Question

A bullet of mass 25 grams is horizontally fired with a velocity of 150 m/s^-1 from a gun of mass 4kg what is the recoil velocity of the gun ?

Answer 1

According to law of conservation of linear momentum,

MV = mv.

M is mass of pistol = 4 kg.

m is mass of bullet = 25 g = 25 (10 ^-3) kg.

v = velocity of bullet = 150 m/s.

V is recoil velocity of pistol . V= (m/M) v = [ 25(10^-3)/(4)] (150) = 0.9375 m/s

PLEASE MARK IT AS BRAINLIEST!

Answer 2

mass of bullet,

m1 = 25g

= 25/1000 kg

= 0.025 kg

velocity, v = 150 m/s

mass of gun, m2 = 4kg

Using momentum conservation along horizontal line

m1u1 + m2v2 = m1v1 + m2v2

( 0.025 ) ( 0 ) + ( 4 ) ( 0 ) = 0.025 ( 150 ) + 4 ( V2 )

0 = 0.25 × 15 + 4V2

– 3.75 = 4V2

V2 = – 3.75/4

V2 = – 0.9375 m/s