Question

20 g bullet fired horizontally towards a wooden block with a mass of 4 kg . The velocity of the bullet is 100 ms−1 . After the bullet has embedded itself in the wooden block , both of them move together . What is their common velocity after the impact ?

Answer 1

We can find the common velocity of the bullet and block after the impact from the law of conservation of momentum:

$$m_1v_1+m_2v_2=(m_1+m_2)v_c,$$ $$v_c=\dfrac{m_1v_1+m_2v_2}{m_1+m_2},$$ $$v_c=\dfrac{20\cdot10^{-3}\ kg\cdot100\ \dfrac{m}{s}+0}{20\cdot10^{-3}\ kg+4\ kg}=0.5\ \dfrac{m}{s}.$$