Question

A bullet of mass m having a horizontal velocity of 500 m/s hits a stationary block of mass 6.15 kg. The block breaks into two parts viz. Q (mass of 3 kg) and R (mass of 3.15 kg) with the bullet embedded in R. The parts Q and R travel in the direction of initial velocity of the bullet. If the velocity of Q is 3 m/s and the velocity of R is 5 m/s the massof the bullet m is

• A. 5 kg
• B. 0.5kg
• C. 0.05 kg
• D. 0.005 kg

Answer 1

The correct option is C 0.05 kg

Let us consider the bullet and the block as a system.

$\sum \text{(initial momentum of system)}=\sum \text{(final momentum of system)}$

$m\times 500=(3\times 3)+(3.15+m)5$

$500m=9+15.75+5m$

$\therefore 495m=24.75$

$\therefore m=0.05kg$