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In the figure given, the position-time graph of a particle of mass $0.1 \mathrm{~kg}$ is shown.
The impulse at $t=2 \mathrm{sec}$ is
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The impulse at $t=2 \mathrm{sec}$ is
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Verified Answer
The correct answer is:
$0.2 \mathrm{~kg} \mathrm{~m} \mathrm{sec}^{-1}$
Impulse $=$ change in momentum
$$
=m \Delta v=\frac{m \Delta x}{\Delta t}=0.1 \times \frac{4-0}{2-0}=0.2 \mathrm{~kg} \mathrm{msec}^{-1} \text {. }
$$
$$
=m \Delta v=\frac{m \Delta x}{\Delta t}=0.1 \times \frac{4-0}{2-0}=0.2 \mathrm{~kg} \mathrm{msec}^{-1} \text {. }
$$
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