Answer to Question #248054 in Molecular Physics | Thermodynamics for Dani

Question

A 50n gun fires a 1.0 gram bullet at a speed pf 2200 m/s. what is the speed pf gun recoil?

Answer 1

Let's first find the mass of the gun:

$m_g = \frac{50N}{g} = \frac{50N}{10m/s^2} = 5kg$

where $g=10m/s^2$ is the gravitatinal acceleration.

The momentum of the bullet is:

$p_b = m_bv_b = 0.001kg \times 2200m/s = 2.2kg⋅m/s$

Then, according to the momentum conservation law, the momentum of the gun has the same magnitude:

$p_g = p_b = 2.2 kg⋅m/s$

The speed of gun then:

$v_g = \frac{p_g}{m_g} = \frac{2.2\space kg\cdot m/s }{5kg}=0.44m/s$

Ans. 0.44m/s