Q: Two wooden blocks of mass M1 = 1 kg, M2 = 2.98 kg lie separately side by side smooth surface. A bullet of mass m = 20 gm strikes the block M1 and pierces through it, then strikes the second block and sticks to it. Consequently both the blocks move with equal velocities. Find the percentage change in speed of the bullet when it escapes from the first block.

Sol: Conservation of linear momentum of the system yields

$\displaystyle m v_0 = M_1 v + m v’ = (M_2 + m ) v $

$\displaystyle v = \frac{m v_0}{M_2 + m}$

and m_{v’} = m v_{0} – M_{1} v

$\displaystyle m v’ = m v_0 – \frac{M_1 m v_0}{M_2 + m}$

$\displaystyle m v’ = m v_0(1 – \frac{M_1}{M_2 + m}) $

Now find , $\displaystyle \frac{|v’-v_0|}{v_0} \times 100 = \frac{M_2 – M_1 + m}{M_2 + m} \times 100 $