Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

81 of 130

Question 81 of 130

The figure shows the position - time (x−t)(x-t) graph of one-dimensional motion of a body of mass 0.4 kg0.4 \mathrm{~kg}. The magnitude of each impulse is

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Answer: (B) 0.8Ns0.8 \mathrm{Ns}

From the graph, it is a straight line so, uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
 Initial velocity =22=1 m/s Final velocity =−22=−1 m/sPˉi=0.4 N−sP‾i=−0.4 N−sJ→=P‾i−P‾i=−0.4−0.4=−0.8 N−s(J→= impulse )∣J→∣=0.8 N−s\begin{aligned} & \text { Initial velocity }=\frac{2}{2}=1 \mathrm{~m} / \mathrm{s} \\ & \text { Final velocity }=-\frac{2}{2}=-1 \mathrm{~m} / \mathrm{s} \\ & \qquad \bar{P}_{\mathrm{i}}=0.4 \mathrm{~N}-\mathrm{s} \\ & \qquad \overline{\mathrm{P}}_{\mathrm{i}}=-0.4 \mathrm{~N}-\mathrm{s} \\ & \overrightarrow{\mathrm{J}}=\overline{\mathrm{P}}_{\mathrm{i}}-\overline{\mathrm{P}}_{\mathrm{i}}=-0.4-0.4=-0.8 \mathrm{~N}-\mathrm{s}(\overrightarrow{\mathrm{J}}=\text { impulse }) \\ & |\overrightarrow{\mathrm{J}}|=0.8 \mathrm{~N}-\mathrm{s} \end{aligned}

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