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a 100kg gun fires a ball of 1kg horizontally from a cliff of height 500m. If falls on the ground at a distance of 400m from the bottom of the cliff. The recoil velocity of the gun is (Take g: `10ms^(-2)`A. `0.2"ms"^(-1)`B. `0.4"ms"^(-1)`C. `0.6"ms"^(-1)`D. `0.8"ms"^(-1)` |
Answer» Correct Answer - B Here, Mass of the gun, M=100kg Mass of the ball, M=1kg Height of the cliff, h=500m g=`10ms^(-2)` Time taken by the ball to reach the ground is `t=sqrt((2h)/(g))=sqrt((2xx500m)/(10ms^(-2)))=10s` Horizontal distance covered=ut `therefore 400=u xx 10` Where u is the velocity of the ball, `u=40 ms^(-1)` According to the law of conservation of linear momentum, we get, 0=Mv+mu `v=-("mu")/(M)=-((1kg)(40ms^(-1)))/(100kg)=-0.4ms^(-1)` -ve sign shows that the direction of recoil of hte gun is oppositre to that of the ball. |
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