When a bullet of mass m is fired with a velocity V from a gun of mass m then the recoil velocity of the gun is?

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CONCEPT:

• Momentum (P): The product of mass and velocity is called a momentum of the system.

Momentum (P) = Mass (m) × Velocity (V)

• Conservation of momentum: Whenever there is no net external force on the system then the total momentum of the system remains constant.

Initial momentum (P1) = Final momentum (P2)

CALCULATION:

• As there is no external force on the pistol and bullet system. So the total momentum of the system will remain constant.

Initially the system was in rest so velocity (V) = 0

Initial momentum (P1) = mass (m) × Velocity (V) = m × 0 = 0

After firing,

Velocity of bullet (Vb) = 100 m/s

Mass of bullet (mb) = 20 g = 20/1000 = 0.02 kg

Momentum of bullet (Pb) = mb × Vb = 0.02 × 100 = 2 kg m/s

Let recoil velocity of pistol = V m/s

Mass of rifle (mr) = 8 kg

Momentum of rifle (Pr) = mr × V = 8 × V = 8V kg m/s

Total final momentum of the system (P2) = Momentum of rifle + momentum of bullet = (8V + 2) kg m/s

According to conservation of momentum;

Initial momentum (P1) = Final momentum (P2)

0 = 8V + 2

V = - 2/8 = - 1/4 = - 0.25  m/s

The negative sign denotes opposite direction. So the recoil velocity of pistol is 0.25 m/s.

Hence option 4 is correct.

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We don’t have your requested question, but here is a suggested video that might help.

A bullet of mass 20g is fired horizontally with velocity 100 m/s from a pistol of mass 1.5kg calculate the recoil velocity of the pistol ?

If bullet of mass 'm1' is fired from a gun of mass 'm2' with a speed of 'V1', then the recoil velocity of gun is ______.

• -`("m"_1"V"_1)/"m"_2`

• -`("m"_2)/("m"_1"V"_1)`

• -`("m"_1)/("m"_2"V"_1)`

• -`("m"_2"V"_1)/"m"_1`

If bullet of mass 'm1' is fired from a gun of mass 'm2' with a speed of 'V1', then the recoil velocity of gun is `underline(-("m"_1"V"_1)/"m"_2)`.

Explanation:

m1V1 - m2V2 = 0

`therefore "V"_2 = (- "m"_1"V"_1)/"m"_2`

Concept: Principle of Conservation of Linear Momentum

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