Elastic Collision Questions in English

(D) For a one-dimensional elastic collision,the velocity of the first body $(m_1 = m)$ after the collision $(v_1)$ is given by the formula:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$
Given $m_1 = m$,$m_2 = 2m$,$u_1 = v$,and $u_2 = 0$:
$v_1 = \left( \frac{m - 2m}{m + 2m} \right) v + 0 = \left( \frac{-m}{3m} \right) v = -\frac{v}{3}$
Initial kinetic energy $(KE_i)$ = $\frac{1}{2} mv^2$
Final kinetic energy $(KE_f)$ = $\frac{1}{2} m v_1^2 = \frac{1}{2} m \left( -\frac{v}{3} \right)^2 = \frac{1}{2} m \left( \frac{v^2}{9} \right) = \frac{1}{18} mv^2$
The ratio of $KE_i$ to $KE_f$ is:
$\frac{KE_i}{KE_f} = \frac{\frac{1}{2} mv^2}{\frac{1}{18} mv^2} = \frac{1/2}{1/18} = \frac{18}{2} = 9$
Thus,the ratio is $9 : 1$.

(A) For a head-on elastic collision between two bodies of equal masses $(m_1 = m_2 = m)$,the bodies simply exchange their velocities.
Let the initial velocities be $u_1$ and $u_2$,and the final velocities be $v_1$ and $v_2$.
According to the conservation of momentum: $m u_1 + m u_2 = m v_1 + m v_2$,which simplifies to $u_1 + u_2 = v_1 + v_2$.
According to the property of elastic collision for equal masses: $v_1 = u_2$ and $v_2 = u_1$.
Given the final velocities are $v_1 = +3\,m/s$ and $v_2 = -2\,m/s$.
Therefore,the initial velocities must be $u_1 = v_2 = -2\,m/s$ and $u_2 = v_1 = +3\,m/s$.

(C) $1$. In the first elastic collision between block $A$ and block $B$:
$V_A = \left(\frac{M-m}{M+m}\right)u$
$V_B = \left(\frac{2M}{M+m}\right)u$
$2$. After the collision,block $B$ moves towards block $C$ and compresses the spring. At the time of maximum compression,the velocities of blocks $B$ and $C$ become equal (let this velocity be $v$).
$3$. By the law of conservation of linear momentum for the system of blocks $B$ and $C$:
$m V_B = (m + m)v$
$m \left(\frac{2M}{M+m}\right)u = 2mv$
$v = \left(\frac{M}{M+m}\right)u$
$4$. The velocity of block $C$ with respect to block $A$ is:
$V_{CA} = V_C - V_A = v - V_A$
$V_{CA} = \left(\frac{M}{M+m}\right)u - \left(\frac{M-m}{M+m}\right)u$
$V_{CA} = \left(\frac{M - M + m}{M+m}\right)u = \left(\frac{m}{M+m}\right)u$

(B) In a one-dimensional elastic collision between identical masses,the colliding body comes to rest and transfers its velocity to the stationary body if the number of incoming bodies equals the number of outgoing bodies.
Here,$2$ balls are moving with velocity $v$ and collide with a row of $6$ identical stationary balls.
By the principle of conservation of linear momentum and kinetic energy for elastic collisions of identical masses,the $2$ incoming balls will come to rest,and $2$ balls from the right end of the row will move out with the same velocity $v$.
Therefore,the correct option is $B$.

(D) In an elastic collision,both the total linear momentum and the total kinetic energy of the system are conserved.
While the total kinetic energy of the system remains constant,the kinetic energy of individual particles may change due to the exchange of energy during the collision.
Therefore,the total kinetic energy of both particles is the conserved quantity.

(A) For a head-on elastic collision between two bodies of equal masses $(m_1 = m_2 = m)$,the velocities are exchanged.
Let the initial velocities be $u_1$ and $u_2$,and the final velocities be $v_1$ and $v_2$.
According to the properties of a one-dimensional elastic collision with equal masses,$v_1 = u_2$ and $v_2 = u_1$.
Given the final velocities are $v_1 = +3 \, m/s$ and $v_2 = -2 \, m/s$.
Therefore,the initial velocities must be $u_1 = v_2 = -2 \, m/s$ and $u_2 = v_1 = +3 \, m/s$.
Thus,the original velocities are $-2 \, m/s$ and $+3 \, m/s$.

(C) The speed of the ball just before striking the lift is given by $v = \sqrt{2gh}$.
Taking $g = 9.8\, m/s^2$ and $h = 10\, m$,we get $v = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14\, m/s$.
Let the downward direction be positive. The velocity of the ball before collision is $u_1 = 14\, m/s$ and the velocity of the lift is $u_2 = 1\, m/s$.
Since the mass of the lift is much greater than the mass of the ball $(M_{lift} \gg M_{ball})$,the collision is effectively an elastic collision with a stationary wall moving at velocity $u_2$.
The velocity of the ball after collision $v_1$ is given by the formula $v_1 = 2u_2 - u_1$ (relative to the ground).
Substituting the values: $v_1 = 2(1) - 14 = 2 - 14 = -12\, m/s$.
The negative sign indicates that the ball is moving upwards with a speed of $12\, m/s$.

(D) The velocity of the first body after a one-dimensional elastic collision is given by the formula:
$v_{1} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1} + \frac{2 m_{2} u_{2}}{m_{1}+m_{2}}$
Given $m_{1} = m$,$m_{2} = 2m$,$u_{1} = v$,and $u_{2} = 0$:
$v_{1} = \left(\frac{m - 2m}{m + 2m}\right) v + \frac{2(2m)(0)}{3m} = \left(\frac{-m}{3m}\right) v = -\frac{v}{3}$
Initial kinetic energy of the first body: $KE_{i} = \frac{1}{2} mv^{2}$
Final kinetic energy of the first body: $KE_{f} = \frac{1}{2} m\left(-\frac{v}{3}\right)^{2} = \frac{1}{2} m \left(\frac{v^{2}}{9}\right) = \frac{KE_{i}}{9}$
The ratio of $KE_{i}$ to $KE_{f}$ is:
$\frac{KE_{i}}{KE_{f}} = \frac{KE_{i}}{KE_{i}/9} = 9$
Thus,the ratio is $9 : 1$.

(B) Let $m_1$ be the mass of the heavy body and $m_2$ be the mass of the small object. Given $m_1 \gg m_2$.
Since the collision is elastic and the body $m_1$ is very heavy,it continues to move with its initial velocity $u_1 = 30\, m/s$ after the collision.
For an elastic collision,the coefficient of restitution $e = 1$.
The formula for the coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$,where $u_1 = 30\, m/s$,$u_2 = 0$,$v_1 = 30\, m/s$,and $v_2 = v$.
Substituting the values: $1 = \frac{v - 30}{30 - 0}$.
$1 = \frac{v - 30}{30} \Rightarrow v - 30 = 30 \Rightarrow v = 60\, m/s$.

(A) Let $v_{1}$ be the velocity just before the collision and $v_{2}$ be the velocity just after the collision.
Using the principle of conservation of energy,the velocity of a body falling from height $h$ is given by $v = \sqrt{2gh}$.
Velocity just before collision: $v_{1} = \sqrt{2gh_{1}} = \sqrt{2 \times g \times 10}$.
Velocity just after collision: $v_{2} = \sqrt{2gh_{2}} = \sqrt{2 \times g \times 2.5}$.
The ratio of velocities is $\frac{v_{1}}{v_{2}} = \frac{\sqrt{2gh_{1}}}{\sqrt{2gh_{2}}} = \sqrt{\frac{h_{1}}{h_{2}}}$.
Substituting the values: $\frac{v_{1}}{v_{2}} = \sqrt{\frac{10}{2.5}} = \sqrt{4} = 2$.
Thus,the ratio of the velocities is $2$.

(B) Let the velocity of the ball just before collision be $u_1$ and the velocity of the lift be $u_2$.
$1$. Calculate the velocity of the ball before collision:
Using $v^2 - u^2 = 2gh$,where $u=0$,$g=10\,m/s^2$,and $h=5\,m$:
$u_1 = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\,m/s$ (downwards).
$2$. Define the coordinate system:
Let upwards be positive $(+)$ and downwards be negative $(-)$.
So,$u_1 = -10\,m/s$ and $u_2 = +1\,m/s$.
$3$. Collision analysis:
Since the lift is much heavier than the ball $(m_{lift} \gg m_{ball})$,the lift's velocity remains unchanged after the elastic collision $(e=1)$.
Using the relative velocity formula for elastic collision: $v_1 - v_2 = -e(u_1 - u_2)$.
Since $v_2 = u_2 = 1\,m/s$:
$v_1 - 1 = -1(-10 - 1)$
$v_1 - 1 = -1(-11)$
$v_1 - 1 = 11$
$v_1 = 12\,m/s$.
Since the result is positive,the ball rebounds upwards with a velocity of $12\,m/s$.

(A) $1$. First,calculate the velocity of the sphere just before the collision using the principle of conservation of energy. The potential energy at the suspension height is converted into kinetic energy at the bottom: $mgh = \frac{1}{2}mv^2$.
$2$. Given $h = 1\,m$ and $g = 10\,m/s^2$,the velocity $v$ is $\sqrt{2gh} = \sqrt{2 \times 10 \times 1} = \sqrt{20}\,m/s$.
$3$. The collision is perfectly elastic between two bodies of equal mass $(m_1 = m_2 = 0.1\,kg)$. In a one-dimensional elastic collision between two equal masses,the bodies exchange their velocities.
$4$. Since the sphere had velocity $v$ and the block was at rest,after the collision,the sphere comes to rest and the block moves with velocity $v$.
$5$. The kinetic energy of the block after the collision is $K.E. = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.1 \times (\sqrt{20})^2 = 0.05 \times 20 = 1\,J$.

(C) In a head-on elastic collision,the fraction of kinetic energy transferred from a body of mass $m_1$ to a body of mass $m_2$ (initially at rest) is given by the formula:
$f = \frac{4m_1m_2}{(m_1 + m_2)^2}$
Here,$m_1 = m$ and $m_2 = nm$.
Substituting these values into the formula:
$f = \frac{4(m)(nm)}{(m + nm)^2}$
$f = \frac{4nm^2}{m^2(1 + n)^2}$
$f = \frac{4n}{(1 + n)^2}$

(D) Let the initial velocity of the proton be $u$ and the final velocities of the proton and the particle be $v_1$ and $v_2$ respectively. Let the proton move at an angle $\theta$ with the initial direction, then the particle moves at an angle $(90^\circ - \theta)$.
By conservation of linear momentum:
Along the initial direction ($x$-axis): $mu = mv_1 \cos \theta + Mv_2 \cos(90^\circ - \theta) = mv_1 \cos \theta + Mv_2 \sin \theta$ ...$(i)$
Along the perpendicular direction ($y$-axis): $0 = mv_1 \sin \theta - Mv_2 \sin(90^\circ - \theta) = mv_1 \sin \theta - Mv_2 \cos \theta$ ...$(ii)$
From $(ii)$, $Mv_2 \cos \theta = mv_1 \sin \theta$, so $Mv_2 = mv_1 \tan \theta$.
Since the collision is elastic, kinetic energy is conserved:
$\frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2$
$mu^2 = mv_1^2 + Mv_2^2$ ...$(iii)$
Squaring and adding equations $(i)$ and $(ii)$:
$(mu)^2 = (mv_1 \cos \theta + Mv_2 \sin \theta)^2 + (mv_1 \sin \theta - Mv_2 \cos \theta)^2$
$m^2u^2 = m^2v_1^2 + M^2v_2^2$ ...$(iv)$
Comparing $(iii)$ and $(iv)$:
$m(mv_1^2 + Mv_2^2) = m^2v_1^2 + M^2v_2^2$
$m^2v_1^2 + m M v_2^2 = m^2v_1^2 + M^2v_2^2$
$m M v_2^2 = M^2v_2^2$
$m = M$
Thus, the mass of the unknown particle is $m$.

(C) For a head-on elastic collision between a particle of mass $m_1$ moving with velocity $u_1$ and a particle of mass $m_2$ at rest $(u_2 = 0)$,the final velocity $v_1$ of the first particle is given by:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1$
The fraction of kinetic energy retained by the first particle is:
$\frac{K_f}{K_i} = \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_1 u_1^2} = \left( \frac{v_1}{u_1} \right)^2 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2$
Given $m_1 = m$ and $m_2 = 2m$:
$\frac{K_f}{K_i} = \left( \frac{m - 2m}{m + 2m} \right)^2 = \left( \frac{-m}{3m} \right)^2 = \left( -\frac{1}{3} \right)^2 = \frac{1}{9}$
The fraction of kinetic energy lost is:
$\frac{\Delta K}{K_i} = 1 - \frac{K_f}{K_i} = 1 - \frac{1}{9} = \frac{8}{9}$
Percentage loss in energy:
$\text{Percentage loss} = \frac{8}{9} \times 100 \approx 88.89\% \approx 90\%$

(C) For small angles,$\cos \theta \approx 1 - \frac{\theta^2}{2}$. The velocity of the bob of mass $m$ at the lowest point is $v = \sqrt{2gl(1 - \cos \theta_0)} \approx \sqrt{2gl(\frac{\theta_0^2}{2})} = \theta_0 \sqrt{gl}$.
After the elastic collision,the bob bounces back with velocity $v'$ and reaches angle $\theta_1$,so $v' = \theta_1 \sqrt{gl}$.
In an elastic collision with a stationary block of mass $M$,the velocity of the bob $m$ after collision is $v' = \left( \frac{m - M}{m + M} \right)v$.
Since the bob bounces back,its velocity is negative relative to the initial direction,so $v' = -\theta_1 \sqrt{gl}$.
Thus,$-\theta_1 \sqrt{gl} = \left( \frac{m - M}{m + M} \right) \theta_0 \sqrt{gl}$.
$-\theta_1 = \frac{m - M}{m + M} \theta_0 \implies -\theta_1(m + M) = \theta_0(m - M)$.
$-m\theta_1 - M\theta_1 = m\theta_0 - M\theta_0$.
$M(\theta_0 - \theta_1) = m(\theta_0 + \theta_1)$.
$M = m\left( \frac{\theta_0 + \theta_1}{\theta_0 - \theta_1} \right)$.

(D) Let the initial velocity of the alpha-particle be $v_0$ and its final velocity be $-v_1$ (since it is scattered backwards).
Let the mass of the nucleus be $M$ and its final velocity be $v_2$.
According to the conservation of linear momentum:
$mv_0 = -mv_1 + Mv_2$ --- $(1)$
According to the property of $1$-dimensional elastic collision:
$v_0 = v_1 + v_2$ --- $(2)$
From $(2)$,$v_1 = v_0 - v_2$. Substituting this into $(1)$:
$mv_0 = -m(v_0 - v_2) + Mv_2$
$mv_0 = -mv_0 + mv_2 + Mv_2$
$2mv_0 = (m + M)v_2 \Rightarrow v_2 = \frac{2mv_0}{m + M}$
The final kinetic energy of the alpha-particle is $K_f = \frac{1}{2}mv_1^2$.
Given that it loses $64\%$ of its initial kinetic energy $(K_i = \frac{1}{2}mv_0^2)$,the remaining kinetic energy is $36\%$ of $K_i$:
$K_f = 0.36 K_i \Rightarrow \frac{1}{2}mv_1^2 = 0.36 \times \frac{1}{2}mv_0^2$
$v_1^2 = 0.36 v_0^2 \Rightarrow v_1 = 0.6 v_0$
Using the formula for final velocity in elastic collision: $v_1 = \left( \frac{m - M}{m + M} \right) v_0$ (where $v_1$ is in the direction of $v_0$).
Since it is scattered backwards,$v_1 = -0.6 v_0$,so $\frac{m - M}{m + M} = -0.6$.
$m - M = -0.6m - 0.6M$
$1.6m = 0.4M \Rightarrow M = 4m$.

(B) Let the initial velocity of the $2\,kg$ body be $v_0$ and the final velocity be $v_0/4$. Let the mass of the second body be $m$ and its final velocity be $v$.
By the law of conservation of linear momentum:
$2v_0 = 2(v_0/4) + mv$
$2v_0 = v_0/2 + mv$
$mv = 3v_0/2$ --- $(1)$
Since the collision is perfectly elastic,the coefficient of restitution $e = 1$:
$e = (v_{2} - v_{1}) / (u_{1} - u_{2}) = 1$
$v - v_0/4 = v_0 - 0$
$v = v_0 + v_0/4 = 5v_0/4$
Substitute $v$ into equation $(1)$:
$m(5v_0/4) = 3v_0/2$
$m = (3/2) * (4/5) = 12/10 = 1.2\,kg$.

(A) For a one-dimensional elastic collision,the final velocity $v_1$ of the first particle (mass $m_1$) colliding with a stationary particle (mass $m_2$) is given by:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1$
Given $m_1 = m$,$m_2 = 2m$,and initial kinetic energy $E = \frac{1}{2} m u_1^2$.
The final kinetic energy $E'$ of the first particle is:
$E' = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m \left( \frac{m - 2m}{m + 2m} \right)^2 u_1^2$
$E' = \left( \frac{-m}{3m} \right)^2 \left( \frac{1}{2} m u_1^2 \right)$
$E' = \left( -\frac{1}{3} \right)^2 E = \frac{1}{9} E$
Thus,the kinetic energy of the first particle after the collision is $\frac{E}{9}$.

(D) In a one-dimensional elastic collision between two bodies of equal mass,where one body is initially at rest,the bodies exchange their velocities after the collision.
Since bob $A$ and bob $B$ have the same mass and the collision is elastic,bob $A$ transfers its entire velocity to bob $B$.
Consequently,bob $A$ comes to rest immediately after the collision,and bob $B$ starts moving with the velocity that bob $A$ had just before the collision.

(C) Newton's law of restitution states that the coefficient of restitution $e$ is the ratio of the relative velocity of separation to the relative velocity of approach.
$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}$
From the given diagram,the velocities just before impact are $u_1 = 3u$ and $u_2 = u$.
Therefore,the relative velocity of approach is $u_1 - u_2 = 3u - u = 2u$.
The velocities just after impact are $v_1$ and $v_2$.
Therefore,the relative velocity of separation is $v_2 - v_1$.
Substituting these into the formula:
$e = \frac{v_2 - v_1}{2u}$
$e \times 2u = v_2 - v_1$

(B) For an elastic collision between two bodies of masses $m_{1}$ and $m_{2}$,the law of conservation of momentum states:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})$ $...(i)$
The law of conservation of kinetic energy $(KE)$ states:
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$m_{1}(u_{1}^{2} - v_{1}^{2}) = m_{2}(v_{2}^{2} - u_{2}^{2})$ $...(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$u_{1} + v_{1} = v_{2} + u_{2}$
$(u_{1} - u_{2}) = (v_{2} - v_{1})$
$\frac{v_{2} - v_{1}}{u_{1} - u_{2}} = 1$
This ratio is the coefficient of restitution $e$. For a perfectly elastic collision,$e = 1$. By definition,the terms 'elastic' and 'perfectly elastic' refer to the same physical process where both momentum and kinetic energy are conserved.

(A) When a ball is dropped from a height $h$,its velocity just before the first impact is $v_0 = \sqrt{2gh}$.
After the first impact with a coefficient of restitution $e$,the velocity becomes $v_1 = ev_0 = e\sqrt{2gh}$.
The height reached after the first jump is $h_1 = \frac{v_1^2}{2g} = e^2h$.
After the second impact,the velocity becomes $v_2 = ev_1 = e^2\sqrt{2gh}$.
The height reached after the second jump is $h_2 = \frac{v_2^2}{2g} = e^4h$.
In general,the height attained after $n$ jumps is given by $h_n = e^{2n}h$.
For $n = 2$,the height is $h_2 = e^{2(2)}h = e^4h$.

(A) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $m_1 = 5\, kg$ be the mass of the first body and $u_1 = 10\, m/s$ be its initial velocity.
Let $m_2 = 20\, kg$ be the mass of the second body and $u_2 = 0\, m/s$ be its initial velocity.
After the collision,the first body comes to rest,so $v_1 = 0\, m/s$.
Let $v_2$ be the final velocity of the second body.
The conservation equation is: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
Substituting the values: $(5 \times 10) + (20 \times 0) = (5 \times 0) + (20 \times v_2)$.
$50 + 0 = 0 + 20 v_2$.
$50 = 20 v_2$.
$v_2 = \frac{50}{20} = 2.5\, m/s$.

(A) When two objects of equal mass undergo a perfectly elastic head-on collision,they exchange their velocities.
Let the masses of the four balls be $m_1 = m_2 = m_3 = m_4 = m$.
The initial velocity of the first ball is $v_1 = 0.4\, m/s$,and the others are at rest $(v_2 = v_3 = v_4 = 0)$.
$1$. Collision between ball $1$ and ball $2$: Ball $1$ comes to rest,and ball $2$ acquires velocity $0.4\, m/s$.
$2$. Collision between ball $2$ and ball $3$: Ball $2$ comes to rest,and ball $3$ acquires velocity $0.4\, m/s$.
$3$. Collision between ball $3$ and ball $4$: Ball $3$ comes to rest,and ball $4$ acquires velocity $0.4\, m/s$.
Thus,the velocity of the last ball is $0.4\, m/s$.

(B) Let the mass of the first body be $m_1 = 2 \ kg$ and its initial velocity be $u_1$. Let the mass of the second body be $m_2$,which is initially at rest $(u_2 = 0)$.
After an elastic collision,the final velocity $v_1$ of the first body is given by the formula:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1$
According to the problem,the first body continues to move in the original direction with one-fourth of its original speed,so $v_1 = \frac{u_1}{4}$.
Equating the two expressions for $v_1$:
$\frac{u_1}{4} = \left( \frac{2 - m_2}{2 + m_2} \right) u_1$
$\frac{1}{4} = \frac{2 - m_2}{2 + m_2}$
$2 + m_2 = 4(2 - m_2)$
$2 + m_2 = 8 - 4m_2$
$5m_2 = 6$
$m_2 = \frac{6}{5} = 1.2 \ kg$.

(B) In an elastic collision between two particles,the impulsive force acts between the particles during the short duration of the collision.
According to Newton's second law,the impulsive force changes the momentum of each individual particle.
However,for the system as a whole,the impulsive forces are internal forces.
According to the law of conservation of linear momentum,the total momentum of the system remains constant in the absence of any external force.
Therefore,the impulsive force cannot change the total momentum of the system.

(B) Let the mass of the heavy body be $M$ and the mass of the light body be $m = M/2$. Let the initial velocity of the heavy body be $u_1 = 6\,ms^{-1}$ and the light body be $u_2 = 0$.
For a one-dimensional elastic collision,the final velocity $v_2$ of the second body (light body) is given by the formula:
$v_2 = \frac{2M_1 u_1}{M_1 + M_2} + \frac{(M_2 - M_1) u_2}{M_1 + M_2}$
Substituting $M_1 = M$,$M_2 = M/2$,$u_1 = 6$,and $u_2 = 0$:
$v_2 = \frac{2M(6)}{M + M/2} + 0$
$v_2 = \frac{12M}{1.5M} = \frac{12}{1.5} = 8\,ms^{-1}$.

(C) Let the mass of each ball be $m$. The initial velocity of the first ball is $u$ and the second ball is at rest $(0)$.
The initial kinetic energy of the system is $KE_i = \frac{1}{2}mu^2$.
The loss in kinetic energy during a perfectly inelastic collision between two equal masses is given by $\Delta KE = \frac{1}{4}m(1-e^2)u^2$.
Given that the loss in kinetic energy is $\frac{1}{4}$ of the initial kinetic energy:
$\Delta KE = \frac{1}{4} KE_i = \frac{1}{4} (\frac{1}{2}mu^2) = \frac{1}{8}mu^2$.
Equating the two expressions for $\Delta KE$:
$\frac{1}{4}m(1-e^2)u^2 = \frac{1}{8}mu^2$.
Dividing both sides by $\frac{1}{4}mu^2$:
$1-e^2 = \frac{1}{2}$.
$e^2 = 1 - \frac{1}{2} = \frac{1}{2}$.
$e = \frac{1}{\sqrt{2}}$.

(C) When two particles of equal mass undergo a head-on perfectly elastic collision,they exchange their velocities.
In this scenario,the two moving balls collide with the first ball of the stationary row. Because the masses are identical and the collision is perfectly elastic,the momentum and kinetic energy are transferred through the chain.
The first moving ball transfers its velocity $V$ to the first stationary ball,which then transfers it to the next,and so on,until the velocity reaches the end of the row.
Since there are two incoming balls,this process happens for both. Consequently,the two balls at the extreme right end of the row will move off with speed $V$ each,while the remaining balls (the original six plus the two incoming ones) will come to rest.

(B) Let the mass of the neutron be $m$. The mass of the atom is $Am$.
Using the conservation of linear momentum: $mu + Am(0) = mv_1 + Amv_2$, which simplifies to $u = v_1 + Av_2$ (Equation $1$).
For an elastic head-on collision, the coefficient of restitution $e = 1$, so $v_2 - v_1 = u$ (Equation $2$).
From Equation $2$, $v_2 = u + v_1$. Substituting this into Equation $1$:
$u = v_1 + A(u + v_1) = v_1 + Au + Av_1 = v_1(1 + A) + Au$.
$v_1(1 + A) = u - Au = u(1 - A)$.
$v_1 = u \frac{1 - A}{1 + A}$.
The initial kinetic energy is $E = \frac{1}{2}mu^2$.
The final kinetic energy of the neutron is $E' = \frac{1}{2}mv_1^2 = \frac{1}{2}m \left( u \frac{1 - A}{1 + A} \right)^2$.
$E' = \left( \frac{1}{2}mu^2 \right) \left( \frac{1 - A}{1 + A} \right)^2 = E \left( \frac{A - 1}{A + 1} \right)^2$.

(D) In an elastic collision,the total kinetic energy is conserved throughout the process,including the time of contact. During the collision,kinetic energy is temporarily converted into elastic potential energy and then back into kinetic energy,so the total mechanical energy remains constant. Thus,the $Assertion$ is incorrect because kinetic energy is not conserved *during* the contact time (it is converted to potential energy).
The $Reason$ is also incorrect because the law of conservation of energy is a universal law; energy spent against friction is converted into heat or sound,and the total energy of the system (including heat/sound) remains conserved.

(A) In an elastic collision,kinetic energy is conserved,therefore the ball rebounds with the same velocity $u$ in the opposite direction.
According to Newton's second law,the force $F$ exerted by the surface is equal to the rate of change of momentum.
The change in momentum for one ball is $\Delta p = m(u - (-u)) = 2mu$.
Since $n$ balls collide per second,the total change in momentum per second is $n \times 2mu = 2mnu$.
Therefore,the force experienced by the surface is $F = 2mnu$.
Both the $Assertion$ and $Reason$ are correct,and the $Reason$ explains why the change in momentum is $2mu$ per ball.

(A) During the collision of two elastic bodies,the bodies undergo deformation.
As the bodies deform,the intermolecular distance between the particles decreases,leading to an increase in the elastic potential energy of the system.
According to the law of conservation of energy,the total energy of the system remains constant.
Since the elastic potential energy increases during the deformation phase of the collision,the kinetic energy $(K.E.)$ of the system must decrease to compensate for this change.
Therefore,both the $Assertion$ and the $Reason$ are correct,and the $Reason$ provides the correct explanation for the $Assertion$.

(B) For a one-dimensional elastic collision where body $B$ is initially at rest,the final velocity $v_1$ of body $A$ is given by:
$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u$
Substituting $m_1 = 4m$ and $m_2 = 2m$:
$v_1 = \left( \frac{4m - 2m}{4m + 2m} \right) u = \left( \frac{2m}{6m} \right) u = \frac{u}{3}$
Initial kinetic energy of body $A$ is $K_i = \frac{1}{2}(4m)u^2 = 2mu^2$.
Final kinetic energy of body $A$ is $K_f = \frac{1}{2}(4m)v_1^2 = \frac{1}{2}(4m)\left(\frac{u}{3}\right)^2 = \frac{2mu^2}{9}$.
The energy lost by body $A$ is $\Delta K = K_i - K_f = 2mu^2 - \frac{2mu^2}{9} = \frac{16mu^2}{9}$.
The fraction of energy lost is $\frac{\Delta K}{K_i} = \frac{16mu^2 / 9}{2mu^2} = \frac{16}{18} = \frac{8}{9}$.

(D) By the law of conservation of linear momentum:
$m \vec{u}_A + m \vec{u}_B = m \vec{v}_A + m \vec{v}_B$
Given $m = 0.1 \; kg$,$\vec{u}_A = 3 \hat{i} \; ms^{-1}$,$\vec{u}_B = 5 \hat{j} \; ms^{-1}$,and $\vec{v}_A = 4(\hat{i} + \hat{j}) \; ms^{-1}$.
Substituting the values:
$0.1(3 \hat{i}) + 0.1(5 \hat{j}) = 0.1(4 \hat{i} + 4 \hat{j}) + 0.1 \vec{v}_B$
Dividing by $0.1$:
$3 \hat{i} + 5 \hat{j} = 4 \hat{i} + 4 \hat{j} + \vec{v}_B$
$\vec{v}_B = (3-4) \hat{i} + (5-4) \hat{j} = -\hat{i} + \hat{j} \; ms^{-1}$.
The speed of $B$ after the collision is $|\vec{v}_B| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2} \; ms^{-1}$.
The kinetic energy of $B$ after the collision is $K_B = \frac{1}{2} m |\vec{v}_B|^2$.
$K_B = \frac{1}{2} (0.1) (\sqrt{2})^2 = \frac{1}{2} (0.1) (2) = 0.1 \; J$.
Since $K_B = \frac{x}{10} \; J$,we have $0.1 = \frac{x}{10}$,which implies $x = 1$.

$(53^{\circ})$ From the law of conservation of linear momentum, since the masses are equal $(m_{1} = m_{2})$, we have:
$\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}$
Taking the dot product of the velocity vector with itself:
$v_{1i}^{2} = (\vec{v}_{1f} + \vec{v}_{2f}) \cdot (\vec{v}_{1f} + \vec{v}_{2f})$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2} + 2\vec{v}_{1f} \cdot \vec{v}_{2f}$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2} + 2v_{1f}v_{2f} \cos(\theta_{1} + \theta_{2})$
Since the collision is elastic and the masses are equal, the conservation of kinetic energy implies:
$\frac{1}{2}m v_{1i}^{2} = \frac{1}{2}m v_{1f}^{2} + \frac{1}{2}m v_{2f}^{2}$
$v_{1i}^{2} = v_{1f}^{2} + v_{2f}^{2}$
Comparing the two expressions for $v_{1i}^{2}$, we get:
$2v_{1f}v_{2f} \cos(\theta_{1} + \theta_{2}) = 0$
Since $v_{1f} \neq 0$ and $v_{2f} \neq 0$, we must have $\cos(\theta_{1} + \theta_{2}) = 0$, which implies $\theta_{1} + \theta_{2} = 90^{\circ}$.
Given $\theta_{2} = 37^{\circ}$, we find:
$\theta_{1} = 90^{\circ} - 37^{\circ} = 53^{\circ}$.
Thus, the balls move at right angles to each other after the collision.

(B) It can be observed that the total momentum before and after the collision in each case is constant.
For an elastic collision,the total kinetic energy of a system remains conserved before and after the collision.
For mass of each ball bearing $m,$ we can write:
Total kinetic energy of the system before collision:
$= \frac{1}{2} m V^{2} + \frac{1}{2}(2m)(0)^{2} = \frac{1}{2} m V^{2}$
Case $(i)$:
Total kinetic energy of the system after collision
$= \frac{1}{2} m (0)^{2} + \frac{1}{2}(2m) \left( \frac{V}{2} \right)^{2} = \frac{1}{4} m V^{2}$
Since $\frac{1}{4} m V^{2} \neq \frac{1}{2} m V^{2}$,the kinetic energy is not conserved.
Case $(ii)$:
Total kinetic energy of the system after collision
$= \frac{1}{2}(2m)(0)^{2} + \frac{1}{2} m V^{2} = \frac{1}{2} m V^{2}$
Since $\frac{1}{2} m V^{2} = \frac{1}{2} m V^{2}$,the kinetic energy is conserved.
Case $(iii)$:
Total kinetic energy of the system after collision
$= \frac{1}{2}(3m) \left( \frac{V}{3} \right)^{2} = \frac{1}{6} m V^{2}$
Since $\frac{1}{6} m V^{2} \neq \frac{1}{2} m V^{2}$,the kinetic energy is not conserved.
Thus,only case $(ii)$ represents a possible result for an elastic collision.

(A) Bob $A$ will not rise at all.
In an elastic collision between two equal masses in which one is stationary,while the other is moving with some velocity,the stationary mass acquires the same velocity,while the moving mass immediately comes to rest after the collision. In this case,a complete transfer of momentum and kinetic energy takes place from the moving mass to the stationary mass.
Hence,bob $A$ of mass $m$,after colliding with bob $B$ of equal mass,will come to rest,while bob $B$ will move with the velocity of bob $A$ at the instant of collision. Since bob $A$ comes to rest at the lowest point,it will not rise at all.

(A) For an elastic collision between two hard billiard balls,the potential energy $V(r)$ must satisfy two conditions:
$1$. When the distance $r$ between the centers is greater than $2R$ (where $R$ is the radius of each ball),the balls do not interact,so the potential energy $V(r) = 0$.
$2$. When the distance $r$ is less than $2R$,the balls are in contact and undergo deformation,leading to a rapid increase in potential energy. At the point of contact $r = 2R$,the potential energy should be zero,and for $r < 2R$,it should be positive and increasing.
Looking at the given curves:
- Curve $(v)$ shows $V(r) = 0$ for $r \ge 2R$ and $V(r) > 0$ for $r < 2R$,which correctly describes the interaction.
- Curves $(i), (ii), (iii), (iv),$ and $(vi)$ do not satisfy these physical requirements. For example,$(ii)$ shows potential energy increasing with distance,which is incorrect,and $(i)$ and $(vi)$ show non-zero potential energy for $r > 2R$.
Therefore,the curves that cannot describe the elastic collision are $(i), (ii), (iii), (iv),$ and $(vi)$.

(N/A) Consider two bodies of masses $m_1$ and $m_2$ moving in a straight line along the $X$-direction. Let their initial velocities be $v_{1i}$ and $v_{2i}$ respectively,with $v_{1i} > v_{2i}$.
After the collision,let their final velocities be $v_{1f}$ and $v_{2f}$. In an elastic collision,both linear momentum and kinetic energy are conserved.
Conservation of linear momentum:
$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$
$m_1(v_{1i} - v_{1f}) = m_2(v_{2f} - v_{2i})$ ---$(1)$
Conservation of kinetic energy:
$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$
$m_1(v_{1i}^2 - v_{1f}^2) = m_2(v_{2f}^2 - v_{2i}^2)$
$m_1(v_{1i} - v_{1f})(v_{1i} + v_{1f}) = m_2(v_{2f} - v_{2i})(v_{2f} + v_{2i})$ ---$(2)$
Dividing equation $(2)$ by equation $(1)$:
$v_{1i} + v_{1f} = v_{2f} + v_{2i}$
$v_{1f} = v_{2f} + v_{2i} - v_{1i}$ ---$(3)$
Substituting $(3)$ into $(1)$:
$m_1(v_{1i} - (v_{2f} + v_{2i} - v_{1i})) = m_2(v_{2f} - v_{2i})$
$m_1(2v_{1i} - v_{2i} - v_{2f}) = m_2(v_{2f} - v_{2i})$
$2m_1 v_{1i} - m_1 v_{2i} - m_1 v_{2f} = m_2 v_{2f} - m_2 v_{2i}$
$2m_1 v_{1i} + (m_2 - m_1)v_{2i} = (m_1 + m_2)v_{2f}$
$v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} + \frac{m_2 - m_1}{m_1 + m_2}v_{2i}$
Similarly,for $v_{1f}$:
$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} + \frac{2m_2}{m_1 + m_2}v_{2i}$

(A) Case $1$: If the two masses are equal $(m_{1} = m_{2} = m)$:
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} + \left(\frac{2m_{2}}{m_{1}+m_{2}}\right) v_{2i}$. Assuming $v_{2i} = 0$,we get $v_{1f} = 0$ and $v_{2f} = v_{1i}$. The masses exchange their velocities.
Case $2$: If $m_{2} \gg m_{1}$ (a light body strikes a very heavy stationary body):
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{-m_{2}}{m_{2}}\right) v_{1i} = -v_{1i}$. The light body rebounds with the same speed.
$v_{2f} = \left(\frac{2m_{1}}{m_{1}+m_{2}}\right) v_{1i} \approx 0$. The heavy body remains practically at rest.
Case $3$: If $m_{1} \gg m_{2}$ (a heavy body strikes a light stationary body):
$v_{1f} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{m_{1}}{m_{1}}\right) v_{1i} = v_{1i}$. The heavy body continues to move with almost the same velocity.
$v_{2f} = \left(\frac{2m_{1}}{m_{1}+m_{2}}\right) v_{1i} \approx \left(\frac{2m_{1}}{m_{1}}\right) v_{1i} = 2v_{1i}$. The light body moves with twice the velocity of the heavy body.

(N/A) As shown in the figure,suppose a ball of mass $m_{1}$ moving in the $X$-direction with speed $v_{1i}$ collides elastically with a stationary ball of mass $m_{2}$.
After the collision,these balls move in directions making angles $\theta_{1}$ and $\theta_{2}$ with the $X$-axis with velocities $v_{1f}$ and $v_{2f}$ respectively.
Momentum is conserved in the collision,so the total momentum before collision equals the total momentum after collision.
Taking $X$-components of momentum:
$m_{1} v_{1i} = m_{1} v_{1f} \cos \theta_{1} + m_{2} v_{2f} \cos \theta_{2} \quad \dots (1)$
Taking $Y$-components of momentum:
$0 = m_{1} v_{1f} \sin \theta_{1} - m_{2} v_{2f} \sin \theta_{2} \quad \dots (2)$
Since the collision is elastic,kinetic energy is conserved:
$\frac{1}{2} m_{1} v_{1i}^{2} = \frac{1}{2} m_{1} v_{1f}^{2} + \frac{1}{2} m_{2} v_{2f}^{2} \quad \dots (3)$
Here,we have three independent equations $(1)$,$(2)$,and $(3)$. Typically,$m_{1}, m_{2},$ and $v_{1i}$ are known,while the four variables $v_{1f}, v_{2f}, \theta_{1},$ and $\theta_{2}$ are unknown. To solve for all unknowns,at least one of these four quantities must be known,as three equations can only determine three unknown quantities.

(N/A) head-on collision is a type of collision where the velocities of the colliding bodies are directed along the line joining their centers of mass. In such a collision,the bodies move along the same straight line before and after the impact.

(B) In a perfectly elastic collision between two identical bodies (masses $m_1 = m_2 = m$),the conservation of momentum and kinetic energy leads to the result that the bodies exchange their velocities after the collision.
If body $1$ has velocity $v_1$ and body $2$ has velocity $v_2$,after the collision,body $1$ will have velocity $v_2$ and body $2$ will have velocity $v_1$.

(N/A) collision is said to be a head-on collision (or one-dimensional collision) if the colliding bodies move along the same straight line before and after the collision.

(A) In a one-dimensional elastic collision between two bodies of masses $m_1$ and $m_2$,the kinetic energy transferred from the first body to the second is given by $\Delta K = \frac{4m_1 m_2}{(m_1 + m_2)^2} K_i$,where $K_i$ is the initial kinetic energy of the first body.
This transfer is maximum when the denominator $(m_1 + m_2)^2$ is minimized relative to the numerator,which occurs when $m_1 = m_2$.
Therefore,when two objects of equal mass undergo a head-on elastic collision,they exchange their velocities completely,resulting in the maximum possible energy transfer.

(C) In an elastic collision between two bodies of equal mass where one body is initially at rest,the conservation of linear momentum and kinetic energy leads to the condition:
$\vec{p}_1 = \vec{p}_1' + \vec{p}_2'$
Since the collision is elastic,$K_1 = K_1' + K_2'$,which implies $\frac{p_1^2}{2m} = \frac{p_1'^2}{2m} + \frac{p_2'^2}{2m}$.
This simplifies to $p_1^2 = p_1'^2 + p_2'^2$.
Comparing this with the vector equation $\vec{p}_1 = \vec{p}_1' + \vec{p}_2'$,we can see that this follows the Pythagorean theorem $(a^2 + b^2 = c^2)$.
Therefore,the vectors $\vec{p}_1'$ and $\vec{p}_2'$ must be perpendicular to each other.
Thus,the angle between the two bodies after the collision is $90^{\circ}$.