A moving body with a mass m_1 and velocity u | vedclass

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(A) By the law of conservation of linear momentum:Total momentum before collision $=$ Total momentum after collision$m_1 u + m_2(0) = m_1 \left(\frac{

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) By the law of conservation of linear momentum:Total momentum before collision $=$ Total momentum after collision$m_1 u + m_2(0) = m_1 \left(\frac{2u}{3}\right) + m_2 v$$m_1 u - \frac{2}{3} m_1 u = m_2 v$$\frac{1}{3} m_1 u = m_2 v$  --- $(i)$Since the collision is perfectly elastic,the coefficient of restitution $e = 1$:$e = \frac{v_2 - v_1}{u_1 - u_2} = 1$$\frac{v - 2u/3}{u - 0} = 1$$v - \frac{2u}{3} = u$$v = u + \frac{2u}{3} = \frac{5u}{3}$Substituting $v$ into Eq. $(i)$:$\frac{1}{3} m_1 u = m_2 \left(\frac{5u}{3}\right)$$\frac{m_1}{m_2} = \frac{5u/3}{u/3} = 5$

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