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4-1.Newton's Laws of Motion
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In the figure, the position-time graph of a particle of mass $0.1\, kg$ is shown. The impulse at $t = 2\, second$ is ......... $kg\,ms^{-1}$
A$0.2$
B$-0.2$
C$0.1$
D$-0.4$
Solution
Velocity between $t=0$ and $t=2 \mathrm{sec}$
implies $v_{1}=\frac{d x}{d t}=\frac{4}{2}=2 m / s$
Velocity at $t=2 \sec , v_{f}=0$
Impulse $=$ Change in momentum $=m\left(v_{f}-v_{t}\right)$
$0.1(0-2)=-0.2 k g m \sec ^{-1}$
implies $v_{1}=\frac{d x}{d t}=\frac{4}{2}=2 m / s$
Velocity at $t=2 \sec , v_{f}=0$
Impulse $=$ Change in momentum $=m\left(v_{f}-v_{t}\right)$
$0.1(0-2)=-0.2 k g m \sec ^{-1}$
Standard 11
Physics
