Question

2. A 0.05Kg bullet is shot horizontally in a wooden block of 15Kg at rest. Calculate the speed of the bullet if the maximum height reached by the block with the bullet in it 0.36m.

Answer 1

PROBLEM 2:

The initial momentum of the system is

$p_i=mv_o$

And the momentum of the system just after the bullet hits the block is

P = (m + M)

Conservation of momentum gives us

$p_f=p_i$

$\Rightarrow (m+M)v=mv_o$

$\Rightarrow v=\frac{mv_o}{(m+M)}$

The kinetic energy just after the bullet hits the block is

$KE=\frac{1}{2}(M+m)v^2=\frac{1}{2}(M+m)\left ( \frac{mv_o}{M+m} \right )^2$

$KE= \frac{m^2v_o^2}{2(M+m)}$

And the potential energy at the highest point is

$PE=(M+m)gh$

Conservation of energy gives us

$KE=PE$

$\Rightarrow \frac{m^2v_o^2}{2(M+m)}=(M+m)gh$

$\Rightarrow m^2v_o^2=2(M+m)^2gh$

$\Rightarrow v_o=\sqrt\frac{2(M+m)^2gh}{m^2}$

$\Rightarrow v_o=\frac{M+m}{m}\sqrt{2gh}$

Putting the values

$\Rightarrow v_o=\frac{15+0.05}{0.05}\sqrt{2\times 9.8\times 0.36}=800\ m/s$

So, the speed of the bullet was 800 m/s.