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. `40 ms^(–1)`
B. `120 m^(–1)`
C. `100 m^(–1)`
D. `80 m^(–1)`

Answer

C

Explanation

`m_1 = 0.2 kg , m_2 = 4kg`
Law of conservation of momentum –
`m_1 v_1 +m_2 v_2=0`
`implies v_1/v_2 = -m_2/m_1 =4/(0.2) =-20`
`implies v_1 =-20 v_2`
Now, Total energy –
`1/2 m_1v_1^2 +1/2 m_2v_2^2 =1.05 xx10^3`
`implies1/2 xx0.2 xxv_1^2+1/2xx4((-v_1)/20)^2 =1050`
`implies 0.1 v_1^2+(2v_1^2)/(400) = 1050`
`implies 0.2 v_1^2+0.005v_1^2 =1050`
`implies v_1^2 =(1050)/(0.105) implies 10 = 100m//s`