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Question:
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A body \(A\) moving with momentum \(P\) collides one-dimensionally with another stationary body \(B\) of same mass. During impact, \(A\) gives impulse \(J\) to \(B\). Then which of the following is/are correct?
(a) The total momentum of \(A\) and \(B\) is \(P\) before and after impact and \((P-J)\) during the impact.
(b) During the impact, \(B\) gives impulse of magnitude \(J\) to \(A\).
(c) The coefficient of restitution is \(\left[\frac{2 J}{P}-1\right]\).
(d) The coefficient of restitution is \(\left[\frac{2 J}{P}+1\right]\).
Options:
(a) The total momentum of \(A\) and \(B\) is \(P\) before and after impact and \((P-J)\) during the impact.
(b) During the impact, \(B\) gives impulse of magnitude \(J\) to \(A\).
(c) The coefficient of restitution is \(\left[\frac{2 J}{P}-1\right]\).
(d) The coefficient of restitution is \(\left[\frac{2 J}{P}+1\right]\).
Solution:
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Verified Answer
The correct answer is:
(b) and (c) are correct.
According to the question, a collision between two bodies (\(A\) and \(B\)) is shown in the figure below,
So, the coefficient of restitution
\(\begin{aligned}
e & =\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }} \\
\Rightarrow \quad e & =\frac{\frac{J-p}{m}+\frac{J}{m}}{\frac{p}{m}-0}=\frac{2 J}{P}-1 \quad\left(\because v=\frac{m v}{m}=\frac{p}{m}\right)
\end{aligned}\)
During the impact, \(B\) gives impulse
\(J_B=p+J-p=J\)
Hence, the statement (b) and (c) are correct.
So, the coefficient of restitution
\(\begin{aligned}
e & =\frac{\text { Relative velocity of separation }}{\text { Relative velocity of approach }} \\
\Rightarrow \quad e & =\frac{\frac{J-p}{m}+\frac{J}{m}}{\frac{p}{m}-0}=\frac{2 J}{P}-1 \quad\left(\because v=\frac{m v}{m}=\frac{p}{m}\right)
\end{aligned}\)
During the impact, \(B\) gives impulse
\(J_B=p+J-p=J\)
Hence, the statement (b) and (c) are correct.
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