Ball \(A\) moving with momentum \(p\) undergoes a one dimensinal collision with a stationary ball \(B\) of…

Ball \(A\) moving with momentum \(p\) undergoes a one dimensinal collision with a stationary ball \(B\) of the same mass. During the collision ball \(B\) imparts an impulse \(I\) to ball \(A\). The coefficient of restitution is
  1. \(\frac{2 I}{p}\)
  2. \(\frac{2 I}{p}-1\)
  3. \(\frac{I}{p}+1\)
  4. \(\frac{2 I}{p}+1\)

Solution

Let $m$ be the mass of each ball. Let $u_{1}$ and $u_{2}$ be the velocities of $A$ and $B$ before collision and $v_{1}$ and $v_{2}$ are after collision. Then $u_{1}=\frac{p}{m}$ and $u_{2}=0$ (given). Impulse = change in momentum. It is given that $I=$ change in momentum of $B$. $\begin{aligned} &=m v_{2}-0 \\ \Rightarrow \quad & v_{2} =\frac{I}{m} \end{aligned}$ Therefore, $v_{1}=\frac{p-I}{m}$. Now $\begin{aligned} e &=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \\ &=\frac{\left(\frac{I}{m}-\frac{p-I}{m}\right)}{\left(\frac{p}{m}-0\right)}=\frac{2 I}{p}-1 \end{aligned}$ So the correct choice is (b).

Asked in: JEE Mains - Rotational Motion - Test 4

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