Question

A 60 – g rifle bullet leaves the muzzle with a velocity of 600 m/s. The 3.0 kg rifle is held firmly by a 60.0 – kg man. With what velocity will the man and rifle recoil?

Answer 1

Let's apply the law of conservation of momentum:

$$m_bv_b+(m_{man}+m_{rifle})v_{recoil}=0,$$ $$(m_{man}+m_{rifle})v_{recoil}=-m_bv_b,$$ $$v_{recoil}=-\dfrac{m_bv_b}{m_{man}+m_{rifle}},$$ $$v_{recoil}=-\dfrac{0.06\ kg\times600\ \dfrac{m}{s}}{60\ kg+3\ kg}=-0.57\ \dfrac{m}{s}.$$

The sign minus means that the recoil velocity of the man and rifle directed in the opposite direction to the motion of the bullet.