Question

A bullet of mass 25 g is horizontally fired with a velocity of 500 m/s from a gun of mass 5 Kg. The magnitude of the recoil velocity of the gun is ?​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

by conservation of momentum we know that

initial momentum = final momentum

let m1 be the mass of bullet and m2 be the mass of the gun.

m1u1 + m2u2= m1v1 + m2v2

also we know that initially the gun as well as the bullet are at rest .....hence

0=m1v1+m2v2

also SI unit is kg

mass of bullet =m1=10 g = 0.01kg

velocity of bullet=v1=150m/s

mass of gun=m2=4kg

so.....

0=(0.01*150)+4 v2

4v2= -1.5

v2=-(1.5)/4

= -0.375m/s

negative sign represents that it will recoil in a direction opposite to that of the bullet fired

hope this helps