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Q.

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)

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a

30^o

b

60^o

c

45^o

d

37^o

answer is C.

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Detailed Solution

From impulse-momentum theorem,
$\begin{array}{l} \vec{J} = {\vec{p}}_{f} - {\vec{p}}_{i} \Rightarrow {\vec{p}}_{f} = \vec{J} + {\vec{p}}_{i} \\ Here p_{i} = 3000 \times 300 = 9 \times 10^{5} kgm / s \\ J = 4000 \times 225 = 9 \times 10^{5} kgm / s \end{array}$

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$\tan θ = \frac{p_{i}}{J} = 1 \Rightarrow θ = 45^{\circ}$
So, the angle by which the space probe rotates is
$α = \frac{π}{2} - θ = 45^{\circ}$

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