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

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Question

(A) For a one-dimensional elastic collision,the velocity of the first body $(m_1 = m)$ after the collision with a second body $(m_2 = 2m)$ initially a

Vedclass · Accepted Answer

$9$

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(A) For a one-dimensional elastic collision,the velocity of the first body $(m_1 = m)$ after the collision with a second body $(m_2 = 2m)$ initially at rest $(u_2 = 0)$ is given by:$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 = \frac{m - 2m}{m + 2m} v = -\frac{v}{3}$The kinetic energy of the first body before the collision is $KE_{before} = \frac{1}{2} m v^2$.The kinetic energy of the first body after the collision is $KE_{after} = \frac{1}{2} m v_1^2 = \frac{1}{2} m (-\frac{v}{3})^2 = \frac{1}{2} m \frac{v^2}{9}$.The ratio of the kinetic energy before to after the collision is:$\frac{KE_{before}}{KE_{after}} = \frac{\frac{1}{2} m v^2}{\frac{1}{2} m \frac{v^2}{9}} = 9 : 1$.

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

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