A spacecraft of mass 100 kg breaks into two when its velocity is `10^(4) "ms"^(-1)`. After the break, a mass of 10 kg of the spacecraft is left stationary. The velocity of the remaining part is

Options

A. `10^(3) "ms"^(-1)`
B. `11.11 xx 10^(3) "ms"^(-1)`
C. `11.11 xx 10^(2) "ms"^(-1)`
D. `10^(4) "ms"^(-1)`

Answer

B

Explanation

Given, `m = 100 "kg" , m_2 = 10 "kg" and u = 10^(4) "ms"^(-1)`
From conservation of momentum,
`m. u = m_1 v_1 + m_2 v_2`
`100 xx 10^(4) = 20 xx 0 + 90 xx v_2`
`100 xx 10^(4) = 0 + 90 xx v_2`
`v_2 = (100 xx 10^4)/( 90)`
`v_2 = 1.1 xx 10^4`
`v_2 = 11. 11 xx 10^3 "ms"^(-1)`