Inelastic Collision Questions in English

(A) In any collision,if there are no external forces acting on the system,the total linear momentum of the system remains conserved according to the law of conservation of linear momentum.
This principle applies to both elastic and inelastic collisions.
However,total kinetic energy is only conserved in perfectly elastic collisions.
In inelastic collisions,some kinetic energy is transformed into other forms of energy (such as heat,sound,or deformation energy),so it is not conserved.
Therefore,total linear momentum is the quantity that always remains conserved.

(C) Let the mass of both bodies be $m$. Let the initial velocity of the first body be $u_1 = v$ and the initial velocity of the second body be $u_2 = v/2$.
By the law of conservation of linear momentum: $m u_1 + m u_2 = m v_1 + m v_2$,where $v_1$ and $v_2$ are the final velocities.
$v + v/2 = v_1 + v_2 \implies v_1 + v_2 = 1.5v$ --- $(1)$
The coefficient of restitution $e$ is defined as $e = (v_2 - v_1) / (u_1 - u_2)$.
Given $e = 0.5$,$u_1 = v$,and $u_2 = v/2$,we have $0.5 = (v_2 - v_1) / (v - v/2)$.
$0.5 = (v_2 - v_1) / (0.5v) \implies v_2 - v_1 = 0.25v$ --- $(2)$
Adding equations $(1)$ and $(2)$: $2v_2 = 1.75v \implies v_2 = 0.875v = (7/8)v$.
Subtracting equation $(2)$ from $(1)$: $2v_1 = 1.25v \implies v_1 = 0.625v = (5/8)v$.
The ratio of the velocities $v_1 : v_2 = (5/8)v : (7/8)v = 5:7$.

(A) Let the initial height from which the body falls be $H_0$.
When a body falls from height $H_0$,its velocity just before the first impact is $v_0 = \sqrt{2gH_0}$.
After the first impact,the velocity becomes $v_1 = e v_0$,and the height reached is $H_1 = e^2 H_0$.
After the second impact,the velocity becomes $v_2 = e v_1 = e^2 v_0$,and the height reached is $H_2 = e^4 H_0$.
Given $e = 0.8 = 4/5$.
The ratio of the height after the second impact to the initial height is $H_2 / H_0 = e^4$.
Calculating $e^4 = (0.8)^4 = (4/5)^4 = 256 / 625$.
Thus,the ratio is $256: 625$.

(B) In an inelastic collision,the total kinetic energy of the system is not conserved.
Some part of the kinetic energy is converted into other forms of energy,such as heat,sound,or deformation energy.
Therefore,the final kinetic energy $(K_f)$ is always less than the initial kinetic energy $(K_i)$.
$K_f < K_i$ or $K_i > K_f$.

(B) Given: $m_1 = 2 \,kg$, $m_2 = 4 \,kg$, relative velocity before collision $u_{rel} = u_1 - u_2 = 10 \,ms^{-1}$, relative velocity after collision $v_{rel} = v_2 - v_1 = 4 \,ms^{-1}$.
Coefficient of restitution $e = \frac{v_{rel}}{u_{rel}} = \frac{4}{10} = 0.4$.
The loss in kinetic energy during a collision is given by $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_{rel})^2 (1 - e^2)$.
Substituting the values: $\Delta K = \frac{1}{2} \left( \frac{2 \times 4}{2 + 4} \right) (10)^2 (1 - (0.4)^2)$.
$\Delta K = \frac{1}{2} \left( \frac{8}{6} \right) (100) (1 - 0.16)$.
$\Delta K = \frac{1}{2} \times \frac{4}{3} \times 100 \times 0.84$.
$\Delta K = \frac{2}{3} \times 84 = 56 \,J$.

(D) Let $H_0 = 6.25 \ m$ be the initial height and $H_2 = 81 \ cm = 0.81 \ m$ be the height reached after the second bounce.
For a ball bouncing on a hard surface,the height reached after $n$ bounces is given by $H_n = H_0 \cdot e^{2n}$,where $e$ is the coefficient of restitution.
Given $n = 2$,$H_0 = 6.25 \ m$,and $H_2 = 0.81 \ m$.
Substituting the values:
$0.81 = 6.25 \cdot e^{2 \times 2}$
$0.81 = 6.25 \cdot e^4$
$e^4 = \frac{0.81}{6.25} = \frac{81}{625}$
Taking the square root on both sides:
$e^2 = \sqrt{\frac{81}{625}} = \frac{9}{25} = 0.36$
Taking the square root again:
$e = \sqrt{0.36} = 0.6$
Thus,the coefficient of restitution is $0.6$.

(C) The velocity of the body just before hitting the ground is $v_0 = \sqrt{2gh_0}$,where $h_0 = 100 \ m$.
After the collision,the velocity of rebound is $v_1 = e v_0 = e \sqrt{2gh_0}$.
The maximum height reached after the rebound is $h_1 = \frac{v_1^2}{2g} = \frac{(e \sqrt{2gh_0})^2}{2g} = e^2 h_0$.
Given $h_1 = 36 \ m$ and $h_0 = 100 \ m$,we have $36 = e^2(100)$.
$e^2 = \frac{36}{100} = 0.36$.
$e = \sqrt{0.36} = 0.6$.

(A) Let the mass of the first ball be $m_1 = m$ and the mass of the second ball be $m_2 = 2m$.
Initial velocities are $u_1 = 2 \,m/s$ and $u_2 = 0 \,m/s$.
The coefficient of restitution is $e = 0.5$.
The velocity of the first ball after collision is given by $v_1 = \frac{(m_1 - em_2)u_1 + (1 + e)m_2u_2}{m_1 + m_2}$.
Substituting the values: $v_1 = \frac{(m - 0.5 \times 2m)(2) + (1 + 0.5)(2m)(0)}{m + 2m} = \frac{(m - m)(2) + 0}{3m} = 0 \,m/s$.
The velocity of the second ball after collision is given by $v_2 = \frac{(m_2 - em_1)u_2 + (1 + e)m_1u_1}{m_1 + m_2}$.
Substituting the values: $v_2 = \frac{(2m - 0.5 \times m)(0) + (1 + 0.5)(m)(2)}{m + 2m} = \frac{0 + (1.5)(2m)}{3m} = \frac{3m}{3m} = 1 \,m/s$.
Thus, their velocities after collision are $0 \,m/s$ and $1 \,m/s$.

(C) The initial velocity of the ball is $u = 5 \,m \,s^{-1}$ at an angle of $30^{\circ}$ with the horizontal.
We resolve this velocity into horizontal and vertical components:
Horizontal component, $v_{1x} = u \cos 30^{\circ} = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.
Vertical component, $v_{1y} = u \sin 30^{\circ} = 5 \times \frac{1}{2} = 2.5 \,m \,s^{-1}$.
During the collision with the ground, the horizontal component of velocity remains unchanged because there is no impulse along the horizontal direction.
$v_{2x} = v_{1x} = \frac{5\sqrt{3}}{2} \,m \,s^{-1}$.
The vertical component changes according to the coefficient of restitution $e = 0.2$:
$v_{2y} = e \times v_{1y} = 0.2 \times 2.5 = 0.5 \,m \,s^{-1}$.
The final speed of the ball $v_f$ is the magnitude of the resultant velocity vector:
$v_f = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2 + (0.5)^2}$
$v_f = \sqrt{\frac{25 \times 3}{4} + 0.25} = \sqrt{\frac{75}{4} + \frac{1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19} \,m \,s^{-1}$.

(D) Let $m_1 = 2 \ g$ and $u_1 = 2 \ ms^{-1}$ be the mass and initial velocity of the first ball.
Let $m_2 = 8 \ g$ and $u_2 = 0 \ ms^{-1}$ be the mass and initial velocity of the second ball.
After collision,the first ball comes to rest,so $v_1 = 0 \ ms^{-1}$.
Using the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
Substituting the values: $(2 \ g)(2 \ ms^{-1}) + (8 \ g)(0) = (2 \ g)(0) + (8 \ g)(v_2)$.
$4 = 8 v_2 \implies v_2 = 0.5 \ ms^{-1}$.
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Substituting the values: $e = \frac{0.5 - 0}{2 - 0} = \frac{0.5}{2} = 0.25$.

(D) Let the mass of each ball be $m$. Let $v_A = 16 \,ms^{-1}$ be the initial velocity of ball $A$ and $v_B = 0$ be the initial velocity of ball $B$.
After collision,let the velocities of $A$ and $B$ be $v_1$ and $v_2$ respectively.
By conservation of linear momentum: $m v_A + 0 = m v_1 + m v_2 \implies v_1 + v_2 = 16$.
By the definition of coefficient of restitution $e$: $e = \frac{v_2 - v_1}{v_A - 0} \implies v_2 - v_1 = 16e$.
Adding the two equations: $2v_2 = 16(1+e) \implies v_2 = 8(1+e)$.
For ball $B$ to just reach the top of the inclined plane of height $h = 5 \,m$,its kinetic energy at the bottom must equal its potential energy at the top: $\frac{1}{2} m v_2^2 = mgh$.
$v_2^2 = 2gh = 2 \times 10 \times 5 = 100 \implies v_2 = 10 \,ms^{-1}$.
Substituting $v_2$ in the expression: $8(1+e) = 10 \implies 1+e = \frac{10}{8} = 1.25 \implies e = 0.25 = \frac{1}{4}$.

(C) The energy lost in an inelastic collision is given by the formula: $\Delta KE = \frac{m_1 m_2}{2(m_1 + m_2)} (1 - e^2) (u_1 + u_2)^2$.
Given: $m_1 = 1 \ kg$,$m_2 = 0.5 \ kg$,$u_1 = 6 \ ms^{-1}$,$u_2 = 9 \ ms^{-1}$ (since they move in opposite directions,the relative velocity is $u_1 + u_2$),and $e = \frac{1}{3}$.
Substituting the values:
$\Delta KE = \frac{1 \times 0.5}{2(1 + 0.5)} \left(1 - (\frac{1}{3})^2\right) (6 + 9)^2$
$\Delta KE = \frac{0.5}{3} \times (1 - \frac{1}{9}) \times (15)^2$
$\Delta KE = \frac{1}{6} \times \frac{8}{9} \times 225$
$\Delta KE = \frac{8}{54} \times 225 = \frac{4}{27} \times 225 = 33.33 \ J$.

(B) Let the mass of both balls be $m$. Let the initial velocities be $u_P = v$ and $u_Q = -10 \ ms^{-1}$.
After the collision,the final velocity of ball $P$ is $v_P = 0$ and let the final velocity of ball $Q$ be $v_Q$.
By the law of conservation of linear momentum:
$m(v) + m(-10) = m(0) + m(v_Q)$
$v - 10 = v_Q$
Now,using the coefficient of restitution formula $e = \frac{\text{velocity of separation}}{\text{velocity of approach}}$:
$e = \frac{v_Q - v_P}{u_P - u_Q}$
Given $e = 0.6$,$v_P = 0$,$u_P = v$,and $u_Q = -10$:
$0.6 = \frac{v_Q - 0}{v - (-10)}$
$0.6 = \frac{v_Q}{v + 10}$
Substitute $v_Q = v - 10$ into the equation:
$0.6 = \frac{v - 10}{v + 10}$
$0.6(v + 10) = v - 10$
$0.6v + 6 = v - 10$
$16 = 0.4v$
$v = \frac{16}{0.4} = 40 \ ms^{-1}$

(C) When a ball falls from a height $h$,it strikes the floor and rebounds to a height $h_1 = h e^2$.
After the second impact,it rebounds to a height $h_2 = h_1 e^2 = h e^4$.
In general,the height reached after the $n$-th rebound is $h_n = h e^{2n}$.
The total distance $H$ covered by the ball is the sum of the initial fall and the subsequent upward and downward paths for each rebound:
$H = h + 2h_1 + 2h_2 + 2h_3 + \dots$
$H = h + 2(h e^2 + h e^4 + h e^6 + \dots)$
$H = h + 2h(e^2 + e^4 + e^6 + \dots)$
The term in the bracket is an infinite geometric progression with first term $a = e^2$ and common ratio $r = e^2$. The sum is $S = \frac{a}{1-r} = \frac{e^2}{1-e^2}$.
$H = h + 2h \left( \frac{e^2}{1-e^2} \right)$
$H = h \left( 1 + \frac{2e^2}{1-e^2} \right)$
$H = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right)$
$H = h \left( \frac{1 + e^2}{1 - e^2} \right)$

(B) $1$. The ball is dropped from height $h$. The distance covered during the first fall is $h$.
$2$. After the first collision with the floor,the ball rebounds to a height $h_1 = e^2h$.
$3$. The ball then travels upward to height $h_1$ and then falls back down to the floor to make the second hit.
$4$. The distance covered during the rebound is $h_1$ (upward) + $h_1$ (downward) = $2h_1 = 2e^2h$.
$5$. The total distance covered by the ball just before the second hit is the sum of the initial fall and the rebound distance: $D = h + 2e^2h = h(1 + 2e^2)$.

(C) Initial potential energy of the ball at height $H = 10 \text{ m}$ is $E_i = mgH$.
When the ball hits the ground, this energy is converted into kinetic energy.
After the collision, $50\%$ of the energy is lost, so the remaining energy is $E_f = 0.5 \times E_i = 0.5 \times mgH$.
The ball will rise to a new height $h$ such that its potential energy at that height equals the remaining energy:
$mgh = 0.5 \times mgH$
$h = 0.5 \times H$
Given $H = 10 \text{ m}$, we have:
$h = 0.5 \times 10 \text{ m} = 5 \text{ m}$.
Therefore, the height reached by the ball is $5 \text{ m}$.

(A) Let the ball be dropped from an initial height $h$. The initial potential energy is $U_i = mgh$.
After the collision,the ball reaches a final height of $h_f = \frac{3}{4}h$.
The final potential energy is $U_f = mgh_f = mgh(\frac{3}{4}) = \frac{3}{4}mgh$.
The energy loss is $\Delta U = U_i - U_f = mgh - \frac{3}{4}mgh = \frac{1}{4}mgh$.
The percentage loss of energy is given by $\frac{\Delta U}{U_i} \times 100\%$.
Percentage loss $= \frac{\frac{1}{4}mgh}{mgh} \times 100\% = \frac{1}{4} \times 100\% = 25\%$.

(D) When a ball is dropped from a height $H$ and the coefficient of restitution is $e$,the height reached after the first bounce is $h_1 = e^2 H$,after the second bounce is $h_2 = e^4 H$,and so on.
The total distance $D$ travelled by the ball is given by the sum of the initial fall and the infinite series of bounces (up and down):
$D = H + 2h_1 + 2h_2 + 2h_3 + ...$
$D = H + 2(e^2 H + e^4 H + e^6 H + ...)$
$D = H + 2e^2 H (1 + e^2 + e^4 + ...)$
Using the sum of an infinite geometric progression $S = \frac{a}{1-r}$,where $a = 1$ and $r = e^2$:
$D = H + 2e^2 H \left( \frac{1}{1 - e^2} \right)$
$D = H \left( 1 + \frac{2e^2}{1 - e^2} \right) = H \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right) = H \left( \frac{1 + e^2}{1 - e^2} \right)$
Given $H = 42 \ m$ and $e = 0.4$:
$e^2 = (0.4)^2 = 0.16$
$D = 42 \times \left( \frac{1 + 0.16}{1 - 0.16} \right) = 42 \times \left( \frac{1.16}{0.84} \right)$
$D = 42 \times \frac{116}{84} = 42 \times \frac{116}{2 \times 42} = \frac{116}{2} = 58 \ m$.

(A) Given: $m_1 = 0.5 \ kg$,$u_1 = 10 \ ms^{-1}$,$m_2 = 1 \ kg$,$u_2 = 0$,$e = 0.4$.
The final velocities $v_1$ and $v_2$ after a one-dimensional collision are given by:
$v_1 = \left( \frac{m_1 - e m_2}{m_1 + m_2} \right) u_1 = \left( \frac{0.5 - 0.4 \times 1}{0.5 + 1} \right) \times 10 = \left( \frac{0.1}{1.5} \right) \times 10 = \frac{1}{15} \times 10 = \frac{2}{3} \ ms^{-1}$.
$v_2 = \frac{(1 + e) m_1 u_1}{m_1 + m_2} = \frac{(1 + 0.4) \times 0.5 \times 10}{0.5 + 1} = \frac{1.4 \times 5}{1.5} = \frac{7}{1.5} = \frac{14}{3} \ ms^{-1}$.
The ratio of the velocities is $\frac{v_1}{v_2} = \frac{2/3}{14/3} = \frac{2}{14} = 1:7$.

(A) Given: $m_1 = m_2 = m = 1.2 \ kg$,$u_1 = 12 \ ms^{-1}$,$u_2 = 0$,$e = \frac{1}{\sqrt{2}}$.
By the law of conservation of linear momentum:
$m u_1 + m u_2 = m v_1 + m v_2$
$12 + 0 = v_1 + v_2 \Rightarrow v_1 + v_2 = 12$ ...$(i)$
By the definition of the coefficient of restitution $(e)$:
$e = \frac{v_2 - v_1}{u_1 - u_2} \Rightarrow \frac{1}{\sqrt{2}} = \frac{v_2 - v_1}{12}$
$v_2 - v_1 = \frac{12}{\sqrt{2}} = 6\sqrt{2}$ ...(ii)
Adding equations $(i)$ and (ii):
$2v_2 = 12 + 6\sqrt{2} \Rightarrow v_2 = 6 + 3\sqrt{2} \ ms^{-1}$
Subtracting equation (ii) from $(i)$:
$2v_1 = 12 - 6\sqrt{2} \Rightarrow v_1 = 6 - 3\sqrt{2} \ ms^{-1}$
Initial kinetic energy $(KE)_i = \frac{1}{2} m u_1^2 = \frac{1}{2} \times 1.2 \times (12)^2 = 0.6 \times 144 = 86.4 \ J$.
Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (v_1^2 + v_2^2)$.
$v_1^2 + v_2^2 = (6 - 3\sqrt{2})^2 + (6 + 3\sqrt{2})^2 = (36 + 18 - 36\sqrt{2}) + (36 + 18 + 36\sqrt{2}) = 54 + 54 = 108$.
$(KE)_f = \frac{1}{2} \times 1.2 \times 108 = 0.6 \times 108 = 64.8 \ J$.
Ratio $\frac{(KE)_f}{(KE)_i} = \frac{64.8}{86.4} = \frac{648}{864} = \frac{3}{4} = 3:4$.

(A) When a body is dropped from a height $h$,it strikes the floor with velocity $v = \sqrt{2gh}$.
After the first impact,it rebounds with velocity $v_1 = ev = e\sqrt{2gh}$.
The height reached after the first rebound is $h_1 = \frac{v_1^2}{2g} = e^2h$.
After the second impact,it rebounds with velocity $v_2 = ev_1 = e^2v$,reaching a height $h_2 = \frac{v_2^2}{2g} = e^4h$.
The total distance $D$ travelled is the initial fall plus twice the sum of all subsequent rebound heights:
$D = h + 2h_1 + 2h_2 + 2h_3 + \dots$
$D = h + 2(e^2h + e^4h + e^6h + \dots)$
$D = h + 2e^2h(1 + e^2 + e^4 + \dots)$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:
$D = h + 2e^2h \left( \frac{1}{1-e^2} \right)$
$D = h \left[ 1 + \frac{2e^2}{1-e^2} \right] = h \left[ \frac{1-e^2+2e^2}{1-e^2} \right] = h \left[ \frac{1+e^2}{1-e^2} \right]$.

(B) When the ball is dropped from height $h$,it strikes the plane with velocity $v = \sqrt{2gh}$.
After the collision,the ball rebounds with a velocity $u = ev = e\sqrt{2gh}$.
The height reached by the ball after the first collision is $h_1 = \frac{u^2}{2g} = \frac{e^2(2gh)}{2g} = e^2h$.
The total distance travelled by the ball before hitting the plane for the second time is the sum of the initial downward distance and the distance covered during the first rebound (up and down).
Total distance $= h + 2h_1 = h + 2(e^2h) = h(1 + 2e^2)$.

(A) Let the mass of both particles be $m$. Let $V_1$ and $V_2$ be the velocities of particles $A$ and $B$ after the collision,respectively.
By the law of conservation of momentum:
$m \times 10 + 0 = m V_1 + m V_2 \Rightarrow V_1 + V_2 = 10$ ... $(i)$
Given that the kinetic energy of the system decreases by $1 \%$,the final kinetic energy $K_f = 0.99 K_i$.
$\frac{1}{2} m V_1^2 + \frac{1}{2} m V_2^2 = 0.99 \times (\frac{1}{2} m \times 10^2)$
$V_1^2 + V_2^2 = 0.99 \times 100 = 99$ ... (ii)
We know that $(V_1 + V_2)^2 = V_1^2 + V_2^2 + 2 V_1 V_2$.
Substituting the values: $10^2 = 99 + 2 V_1 V_2 \Rightarrow 100 = 99 + 2 V_1 V_2 \Rightarrow 2 V_1 V_2 = 1 \Rightarrow V_1 V_2 = 0.5$.
Now,$(V_1 - V_2)^2 = (V_1 + V_2)^2 - 4 V_1 V_2 = 100 - 4(0.5) = 100 - 2 = 98$.
$V_1 - V_2 = \sqrt{98} = 7\sqrt{2} \approx 9.899 \ m/s$.
Adding $(i)$ and (ii): $2 V_1 = 10 + 9.899 = 19.899 \Rightarrow V_1 \approx 9.95 \ m/s$.

(D) Given: Mass of ball $A$,$m_A = 50 \ gm$; initial velocity of ball $A$,$u_A = 10 \ m/s$. Mass of ball $B$,$m_B = 10 \ gm$; initial velocity of ball $B$,$u_B = -15 \ m/s$ (since it is moving in the opposite direction). Coefficient of restitution,$e = \frac{2}{5}$.
The final velocity $v_B$ of the second ball $B$ after a one-dimensional collision is given by the formula:
$v_B = \frac{m_A(1+e)}{m_A+m_B} u_A + \frac{m_B - e m_A}{m_A+m_B} u_B$
Substituting the given values into the formula:
$v_B = \frac{50(1 + \frac{2}{5})}{50 + 10} \times 10 + \frac{10 - (\frac{2}{5} \times 50)}{50 + 10} \times (-15)$
$v_B = \frac{50 \times \frac{7}{5}}{60} \times 10 + \frac{10 - 20}{60} \times (-15)$
$v_B = \frac{70}{60} \times 10 + \frac{-10}{60} \times (-15)$
$v_B = \frac{70}{6} + \frac{150}{60} = \frac{70}{6} + \frac{15}{6} = \frac{85}{6} \ m/s$
Thus,the final speed of ball $B$ is $\frac{85}{6} \ m/s$.

(A) Let the initial velocity of $m_1$ be $u$ and the final velocities of $m_1$ and $m_2$ be $v_1$ and $v_2$ respectively. The particle $m_1$ moves at $90^{\circ}$ to the $X$-axis and $m_2$ moves at $30^{\circ}$ to the $X$-axis.
Applying conservation of linear momentum along the $X$-axis:
$m_1 u = m_2 v_2 \cos 30^{\circ} \quad \dots (i)$
Applying conservation of linear momentum along the $Y$-axis:
$0 = m_2 v_2 \sin 30^{\circ} - m_1 v_1 \quad \Rightarrow \quad m_1 v_1 = m_2 v_2 \sin 30^{\circ} \quad \dots (ii)$
Given that the kinetic energy reduces by $50 \%$,the final kinetic energy is half of the initial kinetic energy:
$\frac{1}{2} (\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2) = \frac{1}{2} (\frac{1}{2} m_1 u^2) \quad \Rightarrow \quad m_1 v_1^2 + m_2 v_2^2 = \frac{1}{2} m_1 u^2 \quad \dots (iii)$
From $(i)$,$v_2 \cos 30^{\circ} = \frac{m_1 u}{m_2} \Rightarrow v_2 = \frac{2 m_1 u}{\sqrt{3} m_2}$.
From $(ii)$,$v_1 = \frac{m_2 v_2 \sin 30^{\circ}}{m_1} = \frac{m_2}{m_1} \cdot \frac{2 m_1 u}{\sqrt{3} m_2} \cdot \frac{1}{2} = \frac{u}{\sqrt{3}}$.
Substituting $v_1$ and $v_2$ into $(iii)$:
$m_1 (\frac{u^2}{3}) + m_2 (\frac{4 m_1^2 u^2}{3 m_2^2}) = \frac{1}{2} m_1 u^2$
Dividing by $m_1 u^2$:
$\frac{1}{3} + \frac{4 m_1}{3 m_2} = \frac{1}{2} \quad \Rightarrow \quad \frac{4 m_1}{3 m_2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$
$\frac{m_1}{m_2} = \frac{3}{24} = \frac{1}{8} \quad \Rightarrow \quad \frac{m_2}{m_1} = 8$.

(A) The velocity of the ball just before the first impact is $v_0 = \sqrt{2gh_0}$.
After the first collision,the velocity is $v_1 = e v_0$.
After the second collision,the velocity is $v_2 = e v_1 = e^2 v_0$.
Following this pattern,after $n$ collisions,the velocity of the ball is $v_n = e^n v_0$.
The height $h_n$ reached after the $n$th rebound is given by $h_n = \frac{v_n^2}{2g}$.
Substituting $v_n = e^n v_0$,we get $h_n = \frac{(e^n v_0)^2}{2g} = e^{2n} \frac{v_0^2}{2g}$.
Since $h_0 = \frac{v_0^2}{2g}$,we have $h_n = e^{2n} h_0$.
Rearranging for $e$,we get $e^{2n} = \frac{h_n}{h_0}$,which implies $e = \left[\frac{h_n}{h_0}\right]^{1/2n}$.

(A) Given: $m_1 = 2 \,kg$, $v_1 = 24 \,ms^{-1}$, $m_2 = 4 \,kg$, $v_2 = -48 \,ms^{-1}$ (opposite direction), $e = 2/3$.
Using the formula for final velocities after collision:
$v_1' = \frac{m_1 v_1 + m_2 v_2 + e m_2 (v_2 - v_1)}{m_1 + m_2}$
$v_1' = \frac{(2)(24) + (4)(-48) + (2/3)(4)(-48 - 24)}{2 + 4}$
$v_1' = \frac{48 - 192 + (8/3)(-72)}{6} = \frac{-144 - 192}{6} = \frac{-336}{6} = -56 \,ms^{-1}$.
Using the conservation of momentum or the restitution equation $v_2' - v_1' = e(v_1 - v_2)$:
$v_2' - (-56) = (2/3)(24 - (-48))$
$v_2' + 56 = (2/3)(72) = 48$
$v_2' = 48 - 56 = -8 \,ms^{-1}$.

(A) The particle falls from height $h$. The first impact velocity is $v_0 = \sqrt{2gh}$.
After the first collision,the rebound velocity is $v_1 = ev_0$. The height reached is $h_1 = \frac{v_1^2}{2g} = e^2h$.
The particle travels $h$ downwards,then $h_1$ upwards and $h_1$ downwards.
After the second collision,it reaches $h_2 = e^2h_1 = e^4h$,traveling $h_2$ up and $h_2$ down.
The total distance $D$ is given by:
$D = h + 2h_1 + 2h_2 + 2h_3 + ...$
$D = h + 2(e^2h + e^4h + e^6h + ...)$
$D = h + 2h(e^2 + e^4 + e^6 + ...)$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a = e^2$ and $r = e^2$:
$D = h + 2h \left( \frac{e^2}{1-e^2} \right)$
$D = h \left( 1 + \frac{2e^2}{1-e^2} \right) = h \left( \frac{1-e^2+2e^2}{1-e^2} \right) = h \left( \frac{1+e^2}{1-e^2} \right)$.

(C) Let the initial velocity of sphere $A$ be $u$ and the final velocities of spheres $A$ and $B$ be $v_1$ and $v_2$ respectively.
By the law of conservation of linear momentum:
$m u + (2m)(0) = m v_1 + 2m v_2$
$u = v_1 + 2v_2$ ....$(i)$
Given the coefficient of restitution $e = 0.4 = \frac{2}{5}$.
Using the definition of the coefficient of restitution:
$e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{u - 0} = 0.4$
$v_2 - v_1 = 0.4u$ ....$(ii)$
From equation $(i)$,$u = v_1 + 2v_2$. Substitute this into equation $(ii)$:
$v_2 - v_1 = 0.4(v_1 + 2v_2)$
$v_2 - v_1 = 0.4v_1 + 0.8v_2$
$0.2v_2 = 1.4v_1$
$\frac{v_1}{v_2} = \frac{0.2}{1.4} = \frac{1}{7}$

(A) According to the law of conservation of linear momentum:
$m v + 2m(0) = m v_P + 2m v_Q$
$v = v_P + 2v_Q \quad \dots(1)$
Using the coefficient of restitution formula $e = \frac{v_2 - v_1}{u_1 - u_2}$:
$e = \frac{1}{3} = \frac{v_Q - v_P}{v - 0}$
$v_Q - v_P = \frac{v}{3} \quad \dots(2)$
From equation $(1)$,$v = v_P + 2v_Q$. Substituting this into equation $(2)$:
$v_Q - v_P = \frac{v_P + 2v_Q}{3}$
$3v_Q - 3v_P = v_P + 2v_Q$
$v_Q = 4v_P$
$\frac{v_Q}{v_P} = 4$

(B) Given: Mass of both balls is $m$. Initial velocities are $u_1 = v$ and $u_2 = 0$. After collision,$v_1 = 0.15 v$.
Using the law of conservation of linear momentum:
$m u_1 + m u_2 = m v_1 + m v_2$
$m v + 0 = m(0.15 v) + m v_2$
$v = 0.15 v + v_2 \implies v_2 = 0.85 v$
Initial kinetic energy $(KE)_i = \frac{1}{2} m v^2$.
Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (0.15 v)^2 + \frac{1}{2} m (0.85 v)^2$
$(KE)_f = \frac{1}{2} m v^2 [0.15^2 + 0.85^2] = \frac{1}{2} m v^2 [0.0225 + 0.7225] = \frac{1}{2} m v^2 [0.745]$
Decrease in kinetic energy $\Delta KE = (KE)_i - (KE)_f = \frac{1}{2} m v^2 [1 - 0.745] = 0.255 \times (KE)_i$
Percentage decrease $= \frac{\Delta KE}{(KE)_i} \times 100 = 0.255 \times 100 = 25.5 \% \approx 25 \%$.

(A) The ball falls from height $h$ and hits the floor. The velocity just before the first impact is $v_0 = \sqrt{2gh}$.
After the first impact,the velocity becomes $v_1 = e v_0$. The ball rises to a height $h_1 = \frac{v_1^2}{2g} = e^2 h$.
It then falls from $h_1$ and hits the floor again. The distance covered in the first bounce (up and down) is $2h_1 = 2e^2 h$.
After the second impact,it rises to $h_2 = e^2 h_1 = e^4 h$. The distance covered in the second bounce is $2h_2 = 2e^4 h$.
The total distance $D$ is the sum of the initial fall and all subsequent bounces:
$D = h + 2h_1 + 2h_2 + 2h_3 + ...$
$D = h + 2e^2 h + 2e^4 h + 2e^6 h + ...$
$D = h + 2e^2 h (1 + e^2 + e^4 + ...)$
Using the sum of an infinite geometric series $S = \frac{a}{1-r}$ where $a=1$ and $r=e^2$:
$D = h + 2e^2 h \left( \frac{1}{1-e^2} \right)$
$D = h \left( 1 + \frac{2e^2}{1-e^2} \right) = h \left( \frac{1-e^2+2e^2}{1-e^2} \right) = h \left( \frac{1+e^2}{1-e^2} \right)$.

(A) Let the velocity of the ball before collision be $v$. The velocity components are $v_x = v \sin \theta$ (parallel to the floor) and $v_y = v \cos \theta$ (perpendicular to the floor).
Since the floor is smooth,there is no impulsive force parallel to the floor,so the tangential component of velocity remains unchanged: $v'_x = v \sin \theta$.
For the normal component,the coefficient of restitution $\varepsilon$ is defined as the ratio of the velocity of separation to the velocity of approach along the normal: $\varepsilon = \frac{v'_y}{v_y}$.
Thus,$v'_y = \varepsilon v_y = \varepsilon v \cos \theta$.
Let $\theta'$ be the angle of reflection with the normal. Then,$\tan \theta' = \frac{v'_x}{v'_y} = \frac{v \sin \theta}{\varepsilon v \cos \theta} = \frac{\tan \theta}{\varepsilon}$.
Therefore,$\theta' = \tan ^{-1}\left(\frac{\tan \theta}{\varepsilon}\right)$.

(D) The time taken for the first fall is $t_0 = \sqrt{\frac{2h}{g}}$.
After the first impact,the ball rises to height $h_1 = e^2 h$ and falls back,taking time $t_1 = 2 \sqrt{\frac{2h_1}{g}} = 2e \sqrt{\frac{2h}{g}}$.
Similarly,for subsequent bounces,the time taken is $t_n = 2e^n \sqrt{\frac{2h}{g}}$.
The total time $T$ is the sum of the initial fall and all subsequent bounces:
$T = \sqrt{\frac{2h}{g}} + 2e \sqrt{\frac{2h}{g}} + 2e^2 \sqrt{\frac{2h}{g}} + \dots = \sqrt{\frac{2h}{g}} \left( 1 + 2e + 2e^2 + \dots \right)$.
Using the sum of a geometric series for $e < 1$,$T = \sqrt{\frac{2h}{g}} \left( 1 + 2e \frac{1}{1-e} \right) = \sqrt{\frac{2h}{g}} \left( \frac{1+e}{1-e} \right)$.
Given $h = 0.4 \text{ m}$,$g = 10 \text{ m/s}^2$,and $T = 10 \text{ s}$:
$10 = \sqrt{\frac{2 \times 0.4}{10}} \left( \frac{1+e}{1-e} \right) = \sqrt{0.08} \left( \frac{1+e}{1-e} \right) = 0.2828 \left( \frac{1+e}{1-e} \right)$.
Actually,using $g = 9.8 \text{ m/s}^2$ or simplifying the expression: $10 = \sqrt{0.08} \frac{1+e}{1-e} \implies \frac{10}{0.2828} = \frac{1+e}{1-e} \approx 35.35$.
Solving for $e$ yields $e \approx 0.944$,which corresponds to $\frac{17}{18} \approx 0.944$.

(A) According to the law of conservation of linear momentum,the net external force on the system is zero,so the momentum along the $y$-axis (perpendicular to the initial direction of motion) must be conserved.
Initially,the system has no momentum along the $y$-axis.
After the collision,the components of momentum along the $y$-axis for balls $A$ and $B$ must be equal and opposite.
Let $m_1 = 4 \ kg$ and $m_2 = 1 \ kg$.
$m_1 v_1 \sin(30^{\circ}) = m_2 v_2 \sin(60^{\circ})$
$4 \cdot v_1 \cdot \frac{1}{2} = 1 \cdot v_2 \cdot \frac{\sqrt{3}}{2}$
$2 v_1 = \frac{\sqrt{3}}{2} v_2$
$\frac{v_1}{v_2} = \frac{\sqrt{3}}{4}$

(B) Let the mass of the bullet be $m = 4.2 \times 10^{-2} \text{ kg}$.
Mass of the block $M = 9m = 9 \times 4.2 \times 10^{-2} \text{ kg}$.
Initial velocity of the bullet $v = 300 \text{ m/s}$.
By the law of conservation of linear momentum, the momentum before collision equals the momentum after collision:
$mv = (m + M)V$
$m(300) = (m + 9m)V$
$300m = 10mV$
$V = 30 \text{ m/s}$.
Heat generated is equal to the loss in kinetic energy:
$\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)V^2$
$\Delta K = \frac{1}{2} \times (4.2 \times 10^{-2}) \times (300)^2 - \frac{1}{2} \times (10 \times 4.2 \times 10^{-2}) \times (30)^2$
$\Delta K = \frac{1}{2} \times 4.2 \times 10^{-2} \times (90000 - 10 \times 900)$
$\Delta K = 2.1 \times 10^{-2} \times (90000 - 9000) = 2.1 \times 10^{-2} \times 81000 = 1701 \text{ J}$.
Since $1 \text{ cal} = 4.2 \text{ J}$, the heat in calories is:
$\text{Heat} = \frac{1701}{4.2} = 405 \text{ cal}$.