Question

a gun of mass 4kg fires a bullet of mass 40kg.the bullet leaves the barrel of the gun at a velocity of 129m/s.if the bullet takes 0.002 sec to move through its barrel,calculaye the force experienced by the gun due to its recoil?

Answer 1

how can be the bullet of 40 kg n gun of 4 kg...
Assuming bullet of 4 kg I m solving the sum.

initial velocity of bullet, u=0m/s
final " ,v=129m/s
time to gain this velocity =0.002s

As we know v-u/t =a
so, acceleration of bullet=129/0.002= 64500m/s^2

Also, Force=m×a
so, force experienced by the bullet= 4×64500= 258000 N

Now applying Newton's 3rd law of motion on the both body ,
Force in the gun =- force on the bullet.
so, force experienced by the gun during recoil = - 258 kN ; where -ve sign signifies that the force is acting in the opposite direction of the bullet.

Answer 2

By using equation,
$(m1 \times u1) + (m2 \times u2) = (m1 \times v1) + (m2 \times v2)$
Where,
m1= 4kg
m2=40kg
u1=u2=0m/s
v1=?????????
v2=129m/s

$(4 \times 0) + (40 \times 0) = (4 \times v1) + (40 \times 129)$
$0 = 4v1 + 5160$