A shell of mass `200g` is ejected from a gun of mass `4 kg` by an explosion that generate `1.05 kJ` of energy. The initial velocity of the shell is

Options

A. `100ms^(-1)`
B. `80ms^(-1)`
C. `40ms^(-1)`
D. `120ms^(-1)`

Answer

a

Explanation

Here, `m_(1) = 200g = (200)/(1000)kg = (1)/(5) kg`
`m_(2) = 4kg` total energy `E = 1.05 kj = 1050 j`
Let `u_(1)` be initial velocity of the shell and `u_(2)` be the initial velocity of the gun
According to principle of conservation of linear momentum `m_(1) u_(1) + m_(2) u_(2) = 0`
`u_(2) = -(m_(1))/(m_(2))ut = (-1)/(5 xx4)u_(1) =-(u_(1))/(20)`
Now, `E = (1)/(2) m_(1)u_(2)^(1) + (1)/(2) m_(2) u_(2)^(2)`
`1050 = (1)/(2) xx (1)/(5) u_(1)^(2) + (1)/(2) xx 4(-(u_(1))/(20))^(2)`
`= (u_(1)^(2))/(10) + (u_(1)^(2))/(200)=(21)/(200)u_(2)^(1)`
`u_(1)^(2)= (200 xx 1050)/(21) = 10000 , `
`u_(1) = sqrt(10000) = 100m//s` .