In figure given below, position-time graph of a particle of mass 0.1 ,Kg is shown. impulse at t = 2s

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4-1.Newton's Laws of Motion

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In the figure given below, the position-time graph of a particle of mass $0.1 \,Kg$ is shown. The impulse at $t = 2\sec $ is .......... $kg\,m\,{\sec ^{ - 1}}$

A

$0.2$

B

$ - 0.2$

C

$0.1$

D

$ - 0.4$

Solution

(b) Velocity between $t = 0$ and $t = 2\sec $

$⇒$ ${v_i} = \frac{{dx}}{{dt}} = \frac{4}{2} = 2\;m/s$

Velocity at $t = 2\sec $, ${v_f} = 0$

Impulse = Change in momentum $ = m({v_f} – {v_i})$

$ = 0.1(0 – 2)$ $ = – 0.2\;kg\;m\;{\sec ^{ – 1}}$

Standard 11

Physics

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