Question

A 70 gram bullet moving east at 400 m/s strikes a 1.3 kg pendulum block on the right and becomes embedded on it. How high will the block-bullet rise above its original position?

Answer 1

According to momentum conservation:

$$mv=(m+M)u,\\\space\\ u=\frac{mv}{m+M}=20.4\text{ m/s}.$$

Now this total kinetic energy is spent to rise the pendulum block:

$$\frac 12(M+m)u^2=(m+M)gh,\\\space\\ h=\frac {u^2}{2g}=21.2\text{ m}.$$