Questions
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)
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/stanθ=piJ=1⇒θ=45∘So, the angle by which the space probe rotates isα=π2−θ=45∘Talk to our academic expert!
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