A ball of mass m moving with speed u undergoe | vedclass

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(D) For a one-dimensional elastic collision, the final velocity $v_{2f}$ of the target mass $m_2$ (initially at rest) is given by: $v_{2f} = \left( \f

Vedclass · Accepted Answer

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

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(D) For a one-dimensional elastic collision, the final velocity $v_{2f}$ of the target mass $m_2$ (initially at rest) is given by: $v_{2f} = \left( \frac{2m_1}{m_1 + m_2} \right) u$.Given $m_1 = m$ and $m_2 = nm$, we have $v_{2f} = \left( \frac{2m}{m + nm} \right) u = \left( \frac{2}{1 + n} \right) u$.The kinetic energy of the target mass after collision is $K_f = \frac{1}{2} (nm) v_{2f}^2 = \frac{1}{2} nm \left( \frac{4u^2}{(1 + n)^2} \right) = \frac{2nmu^2}{(1 + n)^2}$.The initial kinetic energy of the incident ball is $K_i = \frac{1}{2} mu^2$.The fraction of kinetic energy transferred is $f = \frac{K_f}{K_i} = \frac{\frac{2nmu^2}{(1 + n)^2}}{\frac{1}{2} mu^2} = \frac{4n}{(1 + n)^2}$.

$A$ ball of mass $m$ moving with speed $u$ undergoes a head-on elastic collision with a stationary ball of mass $nm$. What is the fraction of the kinetic energy transferred to the heavier ball?

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