Momentum and CollisionHardBloom L1
Question
Two particles of mass m, constrained to move along the circumference of a smooth circular hoop of equal mass m, are initially located at opposite ends of a diameter and given equal velocities v0 shown in the figure. The entire arrangement is located in gravity free space. Their velocity just before collision is
Options
A.$\frac{1}{\sqrt{3}}v_{0}$
B.$\frac{\sqrt{3}}{2}v_{0}$
C.$\frac{2}{\sqrt{3}}v_{0}$
D.$\frac{\sqrt{7}}{3}v_{0}$
Solution
Sol.
3mv2 = 2mv0
v2 = $\frac{2v_{0}}{3}$
$\frac{1}{2} \cdot m \cdot v_{2}^{2} + 2 \times \frac{1}{2}m\left( v_{1}^{2} + v_{2}^{2} \right)$ = $\frac{1}{2}mv_{0}^{2} \times 2$
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