Body A of mass 4m moving with speed u collide | vedclass

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Question

(C) For a one-dimensional elastic collision, the final velocity of body $A$ $(v_1)$ is given by:$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1

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$\frac{8}{9}$

Vedclass · Answer

(C) For a one-dimensional elastic collision, the final velocity of body $A$ $(v_1)$ is given by:$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$Given $m_1 = 4m$, $u_1 = u$, $m_2 = 2m$, and $u_2 = 0$:$v_1 = \left( \frac{4m - 2m}{4m + 2m} \right) u + 0 = \left( \frac{2m}{6m} \right) u = \frac{1}{3} u$Initial kinetic energy of body $A$ is $K_i = \frac{1}{2} (4m) u^2 = 2mu^2$.Final kinetic energy of body $A$ is $K_f = \frac{1}{2} (4m) (\frac{1}{3} u)^2 = 2m (\frac{1}{9} u^2) = \frac{2}{9} mu^2$.Energy lost by body $A$ is $\Delta K = K_i - K_f = 2mu^2 - \frac{2}{9} mu^2 = \frac{16}{9} mu^2$.The fraction of energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{16}{9} mu^2}{2mu^2} = \frac{8}{9}$.

Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2m$ at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body $A$ is:

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