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Question: A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground ...

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. The recoil velocity of the gun is :

(Take g=10ms−2g = 10ms^{- 2})

A

0.2ms−10.2ms^{- 1}

B

0.4ms−10.4ms^{- 1}

C

0.6ms−10.6ms^{- 1}

D

0.8ms−10.8ms^{- 1}

Answer

0.4ms−10.4ms^{- 1}

Explanation

Solution

Here Mass of the gun M=100kgM = 100kg

Mass of the ball m=1kgm = 1kg

Height of the cliff, h=500mh = 500m

g=10ms−2g = 10ms^{- 2}

Time taken by the ball to reach the ground is

t=2hg=2×500m10ms−2=10st = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500m}{10ms^{- 2}}} = 10s

Horizontal distance covered = ut

∴400=u×10\therefore 400 = u \times 10

Where u is the velocity of the ball

u=40ms−1u = 40ms^{- 1}

According to law of conservation of linear momentum we get

0=mv+mu0 = mv + mu

v=−muM=−(1kg)(40ms−1)100kg=−0.4ms−1v = - \frac{mu}{M} = - \frac{(1kg)(40ms^{- 1})}{100kg} = - 0.4ms^{- 1}

-ve sign shows that the direction of recoil of the gun is opposite to that of the ball.