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)`

Options

A. `0.2"ms"^(-1)`
B. `0.4"ms"^(-1)`
C. `0.6"ms"^(-1)`
D. `0.8"ms"^(-1)`

Answer

B

Explanation

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.