Elastic Collision Questions in English

(A) Let the mass of the moving body be $m_1$ and its initial velocity be $u$. The mass of the stationary body is $M_2 = n m_1$ and its initial velocity is $0$.
For a perfectly elastic collision,the law of conservation of momentum gives:
$m_1 u = m_1 v_1 + M_2 v_2$
$m_1 u = m_1 v_1 + n m_1 v_2$
$u = v_1 + n v_2$ ... $(i)$
The coefficient of restitution $e = 1$ for an elastic collision,so:
$v_2 - v_1 = u - 0$
$v_1 = v_2 - u$ ... (ii)
Substituting (ii) into $(i)$:
$u = (v_2 - u) + n v_2$
$2u = (n + 1) v_2$
$v_2 = \frac{2u}{n + 1}$
The initial kinetic energy of the moving body is $K_1 = \frac{1}{2} m_1 u^2$.
The kinetic energy transferred to the stationary body is $K_2 = \frac{1}{2} M_2 v_2^2$.
$K_2 = \frac{1}{2} (n m_1) \left( \frac{2u}{n + 1} \right)^2 = \frac{1}{2} n m_1 \frac{4 u^2}{(n + 1)^2} = \left( \frac{1}{2} m_1 u^2 \right) \frac{4n}{(n + 1)^2}$.
The fraction of kinetic energy transferred is $\frac{K_2}{K_1} = \frac{4n}{(n + 1)^2}$.

(A) Let the radius of the first ball be $r_1 = 2 \,cm$ and the second ball be $r_2 = 4 \,cm$.
Assuming both balls are made of the same material with density $\rho$, the mass $m$ is given by $m = \frac{4}{3} \pi r^3 \rho$.
Thus, the ratio of masses is $\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3} = \left(\frac{2}{4}\right)^3 = \frac{1}{8}$, which implies $m_2 = 8m_1$.
For a one-dimensional elastic collision, the final velocity $v_1$ of the first ball (initially at rest) is given by the formula:
$v_1 = \left(\frac{2m_2}{m_1 + m_2}\right) u_2$, where $u_2 = 81 \,cm \,s^{-1}$ is the initial velocity of the second ball.
Substituting the values:
$v_1 = \left(\frac{2(8m_1)}{m_1 + 8m_1}\right) \times 81$
$v_1 = \left(\frac{16m_1}{9m_1}\right) \times 81$
$v_1 = \frac{16}{9} \times 81 = 16 \times 9 = 144 \,cm \,s^{-1}$.

(D) Let the initial velocity of the first ball be $u_1 = u$ and the second ball be $u_2 = 0$. After the collision, the velocity of the first ball is $v_1 = \frac{u}{3}$.
Since the collision is elastic, the coefficient of restitution $e = 1$.
The formula for the velocity of the first ball after an elastic collision is $v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2$.
Substituting the known values: $\frac{u}{3} = \frac{2 - M}{2 + M} u + 0$.
Dividing by $u$: $\frac{1}{3} = \frac{2 - M}{2 + M}$.
Cross-multiplying: $2 + M = 3(2 - M) = 6 - 3M$.
Rearranging terms: $M + 3M = 6 - 2$, which gives $4M = 4$.
Therefore, $M = 1 \,kg$.

(A) Let the mass of each sphere be $m$. The distance between $A$ and $B$ along the semi-circular path is $\pi r$. Let the initial velocity of $A$ be $v_0$. Since $A$ covers the distance $\pi r$ in time $t$,we have $v_0 = \frac{\pi r}{t}$.
After the first collision,by the law of conservation of momentum and the definition of the coefficient of restitution $e$,the velocities of $A$ and $B$ become $v_A = \frac{v_0(1-e)}{2}$ and $v_B = \frac{v_0(1+e)}{2}$.
The spheres are now moving in the same direction along the circular groove. The relative velocity between them is $v_{rel} = v_B - v_A = v_0 e$.
The distance they must cover to collide again is the full circumference of the groove,which is $2\pi r$.
The time taken for the next collision is $t' = \frac{2\pi r}{v_{rel}} = \frac{2\pi r}{v_0 e}$.
Substituting $v_0 = \frac{\pi r}{t}$,we get $t' = \frac{2\pi r}{(\pi r / t) e} = \frac{2t}{e}$.

(C) Initial kinetic energy of ball $A$ is $K = p^2 / (2m)$,so $p = \sqrt{2mK}$.
Initial velocity of ball $A$ is $u_1 = p/m = \sqrt{2K/m}$.
After collision,ball $A$ moves in the negative $x$-direction with kinetic energy $K' = K/9$.
Let $v_1$ be the final velocity of ball $A$. Then $\frac{1}{2}mv_1^2 = K/9$,which gives $v_1 = \sqrt{2K/(9m)} = \frac{1}{3}\sqrt{2K/m} = u_1/3 = p/(3m)$.
Since the collision is one-dimensional,we apply the law of conservation of linear momentum:
$p_{initial} = p_{final}$
$p = -mv_1 + p_B$
$p_B = p + mv_1$
Substituting $v_1 = p/(3m)$:
$p_B = p + m(p/(3m)) = p + p/3 = 4p/3$.

(A) In an elastic collision between two bodies of equal mass $(m_1 = m_2 = m)$,the final velocities $v_1$ and $v_2$ are given by the conservation of momentum and kinetic energy as:
$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1 + \frac{2m_2}{m_1 + m_2}u_2$
$v_2 = \frac{2m_1}{m_1 + m_2}u_1 + \frac{m_2 - m_1}{m_1 + m_2}u_2$
Since $m_1 = m_2$,these equations simplify to $v_1 = u_2$ and $v_2 = u_1$.
Statement $A$ describes the specific case where $u_2 = 0$,resulting in $v_1 = 0$ and $v_2 = u_1$,which is true.
Statement $B$ is the general case of the same principle,which is also true. Thus,both statements are correct.

(D) Let $v_1$ and $v_2$ be the velocities of the bodies of mass $m$ and $2m$ respectively after the collision.
By the law of conservation of linear momentum: $mv + (2m)(0) = mv_1 + 2mv_2$,which simplifies to $v = v_1 + 2v_2$ (Equation $1$).
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Given $u_1 = v$ and $u_2 = 0$,we have $e = \frac{v_2 - v_1}{v}$,which implies $v_1 = v_2 - ev$ (Equation $2$).
Substitute Equation $2$ into Equation $1$: $v = (v_2 - ev) + 2v_2$.
$v(1+e) = 3v_2$,so $v_2 = \frac{v(1+e)}{3}$.
Now,substitute $v_2$ back into Equation $2$: $v_1 = \frac{v(1+e)}{3} - ev = \frac{v + ev - 3ev}{3} = \frac{v(1-2e)}{3}$.
The ratio of the velocities $v_1/v_2$ is $\frac{v(1-2e)/3}{v(1+e)/3} = \frac{1-2e}{1+e}$.

(B) Let the mass of the moving particle be $m$ and its initial velocity be $u$. The mass of the stationary particle is $m' = \frac{m}{n}$.
Assuming a perfectly elastic head-on collision,we apply the conservation of linear momentum:
$mu = mv_1 + \frac{m}{n}v_2 \Rightarrow u = v_1 + \frac{v_2}{n}$ (Equation $1$)
Using the coefficient of restitution $e=1$ for an elastic collision:
$v_2 - v_1 = u \Rightarrow v_1 = v_2 - u$ (Equation $2$)
Substituting Equation $2$ into Equation $1$:
$u = (v_2 - u) + \frac{v_2}{n} \Rightarrow 2u = v_2(1 + \frac{1}{n}) = v_2(\frac{n+1}{n})$
$v_2 = \frac{2nu}{n+1}$
The kinetic energy transferred to the stationary particle is $K' = \frac{1}{2} m' v_2^2 = \frac{1}{2} (\frac{m}{n}) (\frac{2nu}{n+1})^2 = \frac{1}{2} \frac{m}{n} \frac{4n^2 u^2}{(n+1)^2} = \frac{2mnu^2}{(n+1)^2}$.
The initial kinetic energy is $K = \frac{1}{2} mu^2$.
The fraction of kinetic energy transferred is $\frac{K'}{K} = \frac{\frac{2mnu^2}{(n+1)^2}}{\frac{1}{2} mu^2} = \frac{4n}{(n+1)^2}$.

(A) Let the initial velocity of $m_1$ be $\vec{u}_1$ and final velocities be $\vec{v}_1$ and $\vec{v}_2$. Since the collision is elastic,kinetic energy and momentum are conserved.
Conservation of momentum: $m_1 \vec{u}_1 = m_1 \vec{v}_1 + m_2 \vec{v}_2$.
Squaring both sides: $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2 + 2 m_1 m_2 \vec{v}_1 \cdot \vec{v}_2$.
Since the particles move at $90^{\circ}$ to each other,$\vec{v}_1 \cdot \vec{v}_2 = 0$,so $m_1^2 u_1^2 = m_1^2 v_1^2 + m_2^2 v_2^2$.
Conservation of kinetic energy: $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$,which implies $m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2$.
From energy conservation,$m_1(u_1^2 - v_1^2) = m_2 v_2^2$.
From momentum conservation,$m_1^2(u_1^2 - v_1^2) = m_2^2 v_2^2$.
Dividing the two equations: $m_1 = m_2$. Thus,the ratio $\frac{m_2}{m_1} = 1$.

(A) By the law of conservation of linear momentum:
Total momentum before collision $=$ Total momentum after collision
$m_1 u + m_2(0) = m_1 \left(\frac{2u}{3}\right) + m_2 v$
$m_1 u - \frac{2}{3} m_1 u = m_2 v$
$\frac{1}{3} m_1 u = m_2 v$ --- $(i)$
Since the collision is perfectly elastic,the coefficient of restitution $e = 1$:
$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$
$\frac{v - 2u/3}{u - 0} = 1$
$v - \frac{2u}{3} = u$
$v = u + \frac{2u}{3} = \frac{5u}{3}$
Substituting $v$ into Eq. $(i)$:
$\frac{1}{3} m_1 u = m_2 \left(\frac{5u}{3}\right)$
$\frac{m_1}{m_2} = \frac{5u/3}{u/3} = 5$

(D) Let the initial velocity of the $1 \ kg$ ball be $u$. After the collision,the first ball moves with velocity $v_1$ along the $Y$-axis,and the second ball of mass $m$ moves with velocity $v_2$ at an angle $30^{\circ}$ below the $X$-axis.
Applying the law of conservation of linear momentum along the $X$-axis:
$1 \cdot u = m v_2 \cos(30^{\circ}) \implies u = m v_2 \frac{\sqrt{3}}{2} \implies v_2 = \frac{2u}{m\sqrt{3}} \quad ... (1)$
Applying the law of conservation of linear momentum along the $Y$-axis:
$0 = 1 \cdot v_1 - m v_2 \sin(30^{\circ}) \implies v_1 = m v_2 \sin(30^{\circ}) = m v_2 \cdot \frac{1}{2} \implies v_2 = \frac{2v_1}{m} \quad ... (2)$
From $(1)$ and $(2)$,$\frac{2u}{m\sqrt{3}} = \frac{2v_1}{m} \implies v_1 = \frac{u}{\sqrt{3}}$.
Since the collision is elastic,kinetic energy is conserved:
$\frac{1}{2} (1) u^2 = \frac{1}{2} (1) v_1^2 + \frac{1}{2} m v_2^2$
$u^2 = v_1^2 + m v_2^2$
Substitute $v_1 = \frac{u}{\sqrt{3}}$ and $v_2 = \frac{2v_1}{m} = \frac{2u}{m\sqrt{3}}$:
$u^2 = \left(\frac{u}{\sqrt{3}}\right)^2 + m \left(\frac{2u}{m\sqrt{3}}\right)^2$
$u^2 = \frac{u^2}{3} + m \cdot \frac{4u^2}{3m^2} = \frac{u^2}{3} + \frac{4u^2}{3m}$
$1 = \frac{1}{3} + \frac{4}{3m} \implies \frac{2}{3} = \frac{4}{3m} \implies m = 2 \ kg$.

(A) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
$e = \frac{v_2 - v_1}{u_1 - u_2}$
Given:
$u_1 = 3 \ m/s$,$u_2 = 1.5 \ m/s$
$v_1 = 2.5 \ m/s$,$v_2 = 3.5 \ m/s$
Substituting the values:
$e = \frac{3.5 - 2.5}{3 - 1.5} = \frac{1}{1.5} = \frac{10}{15} = \frac{2}{3} \approx 0.67$
Thus,the coefficient of restitution is approximately $0.67$.

(C) In an elastic collision between two bodies of equal mass where one is initially at rest,the conservation of linear momentum and kinetic energy leads to the equation: $\vec{v}_1 = \vec{v}_1' + \vec{v}_2'$.
Since the collision is elastic and masses are equal,the conservation of kinetic energy gives: $v_1^2 = (v_1')^2 + (v_2')^2$.
Comparing these two equations with the vector relation $\vec{v}_1^2 = (\vec{v}_1' + \vec{v}_2')^2 = (v_1')^2 + (v_2')^2 + 2\vec{v}_1' \cdot \vec{v}_2'$,we find that $2\vec{v}_1' \cdot \vec{v}_2' = 0$.
This implies that the dot product of the final velocity vectors is zero,meaning the angle between the two balls after the collision is $90^{\circ}$ (provided the collision is not head-on).

(A) Let the mass of the first body be $m_1 = 5 \ kg$ and its initial velocity be $u$. Let the mass of the second body be $M$,which is initially at rest $(u_2 = 0)$.
For an elastic collision,the coefficient of restitution $e = 1$.
After the collision,the velocity of the first body becomes $v_1 = \frac{u}{10}$. Let the velocity of the second body be $v_2$.
By the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$5u + M(0) = 5 \left(\frac{u}{10}\right) + M v_2$
$5u = \frac{u}{2} + M v_2 \quad \dots (i)$
Using the property of elastic collision $(v_1 - v_2 = -e(u_1 - u_2))$:
$\frac{u}{10} - v_2 = -1(u - 0)$
$v_2 = \frac{u}{10} + u = \frac{11u}{10} \quad \dots (ii)$
Substituting $v_2$ from Eq. $(ii)$ into Eq. $(i)$:
$5u = \frac{u}{2} + M \left(\frac{11u}{10}\right)$
$5 - 0.5 = M \left(\frac{11}{10}\right)$
$4.5 = M \left(\frac{11}{10}\right)$
$M = \frac{4.5 \times 10}{11} = \frac{45}{11} \approx 4.09 \ kg$.

(C) In an elastic collision,both the total linear momentum and the total kinetic energy of the system are conserved.
Assertion $(A)$ is true because by definition,an elastic collision is one in which there is no loss of kinetic energy.
Reason $(R)$ is false because,during an elastic collision,there is indeed an exchange of energy between the colliding bodies (momentum and kinetic energy are transferred between them),even though the total kinetic energy of the system remains constant.
Therefore,$(A)$ is true but $(R)$ is false.

(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,as no external force acts on the system.
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
Given: $m_A = 20 \ kg$,$u_A = 20 \ m \ s^{-1}$,$m_B = 200 \ kg$,$u_B = 10 \ m \ s^{-1}$.
After collision,$v_A = -10 \ m \ s^{-1}$ (since it bounces back in the opposite direction).
Substituting the values:
$(20 \times 20) + (200 \times 10) = (20 \times -10) + (200 \times v_B)$
$400 + 2000 = -200 + 200 v_B$
$2400 = -200 + 200 v_B$
$2600 = 200 v_B$
$v_B = 13 \ m \ s^{-1}$

(C) For a one-dimensional elastic collision between two masses $m_1$ and $m_2$,where $m_2$ is initially at rest,the velocity $v_2$ of the second mass after the collision is given by $v_2 = \frac{2m_1}{m_1 + m_2} v_1$.
1st collision: Mass $m_1 = m$ hits $m_2 = \frac{m}{2}$ with velocity $v_0$. The velocity of the second ball $v_1$ is:
$v_1 = \frac{2m}{m + \frac{m}{2}} v_0 = \frac{2m}{\frac{3m}{2}} v_0 = \frac{4}{3} v_0$.
2nd collision: Mass $m_2 = \frac{m}{2}$ hits $m_3 = \frac{m}{4}$ with velocity $v_1 = \frac{4}{3} v_0$. The velocity of the third ball $v_2$ is:
$v_2 = \frac{2(\frac{m}{2})}{\frac{m}{2} + \frac{m}{4}} v_1 = \frac{m}{\frac{3m}{4}} v_1 = \frac{4}{3} v_1 = \left(\frac{4}{3}\right)^2 v_0$.
Following this pattern,for the $n^{\text{th}}$ ball,the velocity $v_{n-1}$ after $(n-1)$ collisions will be:
$v_{n-1} = \left(\frac{4}{3}\right)^{n-1} v_0$.

(B) When the spring is at maximum compression,both blocks move with the same velocity $v_{cm}$.
By the law of conservation of linear momentum: $mv + m(0) = (m + m)v_{cm} \Rightarrow v_{cm} = \frac{v}{2}$.
By the law of conservation of energy,the initial kinetic energy of the system equals the sum of the final kinetic energy and the potential energy stored in the spring at maximum compression $x$:
$\frac{1}{2}mv^2 = \frac{1}{2}(2m)v_{cm}^2 + \frac{1}{2}kx^2$
$\frac{1}{2}mv^2 = m(\frac{v}{2})^2 + \frac{1}{2}kx^2$
$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}kx^2$
$\frac{1}{4}mv^2 = \frac{1}{2}kx^2$
$x^2 = \frac{mv^2}{2k} \Rightarrow x = v\sqrt{\frac{m}{2k}}$.