A gun of mass 8kg fires a bullet with a mass 25g at velocity of 320 metre per second what is th recoil velocity of the gun?

8v + .025*320 = 0

assuming the mass of the expelled gases is negligible ...

To find the recoil velocity of the gun, we can use the principle of conservation of momentum. According to this principle, the total momentum before the firing should be equal to the total momentum after the firing.

The momentum of an object is given by the product of its mass and velocity. So, the momentum of the bullet before the firing is:

Momentum of bullet before firing = mass of bullet × velocity of bullet

Given that the mass of the bullet is 25g = 0.025kg, and the velocity of the bullet is 320 m/s, we can calculate the momentum of the bullet before the firing:

Momentum of bullet before firing = 0.025kg × 320 m/s = 8 kg·m/s

Now, let's assume the recoil velocity of the gun is v (in m/s). The momentum of the gun before firing is 0 (since it's initially at rest). Therefore, the momentum of the gun after firing is also 0 (according to the conservation of momentum).

Using the conservation of momentum, we can write the equation:

Momentum of bullet before firing + Momentum of gun before firing = Momentum of bullet after firing + Momentum of gun after firing

0 + 0 = 8 kg·m/s + (8kg × v)

Simplifying the equation, we have:

8 kg × v = -8 kg·m/s

Dividing both sides of the equation by 8 kg, we get:

v = (-8 kg·m/s) / 8 kg = -1 m/s

The negative sign indicates that the gun moves in the opposite direction of the bullet, as expected due to the recoil.

Therefore, the recoil velocity of the gun is 1 m/s in the opposite direction of the bullet.