A body of mass m moving with velocity v colli | vedclass

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(D) Let the mass of the first body be $m_1 = m$ and the second body be $m_2 = 2m$.Initial velocity of the first body is $u_1 = v$ and the second body

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$9$

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(D) Let the mass of the first body be $m_1 = m$ and the second body be $m_2 = 2m$.Initial velocity of the first body is $u_1 = v$ and the second body is $u_2 = 0$.For a head-on elastic collision,the final velocity $v_1$ of the first body is given by:$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{m - 2m}{m + 2m} v = \frac{-m}{3m} v = -\frac{v}{3}$.The kinetic energy before collision is $K_i = \frac{1}{2} m v^2$.The kinetic energy after collision is $K_f = \frac{1}{2} m v_1^2 = \frac{1}{2} m (-\frac{v}{3})^2 = \frac{1}{2} m \frac{v^2}{9} = \frac{K_i}{9}$.The ratio of kinetic energy before and after collision is $\frac{K_i}{K_f} = \frac{K_i}{K_i / 9} = 9:1$.

$A$ body of mass $m$ moving with velocity $v$ collides head-on with another body of mass $2m$ which is initially at rest. The ratio of the kinetic energy of the colliding body before and after the collision will be: (in $:1$)

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