JEE MAIN - Physics (2010 - No. 13)
The figure shows the position$$-$$time $$(x-t)$$ graph of one-dimensional motion of body of mass $$0.4$$ $$kg.$$ The magnitude of each impulse is
$$0.4$$ $$Ns$$
$$0.8$$ $$Ns$$
$$1.6$$ $$Ns$$
$$0.2$$ $$Ns$$
Explanation
Motion is uniform as graph is straight line. In x -t graph, velocity is $${x \over t}$$ and the slope of the graph tells the direction of the velocity. Here because of impulse the direction of velocity or slope of the graph changes.
During each collision
Initial velocity $${v_1} = {2 \over 2} = 1\,m{s^{ - 1}}$$
Final velocity $${v_2} = - {2 \over 2} = - 1\,m{s^{ - 1}}$$
Impulse $$=$$ Change in momentum
$$ = m\left| {{v_2} - {v_1}} \right| = 0.4 \times 2 = 0.8\,\,Ns$$
During each collision
Initial velocity $${v_1} = {2 \over 2} = 1\,m{s^{ - 1}}$$
Final velocity $${v_2} = - {2 \over 2} = - 1\,m{s^{ - 1}}$$
Impulse $$=$$ Change in momentum
$$ = m\left| {{v_2} - {v_1}} \right| = 0.4 \times 2 = 0.8\,\,Ns$$
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