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Question

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


A
0.2 kg m sec1
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B
0.2 kg m sec1
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C
0.1 kg m sec1
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D
0.4 kg m sec1
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Solution

The correct option is B 0.2 kg m sec1
From the xt graph, velocity of particle is given by:
v=slope of x-t curve=dxdt
From t=0 to t=2 s, slope of xt curve is constant, so velocity is,
v=slope of x-t curve=4020=2 m/s
Just after t=2 s, x=constant v=dxdt=0 m/s
At t=2 s, particle is undergoing a transition in velocity from vi=2 m/s to vf=0
From impulse momentum-theorem on particle:
J=ΔP=mvfmvi ...(i)
J=0.1(02)=0.2 kg m sec1

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