Inelastic Collision Questions in English

(C) The velocity of the sphere before impact is $\vec{v}_1 = 3 \hat{i} + \hat{j}$.
The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in the $yz$-plane (normal to the $x$-axis).
The component of velocity parallel to the wall ($\hat{j}$ component) remains unchanged during the collision because the impulse acts only along the normal to the wall.
Thus,$v_{y, \text{after}} = v_{y, \text{before}} = 1 \hat{j}$.
The component of velocity perpendicular to the wall ($\hat{i}$ component) changes according to the coefficient of restitution $e$.
The velocity of approach along the normal is $u_x = 3$.
The velocity of separation along the normal is $v_x = -e \cdot u_x = -\frac{1}{3} \cdot 3 = -1$.
Therefore,the velocity vector after the collision is $\vec{v}_2 = -1 \hat{i} + 1 \hat{j} = -\hat{i} + \hat{j}$.

(B) $1$. During the collision interval $\Delta t = t_2 - t_1$,the impulsive force exerted by the bullet on the bob and vice-versa is internal to the system (bullet + bob). The external forces (gravity and tension) are non-impulsive. Therefore,the total linear momentum of the system is conserved during the interval $\Delta t$.
$2$. Kinetic energy is not conserved because the collision is inelastic (the bullet passes through the bob,causing deformation and heat).
$3$. Mechanical energy is not conserved during the collision due to energy loss. After the collision (from $t_2$ to $t_3$),mechanical energy is conserved for the bob as it swings up,but not for the whole system including the bullet.
$4$. Momentum of the bob alone is not conserved after $t_2$ because gravity and tension act on it.
$5$. Thus,the correct statement is that the total momentum of the bob and the bullet is conserved during the interval $\Delta t = t_2 - t_1$.

(C) In a collision where a projectile of mass $m_1$ strikes a stationary target of mass $m_2$,and the two particles move at an angle of $90^{\circ}$ to each other,the condition is given by $m_2 = m_1 \cdot e$,where $e$ is the coefficient of restitution.
For an inelastic collision,kinetic energy is lost,which implies $e < 1$.
Substituting $e < 1$ into the condition $m_2 = m_1 \cdot e$,we get $m_2 < m_1$,which means $m_1 > m_2$.
Since the target nucleus is ${}^4He$ $(m_2 = 4 \text{ amu})$,the projectile nucleus must have a mass $m_1 > 4 \text{ amu}$.
Among the given options,${}^{28}N$ and ${}^{12}C$ are heavier than ${}^4He$. However,in standard physics problems of this type,the projectile is typically heavier than the target to satisfy the scattering geometry. ${}^{12}C$ is a valid candidate.

(C) Since the balls are identical,let their mass be $m$.
According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.
Initial momentum in $x$-direction: $P_{ix} = m(v_{Px}) + m(v_{Qx}) = m(6) + m(-5) = m \ kg \ m/s$.
Initial momentum in $y$-direction: $P_{iy} = m(v_{Py}) + m(v_{Qy}) = m(0) + m(2) = 2m \ kg \ m/s$.
After the collision,ball $P$ comes to rest,so $v'_{Px} = 0$ and $v'_{Py} = 0$.
Let the velocity components of ball $Q$ after the collision be $v'_{Qx}$ and $v'_{Qy}$.
Final momentum in $x$-direction: $P_{fx} = m(0) + m(v'_{Qx}) = m(v'_{Qx})$.
Final momentum in $y$-direction: $P_{fy} = m(0) + m(v'_{Qy}) = m(v'_{Qy})$.
Equating initial and final momenta:
$m(v'_{Qx}) = m \implies v'_{Qx} = 1 \ m/s$.
$m(v'_{Qy}) = 2m \implies v'_{Qy} = 2 \ m/s$.
Therefore,the velocity components of ball $Q$ after the collision are $v_x = 1 \ m/s$ and $v_y = 2 \ m/s$.

(C) Let $h = 1\,m = 100\,cm$ be the initial height and $h_1 = 25\,cm$ be the height after the first rebound.
When a body is dropped from a height $h$,its velocity just before hitting the floor is $u = \sqrt{2gh}$.
After the rebound,if the body rises to a height $h_1$,its velocity just after the collision is $v = \sqrt{2gh_1}$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.
$e = \frac{v}{u} = \frac{\sqrt{2gh_1}}{\sqrt{2gh}} = \sqrt{\frac{h_1}{h}}$.
Substituting the given values:
$e = \sqrt{\frac{25\,cm}{100\,cm}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

(A) The particle falls from height $h$. After the first collision,it rebounds to height $h_1 = e^2 h$.
After the second collision,it rebounds to height $h_2 = e^2 h_1 = e^4 h$,and so on.
The total distance $S$ traveled is the sum of the initial fall and all subsequent upward and downward paths:
$S = h + 2h_1 + 2h_2 + 2h_3 + \dots$
$S = h + 2(e^2 h) + 2(e^4 h) + 2(e^6 h) + \dots$
$S = h + 2e^2 h (1 + e^2 + e^4 + \dots)$
This is a geometric series with first term $a = 1$ and common ratio $r = e^2$. The sum is $\frac{1}{1 - e^2}$.
$S = h + 2e^2 h \left( \frac{1}{1 - e^2} \right)$
$S = h \left( 1 + \frac{2e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right)$
$S = h \left( \frac{1 + e^2}{1 - e^2} \right)$

(D) The collision occurs along the line joining the centers of the two balls. The angle between the velocity vector $V$ and the line of impact is $30^{\circ}$.
The component of velocity along the line of impact is $u = V \cos 30^{\circ} = \frac{\sqrt{3}V}{2}$.
Let $v_1$ be the velocity of the ball of mass $2m$ and $v_2$ be the velocity of the ball of mass $m$ along the line of impact after the collision.
Applying the law of conservation of linear momentum along the line of impact:
$m u = m v_2 + (2m) v_1$
$u = v_2 + 2v_1$ --- $(1)$
Applying the coefficient of restitution formula $(e = 1/2)$:
$e = \frac{v_1 - v_2}{u}$
$\frac{1}{2} = \frac{v_1 - v_2}{u}$
$u = 2v_1 - 2v_2$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$2u = 4v_1 - v_2 + v_2 = 4v_1$
$v_1 = \frac{u}{2} = \frac{\sqrt{3}V}{2 \times 2} = \frac{\sqrt{3}V}{4}$.

(C) The velocity of the ball just before the collision is given by $u = \sqrt{2gh}$,where $h = 25 \ cm = 0.25 \ m$.
$u = \sqrt{2 \times 9.8 \times 0.25} = \sqrt{4.9} \ m/s$.
The velocity of the ball just after the collision is given by $v = \sqrt{2gh'}$,where $h' = 4 \ cm = 0.04 \ m$.
$v = \sqrt{2 \times 9.8 \times 0.04} = \sqrt{0.784} \ m/s$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach: $e = \frac{v}{u}$.
$e = \sqrt{\frac{2gh'}{2gh}} = \sqrt{\frac{h'}{h}}$.
Substituting the values: $e = \sqrt{\frac{4 \ cm}{25 \ cm}} = \sqrt{\frac{4}{25}} = \frac{2}{5} = 0.4$.

(C) Let $m_b = 0.01\, kg$ be the mass of the bullet,$v_i = 500\, m/s$ be its initial velocity,and $v_f$ be its final velocity.
Let $M = 2\, kg$ be the mass of the block and $V$ be the velocity of the block immediately after the bullet emerges.
Using the principle of conservation of mechanical energy for the block,the kinetic energy gained by the block is equal to the potential energy gained: $\frac{1}{2} M V^2 = Mgh$.
$V = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4\, m/s$.
Now,applying the principle of conservation of linear momentum in the horizontal direction:
$m_b v_i = m_b v_f + M V$.
$0.01 \times 500 = 0.01 \times v_f + 2 \times 1.4$.
$5 = 0.01 v_f + 2.8$.
$0.01 v_f = 5 - 2.8 = 2.2$.
$v_f = \frac{2.2}{0.01} = 220\, m/s$.

(D) The change in momentum during each collision is the impulse imparted to the floor.
For the first impact, the ball arrives with momentum $p$ (downward) and rebounds with $ep$ (upward). The change in momentum is $\Delta p_1 = p - (-ep) = p(1+e)$.
For the second impact, the ball arrives with momentum $ep$ (downward) and rebounds with $e^2p$ (upward). The change in momentum is $\Delta p_2 = ep - (-e^2p) = ep(1+e)$.
For the third impact, the change is $\Delta p_3 = e^2p(1+e)$, and so on.
The total momentum imparted is the sum of all these impulses:
$\Delta p_{total} = p(1+e) + ep(1+e) + e^2p(1+e) + \dots$
$\Delta p_{total} = p(1+e) [1 + e + e^2 + \dots]$
Using the sum of an infinite geometric series $S = \frac{1}{1-e}$ for $|e| < 1$:
$\Delta p_{total} = p(1+e) \cdot \frac{1}{1-e} = p(\frac{1+e}{1-e})$.

(B) Let the velocity of the particle before impact be $u$ and after impact be $v$.
The component of velocity parallel to the ground remains unchanged because there is no impulsive force acting parallel to the surface:
$v \sin \alpha = u \sin \theta$ --- $(1)$
The component of velocity perpendicular to the ground changes according to the coefficient of restitution $e$:
$v \cos \alpha = e u \cos \theta$ --- $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{v \sin \alpha}{v \cos \alpha} = \frac{u \sin \theta}{e u \cos \theta}$
$\tan \alpha = \frac{\tan \theta}{e}$
Rearranging the terms,we get:
$\frac{\tan \theta}{\tan \alpha} = e$

(A) Let the mass of both balls be $m$. Let the initial velocity of ball $P$ be $u_1$ and ball $Q$ be $u_2 = 0$. After the collision,let their velocities be $v_1$ and $v_2$ respectively.
From the conservation of linear momentum: $m u_1 + 0 = m v_1 + m v_2 \implies u_1 = v_1 + v_2$.
From the definition of the coefficient of restitution $e$: $e = \frac{v_2 - v_1}{u_1 - u_2} \implies e u_1 = v_2 - v_1$.
Adding the two equations: $(1+e) u_1 = 2 v_2 \implies v_2 = \frac{1+e}{2} u_1$.
Subtracting the equations: $(1-e) u_1 = 2 v_1 \implies v_1 = \frac{1-e}{2} u_1$.
Given that $v_2 = 2 v_1$,we substitute the expressions:
$\frac{1+e}{2} u_1 = 2 \left( \frac{1-e}{2} \right) u_1$.
$1 + e = 2(1 - e) \implies 1 + e = 2 - 2e$.
$3e = 1 \implies e = 1/3$.

(D) Let $u = 6\, m/s$ be the initial downward velocity and $H = 3.2\, m$ be the height.
First,calculate the velocity $v$ of the ball just before hitting the floor using the equation $v^2 = u^2 + 2gH$:
$v^2 = 6^2 + 2 \times 10 \times 3.2 = 36 + 64 = 100$
$v = 10\, m/s$.
Let $v'$ be the velocity of the ball just after the rebound. Since the ball rebounds to the same height $H$,the velocity required to reach height $H$ is $v' = \sqrt{2gH} = \sqrt{2 \times 10 \times 3.2} = \sqrt{64} = 8\, m/s$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach:
$e = \frac{v'}{v} = \frac{8}{10} = 0.8$.

(B) When a ball strikes a surface,the component of velocity parallel to the surface remains unchanged (assuming no friction),while the component perpendicular to the surface changes according to the coefficient of restitution $e$.
$1$. The horizontal component of velocity remains constant:
$v' \sin \theta = v \sin 45^{\circ}$ --- $(1)$
$2$. The vertical component of velocity after impact is given by $v'_{y} = e v_{y}$:
$v' \cos \theta = e (v \cos 45^{\circ})$ --- $(2)$
$3$. Dividing equation $(1)$ by equation $(2)$:
$\frac{v' \sin \theta}{v' \cos \theta} = \frac{v \sin 45^{\circ}}{e v \cos 45^{\circ}}$
$\tan \theta = \frac{\tan 45^{\circ}}{e}$
$4$. Given $e = 1/\sqrt{3}$ and $\tan 45^{\circ} = 1$:
$\tan \theta = \frac{1}{1/\sqrt{3}} = \sqrt{3}$
$5$. Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(A) When the object hits the ground,its kinetic energy becomes $50\%$ of its initial value. Let $v$ be the velocity just before impact and $v'$ be the velocity just after impact.
$\frac{1}{2}m(v')^2 = \frac{50}{100} \times \frac{1}{2}mv^2 \Rightarrow v' = \frac{v}{\sqrt{2}}$.
The coefficient of restitution $e$ is defined as $e = \frac{v'}{v} = \frac{1}{\sqrt{2}}$.
The total distance $H$ covered by an object dropped from height $h$ undergoing multiple bounces is given by the formula:
$H = h + 2h(e^2) + 2h(e^4) + 2h(e^6) + \dots$
This is a geometric series: $H = h + 2h \left( \frac{e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right) = h \left( \frac{1 + e^2}{1 - e^2} \right)$.
Substituting $e^2 = \frac{1}{2}$:
$H = h \left( \frac{1 + 1/2}{1 - 1/2} \right) = h \left( \frac{3/2}{1/2} \right) = 3h$.

(B) According to the law of conservation of linear momentum in the original direction of motion:
Initial momentum = Final momentum
$m(2v) + (2m)(v) = m(0) + m(v') \cos(45^o) + m(v') \cos(45^o)$
$2mv + 2mv = 0 + 2mv' \cos(45^o)$
$4mv = 2mv' (1 / \sqrt{2})$
$4v = v' \sqrt{2}$
$v' = 4v / \sqrt{2} = 2\sqrt{2}v$
Therefore,the speed of each of the moving particles is $2\sqrt{2}v$.

(D) The initial kinetic energy of the ball just before impact is $K_i = \frac{1}{2} m v_i^2$.
After the collision,the velocity of the ball is $v_f = e v_i$,where $e$ is the coefficient of restitution.
The final kinetic energy is $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (e v_i)^2 = e^2 K_i$.
The loss in kinetic energy is $\Delta K = K_i - K_f = K_i - e^2 K_i = K_i(1 - e^2)$.
The percentage loss in energy is $\frac{\Delta K}{K_i} \times 100\% = (1 - e^2) \times 100\%$.
Given $e = 0.5$,the percentage loss is $(1 - (0.5)^2) \times 100\% = (1 - 0.25) \times 100\% = 0.75 \times 100\% = 75\%$.

(C) Let the initial velocity of the ball just before the first impact be $u = \sqrt{2gh}$.
After the first rebound,the velocity of the ball is $v_1 = eu = e\sqrt{2gh}$.
The height attained after the first rebound is $h_1 = \frac{v_1^2}{2g} = \frac{e^2(2gh)}{2g} = e^2h$.
After the second rebound,the velocity of the ball is $v_2 = ev_1 = e(e\sqrt{2gh}) = e^2\sqrt{2gh}$.
The height attained after the second rebound is $h_2 = \frac{v_2^2}{2g} = \frac{(e^2\sqrt{2gh})^2}{2g} = \frac{e^4(2gh)}{2g} = e^4h$.

(B) The time taken for the ball to reach the surface for the first time is given by $t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 4.9}{9.8}} = \sqrt{1} = 1 \, s$.
After the first impact,the velocity of the ball just before hitting the surface is $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.9} = 9.8 \, m/s$.
The velocity of the ball just after the first bounce is $v' = e \times v = \frac{3}{4} \times 9.8 = 7.35 \, m/s$.
The time taken to reach the maximum height and return to the surface (time of flight for the second bounce) is $t_2 = \frac{2v'}{g} = \frac{2 \times 7.35}{9.8} = 1.5 \, s$.
Thus,the ball hits the surface for the second time $1.5 \, s$ after the first impact.

(C) Let the mass of each ball be $m$. The initial velocity of the first ball is $u_1 = 6\, m/s$ and the second ball is $u_2 = 0$.
Let the final speed of each ball after the collision be $v$.
According to the law of conservation of linear momentum along the original direction of motion ($X$-axis):
$m u_1 + m u_2 = m v \cos 30^{\circ} + m v \cos 30^{\circ}$
$m(6) + m(0) = 2 m v \cos 30^{\circ}$
$6 = 2 v \left( \frac{\sqrt{3}}{2} \right)$
$6 = v \sqrt{3}$
$v = \frac{6}{\sqrt{3}} = 2\sqrt{3}\, m/s$.

(D) In any collision (elastic or inelastic),the total linear momentum of the system remains conserved,provided no external force acts on the system.
In an inelastic collision,the kinetic energy is not conserved,but the linear momentum is always conserved.

(A-D) Let $v_{1}$ and $v_{2}$ be the velocities of the two balls after the collision.
From the law of conservation of linear momentum:
$2mv_0 = mv_1 + mv_2$
$\therefore 2v_0 = v_1 + v_2$
From the definition of the coefficient of restitution $e$:
$e = \frac{v_2 - v_1}{2v_0 - 0} \Rightarrow v_2 - v_1 = 2v_0e$
Adding the two equations:
$(v_1 + v_2) + (v_2 - v_1) = 2v_0 + 2v_0e$
$2v_2 = 2v_0(1 + e) \Rightarrow v_2 = v_0(1 + e)$
Subtracting the equations:
$(v_1 + v_2) - (v_2 - v_1) = 2v_0 - 2v_0e$
$2v_1 = 2v_0(1 - e) \Rightarrow v_1 = v_0(1 - e)$
Since $0 < e < 1$,both $v_1$ and $v_2$ are positive,meaning both balls move forward.
$(b)$ For a general collision,by the law of conservation of linear momentum:
$\vec{p} = \vec{p_1} + \vec{p_2}$
For an inelastic collision,kinetic energy is lost,so:
$\frac{p^2}{2m} > \frac{p_1^2}{2m} + \frac{p_2^2}{2m} \Rightarrow p^2 > p_1^2 + p_2^2$
Using the law of cosines for the vector triangle formed by $\vec{p}, \vec{p_1},$ and $\vec{p_2}$:
$p^2 = p_1^2 + p_2^2 - 2p_1p_2 \cos(180^o - \theta) = p_1^2 + p_2^2 + 2p_1p_2 \cos \theta$
Since $p^2 > p_1^2 + p_2^2$,it follows that $2p_1p_2 \cos \theta > 0$,which implies $\cos \theta > 0$.
Therefore,$\theta < 90^o$.

(N/A) Consider the diagram where bob $B$ is displaced through an angle $\theta$ and released.
At $t=0$,suppose bob $B$ is displaced by $\theta=10^{\circ}$ to the right. It is given potential energy $E_{1}=E$. Energy of $A$,$E_{2}=0$.
When $B$ is released,it strikes $A$ at $t=T/4$. In the head-on elastic collision between identical masses,they exchange velocities. Thus,$B$ comes to rest and $A$ gets the velocity of $B$. Therefore,$E_{1}=0$ and $E_{2}=E$.
At $t=2T/4$,$B$ reaches its extreme right position when $KE$ of $A$ is converted into $PE=E_{2}=E$. Energy of $B$,$E_{1}=0$.
At $t=3T/4$,$A$ reaches its mean position,when its $PE$ is converted into $KE=E_{2}=E$. It collides elastically with $B$ and transfers its entire energy to $B$. Thus,$E_{2}=0$ and $E_{1}=E$. The entire process is repeated.
$(b)$ The values of energies of $B$ and $A$ at different time intervals are tabulated below:

Time $(t)$	Energy of $B$ $(E_{1})$	Energy of $A$ $(E_{2})$
$0$	$E$	$0$
$T/4$	$0$	$E$
$2T/4$	$0$	$E$
$3T/4$	$E$	$0$
$4T/4$	$E$	$0$
$5T/4$	$0$	$E$
$6T/4$	$0$	$E$
$7T/4$	$E$	$0$
$8T/4$	$E$	$0$

(B) According to the principle of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.
Before the collision,the momentum of the system is $p_i = mv$.
After the collision,the particle comes to rest,so its momentum is $0$. Let the velocity of the center of mass of the rod be $V_{cm}$. The momentum of the rod is $MV_{cm}$.
By conservation of linear momentum: $mv = 0 + MV_{cm}$.
Therefore,the velocity of the center of mass of the rod is $V_{cm} = mv/M$.

(D) The initial height is $h = 5\, m$. The ball rises to a fraction $f = \frac{81}{100}$ of its previous height after each bounce. Thus,$e^2 = \frac{81}{100}$,which means the coefficient of restitution $e = 0.9$.
Total distance $S = h + 2(fh) + 2(f^2h) + \dots = h + 2h(f + f^2 + \dots) = h + 2h \left( \frac{f}{1-f} \right) = h \left( 1 + \frac{2f}{1-f} \right) = h \left( \frac{1+f}{1-f} \right)$.
Substituting $h = 5$ and $f = 0.81$: $S = 5 \left( \frac{1.81}{0.19} \right) = 5 \times \frac{181}{19} \approx 47.63\, m$.
Total time $t = \sqrt{\frac{2h}{g}} + 2e\sqrt{\frac{2h}{g}} + 2e^2\sqrt{\frac{2h}{g}} + \dots = \sqrt{\frac{2h}{g}} \left( 1 + 2e + 2e^2 + \dots \right) = \sqrt{\frac{2h}{g}} \left( 1 + \frac{2e}{1-e} \right) = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} \right)$.
Substituting $h = 5, g = 10, e = 0.9$: $t = \sqrt{\frac{2 \times 5}{10}} \left( \frac{1+0.9}{1-0.9} \right) = 1 \times \frac{1.9}{0.1} = 19\, s$.
Average speed $v_{av} = \frac{S}{t} = \frac{5 \times 1.81 / 0.19}{19} = \frac{5 \times 181 / 19}{19} = \frac{905}{361} \approx 2.5\, m/s$.

(D) Let the mass of each ball be $m$. The initial velocity of ball $A$ is $u_1 = 9\, m/s$ and ball $B$ is $u_2 = 0$.
After the collision,let the velocities of balls $A$ and $B$ be $v_1$ and $v_2$ respectively,both at an angle of $30^{\circ}$ with the original direction.
According to the law of conservation of linear momentum along the $y$-axis (perpendicular to the initial direction of motion):
$\sum P_{iy} = \sum P_{fy}$
$0 = m v_1 \sin 30^{\circ} - m v_2 \sin 30^{\circ}$
Since the masses are identical and $\sin 30^{\circ} \neq 0$,we get:
$v_1 \sin 30^{\circ} = v_2 \sin 30^{\circ}$
$v_1 = v_2$
Therefore,the ratio of the velocities $v_1 : v_2$ is $1 : 1$.
Thus,$x = 1$.

(B) In this collision,the electron transfers a portion of its kinetic energy to the molecule to excite it from the ground state to a metastable excited state.
Since some kinetic energy is converted into the internal potential energy of the molecule,the total kinetic energy of the system decreases. Therefore,the collision is inelastic,and $K^{\prime} < K$.
According to the law of conservation of linear momentum,the total momentum of an isolated system remains constant regardless of whether the collision is elastic or inelastic.
Thus,the total initial momentum $p$ must be equal to the total final momentum $p^{\prime}$,so $p = p^{\prime}$.
Combining these,we get $K^{\prime} < K$ and $p = p^{\prime}$.

(B) For a head-on collision between two bodies of equal mass $m$,where the second body is initially at rest $(u_2 = 0)$ and the first body has initial velocity $u_1 = v$:
$1$. Conservation of linear momentum: $mv + 0 = mv_1 + mv_2 \implies v = v_1 + v_2$.
$2$. Coefficient of restitution definition: $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{v - 0} \implies ev = v_2 - v_1$.
$3$. Adding the two equations: $v + ev = (v_1 + v_2) + (v_2 - v_1) = 2v_2 \implies v_2 = \frac{(1+e)v}{2}$.
$4$. Subtracting the equations: $v - ev = (v_1 + v_2) - (v_2 - v_1) = 2v_1 \implies v_1 = \frac{(1-e)v}{2}$.
Thus,the correct option is $B$.

(B) From the figure,the ball approaches the surface at an angle of $30^{\circ}$ with the normal and rebounds at the same angle.
Initial velocity vector: $\vec{v}_i = (20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) \, m/s$
Final velocity vector: $\vec{v}_f = (-20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) \, m/s$
Change in velocity: $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$
$\Delta \vec{v} = (-20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) - (20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j})$
$\Delta \vec{v} = -40 \sin 30^{\circ} \hat{i}$
Magnitude of change in velocity: $|\Delta \vec{v}| = 40 \times \sin 30^{\circ} = 40 \times 0.5 = 20 \, m/s$.
Wait,re-evaluating based on the standard reflection geometry: The component parallel to the surface remains unchanged,while the component perpendicular to the surface reverses.
Let the normal be along the $y$-axis.
$\vec{v}_i = v \sin 30^{\circ} \hat{i} - v \cos 30^{\circ} \hat{j}$
$\vec{v}_f = -v \sin 30^{\circ} \hat{i} - v \cos 30^{\circ} \hat{j}$
$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = -2v \sin 30^{\circ} \hat{i} = -2(20)(0.5) \hat{i} = -20 \hat{i}$.
If the normal is along the $x$-axis:
$\vec{v}_i = -v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$
$\vec{v}_f = v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$
$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 2v \cos 30^{\circ} \hat{i} = 2(20)(\frac{\sqrt{3}}{2}) \hat{i} = 20\sqrt{3} \hat{i}$.
Given the options,the magnitude is $20\sqrt{3}$.

(D) Let the masses of the two balls be $m_1 = m_2 = m$.
Let the initial velocities be $u_1 = 6 \, m/s$ and $u_2 = -6 \, m/s$.
The coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$,where $v_1$ and $v_2$ are the final velocities.
Since the masses are equal and the collision is head-on,the balls will exchange velocities if $e=1$. For a general $e$,the final velocities are given by $v_1 = \frac{(m_1 - em_2)u_1 + m_2(1+e)u_2}{m_1 + m_2}$ and $v_2 = \frac{(m_2 - em_1)u_2 + m_1(1+e)u_1}{m_1 + m_2}$.
Substituting $m_1 = m_2 = m$,$u_1 = 6$,and $u_2 = -6$:
$v_1 = \frac{(m - em)(6) + m(1+e)(-6)}{2m} = \frac{6 - 6e - 6 - 6e}{2} = -6e$.
$v_2 = \frac{(m - em)(-6) + m(1+e)(6)}{2m} = \frac{-6 + 6e + 6 + 6e}{2} = 6e$.
Given $e = 1/3$,the speed of each ball is $|v| = 6 \times (1/3) = 2 \, m/s$.

(A) The surface is smooth and horizontal,so there is no impulsive force acting in the $x$-direction during the collision. Therefore,the velocity component in the $x$-direction remains unchanged.
$v_x = u_x = 2 \, m/s$
In the $y$-direction (normal to the surface),the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach.
$e = \frac{v_y}{u_y}$
Given $e = \frac{1}{2}$ and $u_y = 4 \, m/s$,we have:
$v_y = e \times u_y = \frac{1}{2} \times 4 = 2 \, m/s$
Thus,the components just after the collision are $v_x = 2 \, m/s$ and $v_y = 2 \, m/s$.

(C) $1$. Calculate the initial kinetic energy of the bullet: $KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.01 \,kg \times (20 \,m/s)^2 = 2 \,J$.
$2$. Calculate the final velocity of the system (bullet + ice) using conservation of momentum: $m_b v_b = (m_b + m_i) v_f \Rightarrow 0.01 \times 20 = (0.01 + 0.99) v_f \Rightarrow v_f = 0.2 \,m/s$.
$3$. Calculate the final kinetic energy of the system: $KE_f = \frac{1}{2} (m_b + m_i) v_f^2 = \frac{1}{2} \times 1 \,kg \times (0.2 \,m/s)^2 = 0.02 \,J$.
$4$. Calculate the lost kinetic energy: $\Delta KE = KE_i - KE_f = 2 - 0.02 = 1.98 \,J$.
$5$. Energy available for melting ice: $Q = 50\% \text{ of } \Delta KE = 0.5 \times 1.98 = 0.99 \,J \approx 1 \,J$.
$6$. Amount of ice melted: $m_{melt} = \frac{Q}{L} = \frac{1 \,J}{336 \,J/g} \approx 0.00297 \,g \approx 0.003 \,g$.

(C) When a ball is dropped from a height $h$ and hits a floor with a coefficient of restitution $e$,the height $h^{\prime}$ to which it rebounds is given by the formula: $h^{\prime} = e^2 h$.
Given:
Initial height $h = 20\,m$.
Coefficient of restitution $e = 0.5$.
Substituting the values into the formula:
$h^{\prime} = (0.5)^2 \times 20\,m$.
$h^{\prime} = 0.25 \times 20\,m$.
$h^{\prime} = 5\,m$.
Therefore,the ball rebounds to a height of $5\,m$.

(C) Let $V_1$ be the velocity just before hitting the ground and $V_2$ be the velocity just after hitting the ground.
Given the ratio of velocities is $\frac{V_1}{V_2} = 4$,which implies $V_1 = 4V_2$.
The kinetic energy just before hitting the ground is $KE_{before} = \frac{1}{2}mV_1^2$.
The kinetic energy just after hitting the ground is $KE_{after} = \frac{1}{2}mV_2^2 = \frac{1}{2}m\left(\frac{V_1}{4}\right)^2 = \frac{1}{2}mV_1^2 \times \frac{1}{16}$.
The loss in kinetic energy is $\Delta KE = KE_{before} - KE_{after} = \frac{1}{2}mV_1^2 \left(1 - \frac{1}{16}\right) = \frac{15}{32}mV_1^2$.
The percentage loss in kinetic energy is $\frac{\Delta KE}{KE_{before}} \times 100 = \left(1 - \frac{1}{16}\right) \times 100 = \frac{15}{16} \times 100 = \frac{1500}{16} = \frac{375}{4} \%$.
Comparing this with $\frac{x}{4} \%$,we get $x = 375$.

(C) For a head-on collision between two identical masses $m$ where one is at rest,the loss in kinetic energy $\Delta KE$ is given by:
$\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (1 - e^2) (u_1 - u_2)^2$
Since $m_1 = m_2 = m$,$u_1 = u$,and $u_2 = 0$:
$\Delta KE = \frac{1}{2} \frac{m^2}{2m} (1 - e^2) u^2 = \frac{1}{4} m u^2 (1 - e^2)$
Given that $\Delta KE = \frac{1}{4} KE_{initial} = \frac{1}{4} (\frac{1}{2} m u^2)$:
$\frac{1}{4} m u^2 (1 - e^2) = \frac{1}{4} (\frac{1}{2} m u^2)$
$(1 - e^2) = \frac{1}{2}$
$e^2 = 1 - \frac{1}{2} = \frac{1}{2}$
$e = \frac{1}{\sqrt{2}}$

(B) Let the mass of each ball be $m$. The initial velocity of the first ball is $u = 9 \ m/s$ and the second ball is at rest $(u_2 = 0)$.
According to the Law of Conservation of Linear Momentum $(COLM)$ along the initial direction of motion:
$m \times u = m \times v \cos 30^{\circ} + m \times v \cos 30^{\circ}$
$m \times 9 = 2 \times m \times v \times \cos 30^{\circ}$
$9 = 2 \times v \times \frac{\sqrt{3}}{2}$
$9 = v \sqrt{3}$
$v = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \ m/s$
Thus,the speed of both balls after the collision is $3 \sqrt{3} \ m/s$.

(C) The loss in kinetic energy during a perfectly inelastic collision between two bodies of equal mass $m$,where one is at rest,is given by the formula: $\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (1 - e^2) (u_1 - u_2)^2$.
Given $m_1 = m_2 = m$,$u_1 = u$,and $u_2 = 0$,the formula becomes: $\Delta KE = \frac{1}{2} \frac{m^2}{2m} (1 - e^2) u^2 = \frac{1}{4} m (1 - e^2) u^2$.
We are given that the loss in kinetic energy is $1/4$ of the initial kinetic energy $(KE_i = \frac{1}{2} m u^2)$.
So,$\frac{1}{4} m (1 - e^2) u^2 = \frac{1}{4} (\frac{1}{2} m u^2)$.
Dividing both sides by $\frac{1}{4} m u^2$,we get: $1 - e^2 = \frac{1}{2}$.
Therefore,$e^2 = 1 - 1/2 = 1/2$.
Taking the square root,we get: $e = 1 / \sqrt{2}$.

(A) Let the mass of each ball be $m$. Let the initial velocity of the first ball be $u$ and the second ball be $0$. Let the velocities after the collision be $v_1$ and $v_2$ respectively. According to the problem,$v_2 = 2v_1$.
By the Conservation of Linear Momentum $(COLM)$:
$m(u) + m(0) = m(v_1) + m(v_2)$
$u = v_1 + v_2$
Since $v_2 = 2v_1$,we have $u = v_1 + 2v_1 = 3v_1$,which implies $v_1 = \frac{u}{3}$ and $v_2 = \frac{2u}{3}$.
The coefficient of restitution $e$ is defined as:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Substituting the values:
$e = \frac{2v_1 - v_1}{u - 0} = \frac{v_1}{u}$
Since $v_1 = \frac{u}{3}$,we get:
$e = \frac{u/3}{u} = \frac{1}{3}$

(D) The velocity of the ball before collision is $\vec{v}_{initial} = a \hat{i} - b \hat{j}$.
Since the surface is smooth,there is no friction,so the horizontal component of velocity remains unchanged: $v_{x, final} = v_{x, initial} = a \hat{i}$.
The vertical component of velocity changes due to the collision. According to the definition of the coefficient of restitution $e$,the velocity of separation along the normal is $e$ times the velocity of approach along the normal.
$v_{y, final} = e \times |v_{y, initial}| = e b$.
Since the ball bounces upward,the final vertical velocity is in the positive $y$-direction: $\vec{v}_{y, final} = e b \hat{j}$.
Therefore,the velocity of the ball just after the collision is $\vec{v}_{final} = a \hat{i} + e b \hat{j}$.

(A) Let the mass of the first ball be $m$ and its initial velocity be $u_1 = 4 \ m/s$. The mass of the second ball is $2m$ and its initial velocity is $u_2 = 0 \ m/s$.
According to the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$m(4) + 2m(0) = m v_1 + 2m v_2$
$4 = v_1 + 2v_2$ --- (Equation $1$)
Using the coefficient of restitution $e = 0.2$:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
$0.2 = \frac{v_2 - v_1}{4 - 0}$
$v_2 - v_1 = 0.8$ --- (Equation $2$)
Adding Equation $1$ and Equation $2$:
$(v_1 + 2v_2) + (v_2 - v_1) = 4 + 0.8$
$3v_2 = 4.8 \Rightarrow v_2 = 1.6 \ m/s$
Substituting $v_2$ into Equation $2$:
$1.6 - v_1 = 0.8 \Rightarrow v_1 = 0.8 \ m/s$
Thus,the velocities after collision are $0.8 \ m/s$ and $1.6 \ m/s$.

(C) Let the mass of the first object be $m$ and its initial velocity be $u$. Let the mass of the second object be $M$ and its initial velocity be $0$. After the collision,the first object stops $(v_1 = 0)$.
By the law of conservation of linear momentum:
$m u + M(0) = m(0) + M v_2$
$m u = M v_2$
$v_2 = \frac{m u}{M}$
The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Substituting the values:
$e = \frac{\frac{m u}{M} - 0}{u - 0} = \frac{m u / M}{u} = \frac{m}{M}$

(B) When a body falls from a height $h_1$,its velocity just before impact is $v_b = \sqrt{2gh_1}$.
After the collision,the velocity of the body becomes $v_f = e \cdot v_b$,where $e$ is the coefficient of restitution.
The body then rises to a height $h_2$,where $v_f = \sqrt{2gh_2}$.
Substituting $v_f$ into the equation,we get $\sqrt{2gh_2} = e \sqrt{2gh_1}$.
Squaring both sides,we get $2gh_2 = e^2 (2gh_1)$,which simplifies to $h_2 = e^2 h_1$.
Given $e = 0.6$ and $h_1 = 1 \ m$:
$h_2 = (0.6)^2 \times 1 \ m = 0.36 \ m$.

(C) Let the mass of both spheres be $m$. The initial velocity of the first sphere is $v$ and the second sphere is $0$. Let their final velocities be $V_1$ and $V_2$ respectively.
By the law of conservation of linear momentum:
$m v + m(0) = m V_1 + m V_2$
$v = V_1 + V_2$ --- $(1)$
By the definition of the coefficient of restitution $e$:
$e = \frac{V_2 - V_1}{u_1 - u_2}$
$e = \frac{V_2 - V_1}{v - 0}$
$e v = V_2 - V_1$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$v + e v = (V_1 + V_2) + (V_2 - V_1)$
$v(1 + e) = 2 V_2$
$V_2 = \frac{v(e + 1)}{2}$
The ratio of the final velocity of the second sphere $(V_2)$ to the initial velocity of the first sphere $(v)$ is:
$\frac{V_2}{v} = \frac{e + 1}{2}$

(D) Let $u$ be the initial velocity of mass $m$ and $v_M$ be the velocity of mass $M$ after the collision. The mass $m$ comes to rest,so its final velocity is $0$.
By the law of conservation of linear momentum:
$mu + M(0) = m(0) + Mv_M$
$mu = Mv_M$
$v_M = \frac{m}{M}u$
The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach:
$e = \frac{v_M - 0}{u - 0} = \frac{v_M}{u}$
Substituting the value of $v_M$:
$e = \frac{(m/M)u}{u} = \frac{m}{M}$

(B) Given: Mass of first block $M_1 = m$,initial velocity $u_1 = v$,final velocity $v_1 = 0$.
Mass of second block $M_2 = 4m$,initial velocity $u_2 = 0$,final velocity $v_2 = ?$.
According to the law of conservation of linear momentum:
$M_1 u_1 + M_2 u_2 = M_1 v_1 + M_2 v_2$
$m(v) + 4m(0) = m(0) + 4m(v_2)$
$mv = 4mv_2$
$v_2 = \frac{v}{4}$
The coefficient of restitution $e$ is defined as:
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Substituting the values:
$e = \frac{\frac{v}{4} - 0}{v - 0} = \frac{v/4}{v} = \frac{1}{4} = 0.25$.

(A) Given: Height $h = 20 \,m$, acceleration due to gravity $g = 10 \,m/s^2$, and coefficient of restitution $e = 0.4$.
First, calculate the velocity $v$ of the ball just before it hits the ground using the equation of motion $v^2 = u^2 + 2gh$ (where initial velocity $u = 0$):
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \,m/s$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of separation (rebound velocity $v'$) to the velocity of approach (impact velocity $v$):
$e = \frac{v'}{v}$.
Therefore, the upward rebound velocity $v'$ is:
$v' = e \times v = 0.4 \times 20 = 8 \,m/s$.

(B) First, calculate the velocity of the ball just before it hits the ground using the equation of motion $v^2 = u^2 + 2gh$. Since the ball falls freely, initial velocity $u = 0$.
$v^2 = 0 + 2 \times 10 \times 20 = 400$
$v = \sqrt{400} = 20 \,ms^{-1}$.
The coefficient of restitution $e$ is defined as the ratio of the velocity of rebound $(v')$ to the velocity of impact $(v)$: $e = \frac{v'}{v}$.
Given $e = 0.4$, the velocity after the first rebound is $v' = e \times v$.
$v' = 0.4 \times 20 = 8 \,ms^{-1}$.

(B) Given: $m_1 = 1 \,kg$, $u_1 = 12 \,ms^{-1}$, $m_2 = 2 \,kg$, $u_2 = -24 \,ms^{-1}$ (negative sign indicates opposite direction).
Coefficient of restitution $e = 2/3$.
The formula for the coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Substituting the values: $\frac{2}{3} = \frac{v_2 - v_1}{12 - (-24)} = \frac{v_2 - v_1}{36}$.
Thus, $v_2 - v_1 = 24$ ... $(i)$.
By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
$(1)(12) + (2)(-24) = (1)v_1 + (2)v_2$.
$12 - 48 = v_1 + 2v_2 \Rightarrow v_1 + 2v_2 = -36$ ... (ii).
From $(i)$, $v_2 = v_1 + 24$. Substituting into (ii):
$v_1 + 2(v_1 + 24) = -36$.
$v_1 + 2v_1 + 48 = -36 \Rightarrow 3v_1 = -84 \Rightarrow v_1 = -28 \,ms^{-1}$.
Then $v_2 = -28 + 24 = -4 \,ms^{-1}$.
The velocities are $-28 \,ms^{-1}$ and $-4 \,ms^{-1}$.

(C) According to the law of conservation of linear momentum,if the net external force on a system is zero,the total momentum of the system remains conserved.
When a ball hits the floor,the interaction between the ball and the Earth (floor) involves internal forces. Since there is no external force acting on the system of the ball and the Earth,the total momentum of the system is conserved.
In an inelastic collision,kinetic energy is not conserved as some energy is dissipated as heat,sound,or deformation energy.
Therefore,the mechanical energy of the ball is not conserved,and the total mechanical energy of the ball and the Earth is also not conserved due to the energy loss during the inelastic collision.