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In the adjoining figure, the position time graph of a particle of mass $0.1 \mathrm{~kg}$ is shown. The impulse at $t=2 \mathrm{~s}$ is
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$0.2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
From the graph we can say, upto $\mathrm{t}=2.0 \mathrm{~s}$, the body moves with a constant velocity
Slope of position-time graph $=\frac{4}{2}=2 \mathrm{~m} / \mathrm{s}$
After $\mathrm{t}=2.0 \mathrm{~s}$, position-time graph is parallel to time axis i.e., body comes to rest.
$\therefore$ Change in velocity $=\mathrm{dx}=2 \mathrm{~m} / \mathrm{s}$
Impulse $=$ Change in momentum
$=\mathrm{mdv}=0.1 \times 2=0.2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
Slope of position-time graph $=\frac{4}{2}=2 \mathrm{~m} / \mathrm{s}$
After $\mathrm{t}=2.0 \mathrm{~s}$, position-time graph is parallel to time axis i.e., body comes to rest.
$\therefore$ Change in velocity $=\mathrm{dx}=2 \mathrm{~m} / \mathrm{s}$
Impulse $=$ Change in momentum
$=\mathrm{mdv}=0.1 \times 2=0.2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$
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