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(D) The velocity of the first body after a one-dimensional elastic collision is given by the formula:$v_{1} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\ri

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The velocity of the first body after a one-dimensional elastic collision is given by the formula:$v_{1} = \left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1} + \frac{2 m_{2} u_{2}}{m_{1}+m_{2}}$Given $m_{1} = m$,$m_{2} = 2m$,$u_{1} = v$,and $u_{2} = 0$:$v_{1} = \left(\frac{m - 2m}{m + 2m}\right) v + \frac{2(2m)(0)}{3m} = \left(\frac{-m}{3m}\right) v = -\frac{v}{3}$Initial kinetic energy of the first body: $KE_{i} = \frac{1}{2} mv^{2}$Final kinetic energy of the first body: $KE_{f} = \frac{1}{2} m\left(-\frac{v}{3}\right)^{2} = \frac{1}{2} m \left(\frac{v^{2}}{9}\right) = \frac{KE_{i}}{9}$The ratio of $KE_{i}$ to $KE_{f}$ is:$\frac{KE_{i}}{KE_{f}} = \frac{KE_{i}}{KE_{i}/9} = 9$Thus,the ratio is $9 : 1$.

$A$ 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 $K.E.$ of the colliding body before and after the collision will be: (in $: 1$)

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