Answer to Question #142133 in Physics for Ranxe

Question

A 50 N gun fires a 1.0 gram bullet at a speed of 2200 m/s. What is the speed of gun recoil?

Answer 1

Let's first find the mass of the gun:

$$M=\dfrac{W}{g}=\dfrac{50\ N}{9.8\ \dfrac{m}{s^2}}=5.1\ kg.$$

Then, we can find the recoil speed of gun from the Law of Conservation of Momentum:

$$Mv_{recoil}+mv=0,$$ $$Mv_{recoil}=-mv,$$ $$v_{recoil}=-\dfrac{mv}{M}=-\dfrac{1.0\cdot 10^{-3}\ kg \cdot 2200\ \dfrac{m}{s}}{5.1\ kg}=-0.43\ \dfrac{m}{s}.$$

The sign minus means that the recoil speed of gun directed in the opposite direction to the speed of the bullet.

Answer:

$v_{recoil}=0.43\ \dfrac{m}{s},$ in the opposite direction to the speed of the bullet.