Question

a bullet of mass 0.01 kg is fired from a gun weighing 5 kg . If the initial speed of the bullet is 250 m/s,calculate the speed at which the gun recoils ?

Answer 1

we know that
mass of bullet × velocity of bullet = mass of gun× velocity of gun (recoil)
0.01 × 250 = 5 × velocity of gun
1/100 × 250 = 5 × velocity of gun
250/100 = 5 × velocity of gun
5/2 = 5 × velocity of gun
5/10 = 1/2 m/s => velocity of gun (recoil)

recoil velocity = 1/2m/s
velocity of gun = -1/2 m/s ( negative sign denotes opposite direction)
hope this helps

Answer 2

Hello dear !
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Here Conservation of linear momentum i.e newton' third law of motion used .

It simply means when momentum transferred it remains conserved 

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In this case of question the momentum transferred  on bullet is equal and opposite to the gun. 

so 
let the recoil velocity of gun = 'v'

p(gun) = -p(bullet) 

5 x v    = -0.01 x 250 

5v       = -2.5 

v         = -2.5 / 5 

v         = -0.5 m/s 
here the negative sign shows the direction which is opposite to the path of bullet