A `3000 kg` space probe is moving in a gravity free space at a constant velocity of `300 m//s`. To change the direction of space probe, rockets have been fired in a direction perpendicular to the direction of initial motion of the space probe, the rocket firing exerts a thrust of `4000 N` for `225 s`. The space probe will turn by an angle of (neglect the mass of the rockets fired)

Options

A. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `37^(@)`

Answer

C

Explanation

From impulse momentum theorem,
`vecJ=vecp_(f)-vecp_(i)impliesvecp_(f)=vecJ+vecp_(i)`
Here `p_(i)=3000xx300=9xx10^(5)kgm//s`
`J=4000xx225=9xx10^(5)kgm//s`

`tantheta=p_(i)/J=1impliestheta=45^(@)`
So, the angle by which the space probe rotates is
`alpha=pi/2-theta=45^(@)`