# a bullet of mass 20 grams is horizontally fired with a volicity of 150 mls from pistal of mass 2 kg what is the recoil volicity of pistal

### Asked by gaminggaming64738 | 28th Jun, 2021, 10:30: AM

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Mass of bullet, m_{1} = 20g (= 0.02 kg)

Mass of pistol, m_{2} = 2 kg

Initial velocity of the bullet (u_{1}) and pistol (u_{2}) = 0

Final velocity of the bullet, v_{1} = +150m s^{-1}

Let v be the recoil velocity of the pistol.

The total momentum of the pistol and bullet is zero before the fire. (Since both are at rest)

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s^{-1}) + (2 kg x v m s^{-1})

= (3 + 2v) kg m s^{-1}

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s

Thus, the recoil velocity of the pistol is 1.5 m/s.

Mass of bullet, m_{1} = 20g (= 0.02 kg)

Mass of pistol, m_{2} = 2 kg

Initial velocity of the bullet (u_{1}) and pistol (u_{2}) = 0

Final velocity of the bullet, v_{1} = +150m s^{-1}

Let v be the recoil velocity of the pistol.

The total momentum of the pistol and bullet is zero before the fire. (Since both are at rest)

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s^{-1}) + (2 kg x v m s^{-1})

= (3 + 2v) kg m s^{-1}

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s

Thus, the recoil velocity of the pistol is 1.5 m/s.

### Answered by Shiwani Sawant | 28th Jun, 2021, 11:57: AM

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