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Chain Reaction

Displacement ...

Question

Displacement of a particle of mass $2 k g$ moving in a straight line varies with time at $s = (2 t^{3} + 2) m$ . Impulse of the force acting on the particle over a time interval between $t = 0$ and $t = 1 s$ in

10 N - s

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12 N - s

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8 N - s

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6 N - s

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Solution

The correct option is B $12 N - s$
We have
$s = (2 t^{3} + 2) m$
$v = \frac{d s}{d t} = \frac{d (2 t^{3} + 2)}{d t} = (6 t^{2}) m / s$
At $t = 0 s$
$V = 6 {(0)}^{2} = 0 m / s$
At $t = 1 s$
$V = 6 {(1)}^{2} = 6 m / s$
$Δ V = V (1) - V (0)$
$= 6 - 0 = 6 m / s$
We know that
Impuse $(J) = Δ m V = m Δ V$
$= (2 k g) (6 m / s)$
$12 k g$ $m / s$ or $12 N . s$

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