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

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Question

(C) For a perfectly elastic head-on collision,the fraction of kinetic energy lost by the incident body of mass $m_1$ colliding with a stationary body

Vedclass · Accepted Answer

$8/9$ of initial kinetic energy

Vedclass · Answer

(C) For a perfectly elastic head-on collision,the fraction of kinetic energy lost by the incident body of mass $m_1$ colliding with a stationary body of mass $m_2$ is given by the formula:$\frac{\Delta K}{K} = \frac{4 m_1 m_2}{(m_1 + m_2)^2}$Given $m_1 = m$ and $m_2 = 2m$:$\frac{\Delta K}{K} = \frac{4(m)(2m)}{(m + 2m)^2}$$\frac{\Delta K}{K} = \frac{8m^2}{(3m)^2} = \frac{8m^2}{9m^2} = \frac{8}{9}$Therefore,the body of mass $m$ loses $8/9$ of its initial kinetic energy.

$A$ body of mass $m$ moving with velocity $v$ collides with a stationary body of mass $2m$. The fraction of kinetic energy lost by the body of mass $m$ is:

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