A particle of mass m moving with velocity v = v 0 i ^ collides elastically with a stationary particle of mass 2 m . After the collision, the lighter particle moves with velocity v 1 = 3 v 0 i

Question

A particle of mass m moving with velocity v = v 0 ​ i ^ collides elastically with a stationary particle of mass 2 m . After the collision, the lighter particle moves with velocity v 1 ​ = 3 v 0 ​ ​ i ^ + 3 8 ​ v 0 ​ ​ j ^ ​ . The velocity of the heavier particle after collision is:

TestHub · Accepted Answer

For an elastic collision, both momentum and kinetic energy are conserved. Let the initial velocity of mass $m$ be $v = v_{0} \hat{i}$ and its final velocity be $v_{1} = \frac{v _{0}}{3} \hat{i} + \frac{8 v _{0}}{3} \hat{j}$ . Let the initial velocity of mass $2 m$ be $0$ and its final velocity be $v_{2}$ . By conservation of momentum: $m v + 2 m (0) = m v_{1} + 2 m v_{2}$

$m (v_{0} \hat{i}) = m (\frac{v _{0}}{3} \hat{i} + \frac{8 v _{0}}{3} \hat{j}) + 2 m v_{2}$ Divide by $m$ : $v_{0} \hat{i} = \frac{v _{0}}{3} \hat{i} + \frac{8 v _{0}}{3} \hat{j} + 2 v_{2}$

$2 v_{2} = v_{0} \hat{i} - \frac{v _{0}}{3} \hat{i} - \frac{8 v _{0}}{3} \hat{j} = \frac{2 v _{0}}{3} \hat{i} - \frac{8 v _{0}}{3} \hat{j}$ $v_{2} = \frac{v _{0}}{3} \hat{i} - \frac{8 v _{0}}{6} \hat{j} = \frac{v _{0}}{3} \hat{i} - \frac{2 2 v _{0}}{6} \hat{j} = \frac{v _{0}}{3} \hat{i} - \frac{2 v _{0}}{3} \hat{j}$ We can verify kinetic energy conservation, but it's not strictly necessary to find $v_{2}$ once momentum conservation is applied correctly.

Physics - Center of Mass Question with Solution | TestHub

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