The figure shows the position-time (x-t) grap | vedclass

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(C) The impulse $J$ is equal to the change in linear momentum,$\Delta p = m(v_f - v_i)$.From the graph,the motion consists of segments with constant v

Vedclass · Accepted Answer

$0.8 \; N \cdot s$

Vedclass · Answer

(C) The impulse $J$ is equal to the change in linear momentum,$\Delta p = m(v_f - v_i)$.From the graph,the motion consists of segments with constant velocity.For the interval $t = 0$ to $t = 2 \; s$,the displacement is $2 \; m$. Thus,initial velocity $v_i = \frac{2 \; m}{2 \; s} = 1 \; m/s$.For the interval $t = 2 \; s$ to $t = 4 \; s$,the displacement is $-2 \; m$. Thus,final velocity $v_f = \frac{-2 \; m}{2 \; s} = -1 \; m/s$.The mass of the body is $m = 0.4 \; kg$.The impulse $J$ at each collision (at $t = 2, 6, 10, 14 \; s$) is given by:$J = m(v_f - v_i) = 0.4 \; kg 	imes (-1 \; m/s - 1 \; m/s) = 0.4 	imes (-2) = -0.8 \; kg \cdot m/s$.The magnitude of the impulse is $|J| = |-0.8| = 0.8 \; N \cdot s$.

The figure shows the position-time $(x-t)$ graph of one-dimensional motion of a body of mass $0.4 \; kg$. The magnitude of each impulse is:

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