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Q.

In the figure given the position-time graph of a particle of mass 0.1 kg is shown. The impulse at t = 2 s is :

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a

0.2 kg m s–1

b

0.3 kg m s–1 (2) – 0.2 kg m s–1

c

0.1 kg m s–1

d

– 0.4 kg m s–1

answer is A.

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Detailed Solution

If x=2t⇒v=axdt=2  Impulse = change of momentum = 0.1 × 2 – 0 = 0.2 kg m/s
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