Answer to Question #290397 in Mechanics | Relativity for Lebohang

Question

A 10 g bullet travels horizontally at 2000 m•s-¹ passung through a 2kg pendulum which then moves initially horizontally at 4m•s-¹. Assume that the pendulum loses no mass as the bullet moves through it. Calculate the final speed of the bullet.

Answer 1

Gives

$m_1=0.01kg\\v_1=2000m/sec\\m_2=2kg\\v_2=4m/sec$

We know that

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\frac{{0.01\times2000}+2\times4}{{0.01+2}}=\frac{28}{2.01}=13.93m/sec$