Question

A rifle of mass 5 kg fires a bullet of mass 40 g. The bullet leaves the barrel of the rifle with a velocity 200 m/s. If the bullet takes 0.004 s to leave the barrel, calculate the following: (i) recoil velocity of the rifle and (ii) the force experienced by the rifle due to its recoil.

Answer 1

Mass of the rifle, $m_1$ ​ = 5 kg

Mass of the bullet, $m_2$ ​ = 0.04 kg

Initial velocities, $u_1$ ​ = 0, $u_2$ ​ = 0

After firing, velocity of the bullet, $v_2$ ​ = 200 $\frac{m}{s}$

Velocity of the rifle, $v_1$​ = ?

By the law of conservation of momentum, $m_1$​$u_1$​ + $m_2$​$u_2$​​ = $m_1$​$v_1$​ + $m_2$​$v_2$​​​

$\Rightarrow$ 0 + 0 = (5 $\times$ $v_1$​) + (0.04 $\times$ 200)

$v_1$​​ = -1.6 $\frac{m}{s}$

Initial momentum of the rifle =0

Final momentum of the rifle = 5 kg $\times$ (-1.6 $\frac{m}{s}$) = -8 kg$\frac{m}{s}$

Time interval = 0.004 s

Force = -2000 N