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NCERT Exemplar · Q31

Q.A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 500m. It falls on the ground at a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 1010 m s−2^{-2})

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The ball leaves at 40 m/s40\ \text{m/s} horizontally (from its projectile motion); conserving momentum for the gun-ball system gives a gun recoil of 0.4 m/s0.4\ \text{m/s} in the direction opposite to the ball.

The gun and ball start at rest, so the total horizontal momentum of the system is zero. The firing force is internal, so horizontal momentum is conserved: the forward momentum of the ball is balanced by the backward momentum of the gun.

Ball's horizontal velocity from projectile motion

  1. Time of flight (vertical motion). The ball falls freely from height h=500 mh = 500\ \text{m}:

t=2hg=2×50010=100=10 s.t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500}{10}} = \sqrt{100} = 10\ \text{s}.

  1. Horizontal (muzzle) velocity. It covers the range R=400 mR = 400\ \text{m} at constant horizontal speed:

vball=Rt=40010=40 m/s.v_{\text{ball}} = \frac{R}{t} = \frac{400}{10} = 40\ \text{m/s}.

Conservation of momentum …

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