A bullet of mass $\mathrm{m}$ moving with velocity 'v' is fired into a wooden block of mass 'M'. If the…

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
  1. $\frac{v}{m(M+m)}$
  2. $\frac{m+M}{m}$
  3. $\frac{\mathrm{M}+\mathrm{m}}{\mathrm{mv}}$
  4. $\frac{\mathrm{mv}}{\mathrm{m}+\mathrm{M}}$

Solution

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 $V=\frac{m}{m+M}^{v}$

Asked in: MHT CET 2020 (16 Oct Shift 1)

Practice more Center of Mass Momentum and Collision questions on Aicharya