Position-time graph for a particle of mass 100g is as shown in figure then impulse acting

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4-1.Newton's Laws of Motion

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Position-time graph for a particle of mass $100g$ is as shown in figure then impulse acting on the particle at $t = 5$ second is ............ $N-s$

A

$0.3$

B

$0.1$

C

$0.2$

D

$0.4$

Solution

$\mathrm{I}=\mathrm{m}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right)=0.1[1-(-2)]=0.3 \mathrm{N}-\mathrm{S}$

$\mathrm{V}_{\mathrm{i}}=\frac{10-20}{5}=-2 \mathrm{m} / \mathrm{s}$

$\mathrm{V}_{\mathrm{f}}=\frac{20-10}{10}=1 \mathrm{m} / \mathrm{s}$

Standard 11

Physics

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(AIPMT-2014)