Question

A bullet of mass 20 gram is horizontally fired with a velocity 150 M per second from pistol of mass 2kg what is the recoil velocity of the pistol?​

Answer 1

Explanation:

we have the mass of the bullet 20 gram that is 0.02 kg ...

the mass of the pistil is 2kg initial velocities of the pistil is zero the final velocity of the bullet is 150 metre per second the direction of bullet is taken from left to right let v be the recoil velocity of the pistol.

Solution:

total momentum of the pistol and bullet before the fire when the gun is at rest= (2+0.02)kg×0m/s=0kgm/s

total momentum of the pistol and the bullet after it is fired.

=0.02kga×(150m/s)

2kg×v m/s

(3+2v)kgm/s

according to the law of conservation of momentum

✨ total momentum

after the fire = total momentum before the Fire✨

3+2v=0

v=-1.5m/s

negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet that is right to left.

Answer 2

Answer:

Hey mate

Mass of bullet m1=20g=(0.2kg)

Mass of pistol m2=2 kg

Initial velocity of bullet (u1) and pistol (u2) =0

Find velocity of bullet, v1=150ms^(-1)

Let v be the recoil velocity of pistol

The total Momentum of bullet and pistol is zero(since both are at rest)

Total Momentum of bullet and pistol after it is fired

=[0.02 kg ×150 ms^(-1)] +[ 2 kg v ms^(-1)]

=(3+2v) kg ms^(-1)

Total Momentum after the fire=Total Momentum before the fire

3+2v=0

=> v=1.5 m/s

Thus the recoil velocity of pistol is 1.5 m/s

Hope it will be helpful to you

Thanks ☺️♥️✌️