Question

A ballistic pendulum consists of a large block of wood of mass M=9.99kg. One bullet, of mass m= 10g. traveling horizontally with speed v. hits the pendulum and remains stuck in it. After the collision the pendulum starts to oscillate reaching a maximum height h= 5cm What is the velocity V of the bullet?

Answer 1

Let's first find the velocity of the block immediately after the bullet stucks in it from the law of conservation of energy:

$$PE=KE,$$ $$Mgh=\dfrac{1}{2}Mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot0.05\ m}=0.99\ \dfrac{m}{s}.$$

Finally, we can find the initial velocity of the bullet from the law of conservation of momentum:

$$mv_b=(m+M)v_f,$$ $$v_b=\dfrac{(m+M)v_f}{m},$$ $$v_b=\dfrac{(0.01\ kg+9.99\ kg)\cdot0.99\ \dfrac{m}{s}}{0.01\ kg}=990\ \dfrac{m}{s}.$$