Solve this following

Question:

A bullet of ' $4 \mathrm{~g}^{\prime}$ mass is fired from a gun of mass $4 \mathrm{~kg}$. If the bullet moves with the muzzle speed of $50 \mathrm{~ms}^{-1}$, the impulse imparted to the gun and velocity of recoil of gun are:

  1. $0.4 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.1 \mathrm{~ms}^{-1}$

  2. $0.2 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.05 \mathrm{~ms}^{-1}$

  3. $0.2 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.1 \mathrm{~ms}^{-1}$

  4. $0.4 \mathrm{~kg} \mathrm{~ms}^{-1}, 0.05 \mathrm{~ms}^{-1}$


Correct Option: , 2

Solution:

By momentum conservation

$4 \times 10^{-3}(50-v)-4 v=0$

$\mathrm{v}=\frac{4 \times 10^{-3} \times 50}{4+4 \times 10^{-3}} \approx 0.05 \mathrm{~ms}^{-1}$

Impulse $\mathrm{J}=\mathrm{mv}=4 \times .05=0.2 \mathrm{kgms}^{-1}$

Leave a comment