A ball $P$ moving with a speed of $v \mathrm{~ms}^{-1}$ collides directly with another identical ball $Q$…

A ball $P$ moving with a speed of $v \mathrm{~ms}^{-1}$ collides directly with another identical ball $Q$ moving with a speed $10 \mathrm{~ms}^{-1}$ in the opposite direction. $P$ comes to rest after the collision. If the coefficient of restitution is 0.6 , the value of $v$ is
  1. $30\ ms^{-1}$
  2. $40\ ms^{-1}$
  3. $50\ ms^{-1}$
  4. $60\ ms^{-1}$

Solution

From conservation of moment, $\begin{aligned} & m(v-10)=m v_2 \\ & v_2=(v-10) \\ & \longrightarrow \mathrm{V} \stackrel{10}{\bullet} \mathrm{m} / \mathrm{s} \\ & \mathrm{V}=0 \quad \mathrm{~V}_2 \quad \text { Berore } \end{aligned}$ The coefficient of restitution, $\begin{aligned} \mathrm{e} & =\frac{\text { Velocity of separation }}{\text { Velocity of approach }} \\ & =\frac{v_2-v_1}{u_1+u_2}=\frac{(v-10)-0}{(v+10)} \\ \Rightarrow & 0.6=\frac{v-10}{v+10} \\ \Rightarrow & 0.6 v+6=v-10 \\ \Rightarrow & 0.4 v=16 \Rightarrow v=40 \mathrm{~ms}^{-1} \end{aligned}$

Asked in: AP EAMCET 2015

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