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. `0.2ms^(-1)`
B. `0.4ms^(-1)`
C. `0.6ms^(-1)`
D. `0.8ms^(-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 coverd=ut
:. `400xxuxx10`
where `u` is the velocity of the ball.
`u=10ms^(-1)`
According to law of conservation of linear momentum, we get
`0=Mv+mu`
`v=-(mv)/(M)=-((1kg)(40ms^(-1)))/(100kg)=-0.4ms^(-1)`
-ve sign shows that the direction
-ve sign shows that the direction of recoil of.