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