Question

a bullet of mass 30g is fired horizontally with a velocity of 250 m/s² from a pistol of mass 3kg . what is the recoil velocity of the bullet?​

Answer 1

Answer:-

$2.5 \frac{m}{s}$

Explanation:-

Given:

Mass of gun = 3kg = m1

Mass of bullet= 30g = 0.03kg =  m2

Velocity of bullet=250 m/s = v2

To find:

The velocity of the gun and the force of the gun man

Solution:

According to the law of conservation of momentum we have:

m1v1 = m2v2

Substitute the values of m_{1}, m_{2}, v_{2} in the above expression, we get,

3 × v1 = 0.03 × 250

⇒ v1 = $\frac{0.03*250}{3}$ = 2.5 $\frac{m}{s}$

||MrCuteKing||