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 $v_0$ shown in the

figure. The entire arrangement is located in gravity free space. Their velocity just before collision is :

• A. $\frac{1}{\sqrt{3}}v_0$
• B. $\frac{\sqrt{3}}{2}v_0$
• C. $\frac{2}{\sqrt{3}}v_0$
• D. $\sqrt{\frac{7}{3}}v_0$

Answer 1

Answer: (A)

A

Answer 2

The system consists of two particles and a hoop. Since it's in gravity-free space and the hoop is smooth, both linear momentum and mechanical energy (kinetic energy) are conserved. Angular momentum about the center of the hoop is also conserved.

1. Conservation of Linear Momentum: The initial total linear momentum of the system is zero ($m\vec{v}_0 + m(-\vec{v}_0) + m(0) = 0$). Thus, the center of mass of the system remains at rest, meaning the center of the hoop remains fixed.

2. Conservation of Angular Momentum:

   • Initial angular momentum ($L_i$) about the center of the hoop: Each particle contributes $mRv_0$. So, $L_i = 2mRv_0$.
   • Final angular momentum ($L_f$): Let $\omega_f$ be the final angular velocity of the hoop and $v_{rel}$ be the speed of the particles relative to the hoop (tangential). At the collision point (e.g., top of the hoop), the absolute velocity of a particle is the vector sum of the hoop's point velocity ($R\omega_f$ tangential) and the particle's relative velocity ($v_{rel}$ tangential). By applying $\vec{L}_f = I_{hoop}\vec{\omega}_f + \sum \vec{r}_i \times m\vec{v}_i$, we find $L_f = 3mR^2\omega_f$.
   • Equating $L_i = L_f$: $2mRv_0 = 3mR^2\omega_f \implies \omega_f = \frac{2v_0}{3R}$.

3. Conservation of Kinetic Energy:

   • Initial kinetic energy ($KE_i$): $KE_i = \frac{1}{2}mv_0^2 + \frac{1}{2}mv_0^2 = mv_0^2$.
   • Final kinetic energy ($KE_f$): $KE_f = \frac{1}{2}I_{hoop}\omega_f^2 + \frac{1}{2}m(R\omega_f + v_{rel})^2 + \frac{1}{2}m(R\omega_f - v_{rel})^2$. (Note: One particle's relative velocity is in the opposite direction to the hoop's rotation, so it's $v_{rel}-R\omega_f$ or $R\omega_f-v_{rel}$ depending on which is larger. The sum of squares simplifies this).
   • $KE_f = \frac{1}{2}mR^2\omega_f^2 + \frac{1}{2}m(R^2\omega_f^2 + 2R\omega_f v_{rel} + v_{rel}^2) + \frac{1}{2}m(R^2\omega_f^2 - 2R\omega_f v_{rel} + v_{rel}^2)$
   • $KE_f = \frac{1}{2}m(3R^2\omega_f^2 + 2v_{rel}^2)$.
   • Equating $KE_i = KE_f$: $mv_0^2 = \frac{1}{2}m(3R^2\omega_f^2 + 2v_{rel}^2)$.
   • Substitute $\omega_f = \frac{2v_0}{3R}$ into the energy equation: $2v_0^2 = 3R^2\left(\frac{2v_0}{3R}\right)^2 + 2v_{rel}^2 \implies 2v_0^2 = \frac{4v_0^2}{3} + 2v_{rel}^2$.
   • Solving for $v_{rel}$: $2v_{rel}^2 = 2v_0^2 - \frac{4v_0^2}{3} = \frac{2v_0^2}{3} \implies v_{rel}^2 = \frac{v_0^2}{3} \implies v_{rel} = \frac{v_0}{\sqrt{3}}$.

The phrase "Their velocity just before collision" most likely refers to the speed of the particles relative to the hoop, as the absolute speeds of the two particles are different ($R\omega_f + v_{rel}$ and $|R\omega_f - v_{rel}|$).

Answer: The velocity just before collision (interpreted as speed relative to the hoop) is $\frac{1}{\sqrt{3}}v_0$.