In the figure given below,the position-time g | vedclass

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(B) The impulse acting on a particle is equal to the change in its momentum.Impulse $J = \Delta p = m(v_f - v_i)$.From the position-time graph,the vel

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$-0.2$

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(B) The impulse acting on a particle is equal to the change in its momentum.Impulse $J = \Delta p = m(v_f - v_i)$.From the position-time graph,the velocity $v$ is the slope of the line.For $t = 0$ to $t = 2 \,s$,the velocity $v_i = \frac{\Delta x}{\Delta t} = \frac{4 - 0}{2 - 0} = 2 \,m/s$.For $t > 2 \,s$,the position is constant $(x = 4 \,m)$,so the velocity $v_f = 0 \,m/s$.The mass of the particle is $m = 0.1 \,kg$.Therefore,the impulse at $t = 2 \,s$ is $J = 0.1 \,kg 	imes (0 \,m/s - 2 \,m/s) = -0.2 \,kg \,m \,s^{-1}$.

In the figure given below,the position-time graph of a particle of mass $0.1 \,kg$ is shown. The impulse at $t = 2 \,s$ is .......... $kg \,m \,s^{-1}$.

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