Calculate the bullet speed

Relevant wiki: Analyzing inelastic collisions

Consider $v_{0}$ = the initial speed of the bullet , and $v$ = the speed of the pendulum after the impact . If the bullet remains stuck in the pendulum , then we have to deal with a plastic clush .

From momentum consevation we get $m\cdot v_{0}=(m+M)\cdot v$ .

From the energy conservation of the pendulum we get $(m+M)\cdot \frac{v^2}{2} = (m+M)\cdot g \cdot h$ .

From the two equations we can find $v_{0}$ , which is $\boxed{798 m/s}$

The main insight comes from realizing that the collision between the bullet and the block is completely plastic, therefore the maximum possible amount of energy is lost during the collision. During the collision, we have:

$m_b v_{b} \ = \ (m_b \ + \ M)v_{f}$

Due to conservation of energy:

$\frac{1}{2}(m_b \ + \ M) v_f^2 \ = \ (m_b \ + \ M)gh$

And so we have:

$\frac{1}{2} v_f^2 \ = \ gh \ \Rightarrow \ \frac{m_b^2}{2(m_b \ + \ M)^2} \ v_b^2 \ = \ gh \ \Rightarrow \ v_b \ = \ \sqrt{\frac{2gh(m_b \ + \ M)^2}{m_b^2}} \ = \ \sqrt{2gh} \ \frac{m_b \ + \ M}{m_b} \ = \ \sqrt{2(9.81)(0.19)} \ \frac{4.0097}{0.0097} \ \approx \ 798 \ \text{m}/\text{s}$