Answer to Question #91257 in Mechanics | Relativity for Umar

Question

A bullet of mass 0.045kg is fired from a gun of mass 9kg, it bullet moving with an initial velocity of 200m/s. Find the initial backward velocity of the gun.

Answer 1

Since, the total initial momentum of the gun and the bullet is zero (both are initially at rest), we applying the law of conservation of momentum obtain:

$$p_{total(initial)} = p_{total(final)},$$ $$p_{total(initial)} = 0,$$ $$M_{gun}v_{backward} + m_bv_b = 0.$$

From this equation we can calculate the initial backward velocity of the gun:

$$v_{backward} = - \dfrac{m_bv_b}{M_{gun}} = - \dfrac{0.045 kg \cdot 200 \dfrac{m}{s}}{9 kg} = -1 \dfrac{m}{s}.$$

The sign minus indicates, that the backward velocity of the gun directed in the opposite direction to the velocity of the bullet.

Answer:

$v_{backward} = 1 \dfrac{m}{s}.$