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A 3000 kg space probe is moving in a grayity 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)

a
30o
b
60o
c
45o
d
37o

detailed solution

Correct option is C

From impulse-momentum theorem,J→=p→f−p→i⇒p→f=J→+p→i Here pi=3000×300=9×105kgm/sJ=4000×225=9×105kgm/stan⁡θ=piJ=1⇒θ=45∘So, the angle by which the space probe rotates isα=π2−θ=45∘

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