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Question
A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate:
- the velocity with which the gin recoils.
- the force exerted on gunman due to recoil of the gun
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Solution
Given Data:
- Mass of the gun (M) = 3 kg
- Mass of the bullet (m) = 30 g = 0.03 kg
- Velocity of the bullet (v) = 100 m/s
- Time taken by the bullet to move through the barrel (t) = 0.003 s
1. Velocity with which the gun recoils:
From the law of conservation of momentum:
Total initial momentum = Total final momentum
Initially, both the gun and bullet are at rest, so total momentum is zero. After firing:
M⋅V + m⋅v = 0
Where V is the recoil velocity of the gun. Rearrange the equation to find V:
`V = -(m.v)/M`
`V = - (0.03 xx 100)/3`
−1 m/s
2. Force exerted on the gunman due to recoil:
The force exerted on the gunman can be calculated using Newton's Second Law:
F = M × a
`a = (ΔV)/t`
Here, ΔV = 1 m/s and t = 0.003 s
`a = 1/0.003` = 333.33 m/s2
F = 3 × 333.33 = 1000 N
Final Answers:
- Recoil velocity of the gun = 1 m/s
- Force exerted on the gunman = 1000 N
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