a) A bullet of mass 20 G is horizontally fired with a velocity of 150 metre per second from a pistol of mass 2 kg what is the recoil velocity of the pistol . b) why a gun recoils, when fired?
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Newton’s action-reaction law states that for every action, there is always an equal and opposite reaction that acts on different bodies. This is also known as Newton’s third law of motion .
When a bullet is fired from a gun, the gun recoils that is it moves in a direction opposite to the direction of motion of the bullet. With help of gun powder bullet is fired in forward direction and the gases from the bullet pushes the gun with an equal force in the opposite direction in the form of reaction force. That’s why gun recoils in the backward direction.
Given that:
Mass of bullet, m1 = 20 g (= 0.02 kg)
Mass of the pistol, m2 = 2 kg;
Suppose the initial velocities of the bullet is (u1) and pistol (u2) = 0, .
The final velocity of the bullet, v1 = + 150 m s-1.
Let v be the recoil velocity of the pistol.
Total momenta of the pistol and bullet before the fire, when the gun is at rest
= (2 + 0.02) kg × 0 m s-1
= 0 kg m s-1
Total momenta of the pistol and bullet after it is fired
= 0.02 kg × (+ 150 m s-1)
+ 2 kg × v m s-1
= (3 + 2v) kg m s-1
According to the law of conservation of momentum
Total momenta after the fire = Total momenta before the fire
3 + 2v = 0
v = − 1.5 m s-1.
Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet.
When a bullet is fired from a gun, the gun recoils that is it moves in a direction opposite to the direction of motion of the bullet. With help of gun powder bullet is fired in forward direction and the gases from the bullet pushes the gun with an equal force in the opposite direction in the form of reaction force. That’s why gun recoils in the backward direction.
Given that:
Mass of bullet, m1 = 20 g (= 0.02 kg)
Mass of the pistol, m2 = 2 kg;
Suppose the initial velocities of the bullet is (u1) and pistol (u2) = 0, .
The final velocity of the bullet, v1 = + 150 m s-1.
Let v be the recoil velocity of the pistol.
Total momenta of the pistol and bullet before the fire, when the gun is at rest
= (2 + 0.02) kg × 0 m s-1
= 0 kg m s-1
Total momenta of the pistol and bullet after it is fired
= 0.02 kg × (+ 150 m s-1)
+ 2 kg × v m s-1
= (3 + 2v) kg m s-1
According to the law of conservation of momentum
Total momenta after the fire = Total momenta before the fire
3 + 2v = 0
v = − 1.5 m s-1.
Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet.
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By the law of conservation of linear momentum, we have
Momentum of bullet fired = Momentum of recoiling
p1 = p2m1×v1 = m2×v20.02kg×150 = 2kg×v2(150×2)÷(100×2) = v2300÷200 = v2So, v2 = -3/2 = -1.5 m/s
When a bullet is fired from a gun, the gun recoils that is it moves in a direction opposite to the direction of motion of the bullet. With help of gun powder bullet is fired in forward direction and the gases from the bullet pushes the gun with an equal force in the opposite direction in the form of reaction force. That’s why gun recoils in the backward direction.
Momentum of bullet fired = Momentum of recoiling
p1 = p2m1×v1 = m2×v20.02kg×150 = 2kg×v2(150×2)÷(100×2) = v2300÷200 = v2So, v2 = -3/2 = -1.5 m/s
When a bullet is fired from a gun, the gun recoils that is it moves in a direction opposite to the direction of motion of the bullet. With help of gun powder bullet is fired in forward direction and the gases from the bullet pushes the gun with an equal force in the opposite direction in the form of reaction force. That’s why gun recoils in the backward direction.
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