MHT CET · Physics · Center of Mass Momentum and Collision
A bullet of mass \(\mathrm{m}\) moving with velocity 'v' is fired into a wooden block of mass 'M'. If the bullet remains embedded in the block, the final velocity of the system is
- A \(\frac{V}{m(M+m)}\)
- B \(\frac{m+M}{m}\)
- C \(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{mV}}\)
- D \(\frac{\mathrm{mV}}{\mathrm{m}+\mathrm{M}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{mV}}{\mathrm{m}+\mathrm{M}}\)
Step-by-step Solution
Detailed explanation
Since there is no extra force other than the action and reaction force so the linear momentum should be conserved.
Suppose the system moves with velocity V then momentum before collision is mv and that after collision will be \((M+\) \(\mathrm{m}) \mathrm{V}\)
Equating both we get \(m v=(M+m) V\) or \(V=\frac{m}{m+M}^{v}\)
Suppose the system moves with velocity V then momentum before collision is mv and that after collision will be \((M+\) \(\mathrm{m}) \mathrm{V}\)
Equating both we get \(m v=(M+m) V\) or \(V=\frac{m}{m+M}^{v}\)
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