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(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 velocit

Vedclass · Accepted Answer

$\frac{4 n}{(1+n)^2}$

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(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 $\frac{K_2}{K_1} = \frac{4n}{(n + 1)^2}$.

When a moving body collides with a stationary body of $n$ times its mass,then the amount of kinetic energy transferred to the stationary body is

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