \ ms^{-1}$ and the velocity of the second body is $20 \ ms^{-1}$.

(A) Applying the law of conservation of linear momentum before and after the collision:
$M(150) + 2M(-v) = M(0) + 2M(v')$ (Taking the direction of ball $A$ as positive)
$150 - 2v = 2v'$
$v' = 75 - v \dots (1)$
Now,the coefficient of restitution $e$ is given by:
$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$
$1 = \frac{v' - 0}{150 - (-v)}$
$150 + v = v' \dots (2)$
Equating $(1)$ and $(2)$:
$75 - v = 150 + v$
$2v = -75$
Since we are looking for the speed (magnitude),we consider the relative velocities. Re-evaluating with standard convention: Let $A$ move at $+150$ and $B$ move at $-v$. After collision,$A$ is at $0$ and $B$ moves at $+v'$.
$150M - 2Mv = 2Mv' \Rightarrow 75 - v = v'$
$e = \frac{v' - 0}{150 - (-v)} = 1 \Rightarrow v' = 150 + v$
Solving $75 - v = 150 + v$ gives $v = -37.5$. The magnitude of speed is $37.5 \ m \ s^{-1}$.

(B) In an elastic head-on collision between two particles of equal mass,the particles exchange their velocities. Therefore,the velocity of the incident particle becomes $0$ and the stationary particle moves with velocity $v$.
The change in momentum of the incident particle is $\Delta p = m(v - 0) = mv$.
According to the impulse-momentum theorem,the impulse (area under the $F-t$ graph) is equal to the change in momentum.
$\text{Area} = \Delta p = mv$.
The area under the given trapezoidal $F-t$ graph is:
$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$\text{Area} = \frac{1}{2} \times \left( T + \left( \frac{3T}{4} - \frac{T}{4} \right) \right) \times F_0$
$\text{Area} = \frac{1}{2} \times \left( T + \frac{2T}{4} \right) \times F_0 = \frac{1}{2} \times \left( T + \frac{T}{2} \right) \times F_0 = \frac{1}{2} \times \left( \frac{3T}{2} \right) \times F_0 = \frac{3T F_0}{4}$.
Equating the area to the change in momentum:
$\frac{3T F_0}{4} = mv$
$F_0 = \frac{4mv}{3T}$.

(A) The coefficient of restitution $(e)$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach between two colliding bodies.
Mathematically,$e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}$.
For a perfectly elastic collision,the kinetic energy and momentum are conserved,which implies that the relative velocity of separation is equal to the relative velocity of approach.
Therefore,$e = 1$ for a perfectly elastic collision.
For a perfectly inelastic collision,$e = 0$,and for an inelastic collision,$0 < e < 1$.

(C) In an elastic collision,both the total kinetic energy and the total linear momentum of the system are conserved.
Additionally,the total energy of the system remains conserved in all types of collisions.

(C) Given: $m_1 = 2 \ kg$,$m_2 = 4 \ kg$,$u_1 = 10 \ ms^{-1}$,$u_2 = -10 \ ms^{-1}$.
In a perfectly elastic collision,both linear momentum and kinetic energy are conserved.
By conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$2(10) + 4(-10) = 2v_1 + 4v_2$
$20 - 40 = 2v_1 + 4v_2$
$-20 = 2v_1 + 4v_2 \Rightarrow v_1 + 2v_2 = -10 \quad \dots (i)$
For a perfectly elastic collision,the coefficient of restitution $e = 1$,so the velocity of separation equals the velocity of approach:
$v_2 - v_1 = u_1 - u_2$
$v_2 - v_1 = 10 - (-10) = 20 \quad \dots (ii)$
Adding equations $(i)$ and $(ii)$:
$(v_1 + 2v_2) + (v_2 - v_1) = -10 + 20$
$3v_2 = 10 \Rightarrow v_2 = \frac{10}{3} \ ms^{-1}$
Substituting $v_2$ into equation $(ii)$:
$\frac{10}{3} - v_1 = 20$
$v_1 = \frac{10}{3} - 20 = \frac{10 - 60}{3} = -\frac{50}{3} \ ms^{-1}$
Thus,the final velocities are $v_1 = -\frac{50}{3} \ ms^{-1}$ and $v_2 = \frac{10}{3} \ ms^{-1}$.