A ball P moving with a speed of v ms^{-1} co | vedclass

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(B) Let the mass of both balls be $m$. Let the initial velocities be $u_P = v$ and $u_Q = -10 \ ms^{-1}$.After the collision,the final velocity of bal

Vedclass · Accepted Answer

$40$

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(B) Let the mass of both balls be $m$. Let the initial velocities be $u_P = v$ and $u_Q = -10 \ ms^{-1}$.After the collision,the final velocity of ball $P$ is $v_P = 0$ and let the final velocity of ball $Q$ be $v_Q$.By the law of conservation of linear momentum:$m(v) + m(-10) = m(0) + m(v_Q)$$v - 10 = v_Q$Now,using the coefficient of restitution formula $e = \frac{	ext{velocity of separation}}{	ext{velocity of approach}}$:$e = \frac{v_Q - v_P}{u_P - u_Q}$Given $e = 0.6$,$v_P = 0$,$u_P = v$,and $u_Q = -10$:$0.6 = \frac{v_Q - 0}{v - (-10)}$$0.6 = \frac{v_Q}{v + 10}$Substitute $v_Q = v - 10$ into the equation:$0.6 = \frac{v - 10}{v + 10}$$0.6(v + 10) = v - 10$$0.6v + 6 = v - 10$$16 = 0.4v$$v = \frac{16}{0.4} = 40 \ ms^{-1}$

$A$ ball $P$ moving with a speed of $v \ ms^{-1}$ collides directly with another identical ball $Q$ moving with a speed $10 \ ms^{-1}$ in the opposite direction. $P$ comes to rest after the collision. If the coefficient of restitution is $0.6$,the value of $v$ is (in $ms^{-1}$)

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