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(B) The change in linear momentum $\Delta p$ is equal to the impulse,which is the area under the force-time $(F-t)$ graph.$\Delta p = \int F \, dt = \

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$0 \, N \cdot s$

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(B) The change in linear momentum $\Delta p$ is equal to the impulse,which is the area under the force-time $(F-t)$ graph.$\Delta p = \int F \, dt = 	ext{Area under } F-t 	ext{ graph}$.The graph consists of circular segments. From $t = 0$ to $t = 2$,the area is a quarter-circle below the time axis with radius $r = 2$. Area $= -\frac{1}{4} \pi r^2 = -\frac{1}{4} \pi (2)^2 = -\pi$.From $t = 2$ to $t = 6$,the area is a semi-circle above the time axis with radius $r = 2$. Area $= +\frac{1}{2} \pi r^2 = +\frac{1}{2} \pi (2)^2 = +2\pi$.From $t = 6$ to $t = 8$,the area is a quarter-circle below the time axis with radius $r = 2$. Area $= -\frac{1}{4} \pi r^2 = -\frac{1}{4} \pi (2)^2 = -\pi$.The total area (net impulse) $= -\pi + 2\pi - \pi = 0$.Therefore,the linear momentum gained between $0$ and $8$ seconds is $0 \, N \cdot s$.

$A$ force-time graph for a linear motion is shown in the figure,where the segments are circular. The linear momentum gained between $0$ and $8$ seconds is

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