A spacecraft which is moving with a speed u relative to the earth in the x-direction, enters the gravitational field of a much more massive planet which is moving with a speed 3u in the negative x-direction. The spacecraft exits following the trajectory as shown below.

The speed of the spacecraft with respect to the earth a long time after it has escaped the planet's gravity is given by

Options

A. u
B. 4u
C. 2u
D. 7u

Answer

D

Explanation

from momentum conservation
`-mu+m_13u=m_1v_1+mv_2` ......(1)
from energy conservation
`1/2mu^2+1/2m_19u^2=1/2m_1v_1^2+1/2mv_2^2`
`1/2mu^2+1/2m_1(3u-v_1)(3u+v_1)=1/2mv_2^2`
from equation ...(1)
`rArrm_1(3u-v_1)=m(v_2+u)`
`1/2mu^2+1/2m(v_2+u)(3u+v_1)=1/2mv_2^2`
as `m_1gtgtgtm`, we can assume `v_1~~3u`
`u^2+(v_2+u)(6u)=v_2^2`
`rArrv_2=7u`