A bullet of mass 20 gm is fired in the horizontal direction with a velocity 150 m/s from a pistol of mass 1 kg. Recoil velocity of the pistol is

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Q: 16 (NDA-II/2013)
A bullet of mass 20 gm is fired in the horizontal direction with a velocity 150 m/s from a pistol of mass 1 kg. Recoil velocity of the pistol is

question_subject: 

Science

question_exam: 

NDA-II

stats: 

0,2,9,2,2,6,1

keywords: 

{'recoil velocity': [0, 0, 0, 1], 'bullet': [0, 1, 0, 4], 'velocity': [0, 2, 2, 6], 'pistol': [0, 0, 0, 1], 'horizontal direction': [0, 0, 2, 2], 'mass': [0, 0, 2, 3], 'gm': [1, 1, 1, 3], 'km': [0, 0, 2, 1], 'kg': [0, 1, 9, 24]}

When a bullet is fired from a pistol, according to Newton`s third law of motion, there is an equal and opposite reaction. In this case, the pistol experiences a recoil.

To find the recoil velocity of the pistol, we can use the principle of conservation of momentum. The total momentum before the firing is equal to the total momentum after the firing.

Total momentum before firing = 0 (as the pistol is at rest)

Total momentum after firing = (mass of the bullet * velocity of the bullet) + (mass of the pistol * recoil velocity of the pistol)

Given:

Mass of the bullet = 20 gm = 0.02 kg

Velocity of the bullet = 150 m/s

Mass of the pistol = 1 kg

Using the conservation of momentum equation:

0 = (0.02 kg * 150 m/s) + (1 kg * recoil velocity of the pistol)

Simplifying the equation, we find that the recoil velocity of the pistol is 3 m/s.

Therefore, option 1, 3 m/s, is the correct answer.