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Two particles A and B collide to separate out. Show that their total momentum conserves. |
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Answer» Let initial momenta of the particle A and B be \(\vec{p}_A\) and \(\vec{p}_B\) respectively. The particles collide and separate out with final momentum \(\vec{p}'_A\) and \(\vec{p}'_B\) respectively. Using Newton’s second law of motion, the force of on particle A due to B is given by, \(\vec{F}_{AB}=\frac{\vec{p}_{A'}-\vec{p}_A}{t}\) Where, t is the time interval. Similarly, \(\vec{F}_{BA}=\frac{\vec{p}'_B-\vec{p}_B}{t}\) As per Newton’s third law of motion, \(\vec{F}_{BA}=-\vec{F}_{BA}\) i.e., \(\frac{\vec{p}_{A'}-\vec{p}_A}{t}\) = \(\frac{\vec{p}'_B-\vec{p}_B}{t}\) i.e., \(\vec{p}'_A\) + \(\vec{p}'_B\) = \(\vec{p}_A\) + \(\vec{p}_B\) ∴ Linear momentum is conserved. |
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