MHT CET · Physics · Center of Mass Momentum and Collision
A particle of mass \(m\) collides with another stationary particle of mass M . The particle m stops just after collision. The coefficient of restitution is .
- A \(\frac{m}{M}\)
- B \(\frac{M-m}{M+m}\)
- C 1
- D \(\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}\)
Answer & Solution
Correct Answer
(A) \(\frac{m}{M}\)
Step-by-step Solution
Detailed explanation
Let v be the velocity of mass m and \(\mathrm{v}^{\prime}\) be the velocity of mass M after collision.
By law of conservation of momentum, \(\mathrm{mv}=\mathrm{Mv}^{\prime}\)
\(\therefore \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}\)
Coefficient of restitution \(=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}\)
Here, \(\mathrm{v}_1=0\) and \(\mathrm{u}_2=0\)
\(\therefore \quad e=\frac{v^{\prime}}{v}=\frac{m}{M}\)
By law of conservation of momentum, \(\mathrm{mv}=\mathrm{Mv}^{\prime}\)
\(\therefore \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}\)
Coefficient of restitution \(=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}\)
Here, \(\mathrm{v}_1=0\) and \(\mathrm{u}_2=0\)
\(\therefore \quad e=\frac{v^{\prime}}{v}=\frac{m}{M}\)
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