The position (x) - time (t) graph for a particle of mass $1 k g$ moving along x-axis is shown in figure. Find impulse on particle at $t = 4 s$ .

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Solution

Velocity is nothing but the $s l o p e$ of $x - t$ graph, so the velocity just before $t = 4$ will be
$V_{1} = s l o p e = \frac{(20 - 0) m}{(4 - 0) s e c} = 5 m / s e c$
the velocity just after $t = 4$ will be
$V_{2} = s l o p e = \frac{(0 - 20) m}{(8 - 4) s e c} = - 5 m / s e c$
so the change in momentum at $t = 4 s e c$ will be $Δ P = m (V_{2} - V_{1}) = 1 k g (- 5 - 5) m / s = - 10 K g m / s$
if considering only magnitude the impulse is $10 K g m / s$ or $10 N - s e c$

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