Question

undergoes instantaneous elastic collision with particle 2, at time t₀. Finally, particles 1 and 2 acquire a center of mass speed $v_{cm}$ and oscillate with amplitude b and the same angular frequency ω.

Answer 1

1/2

Answer 2

1. Assume all particles have equal mass $m$ and that particle 1 and particle 2 are initially at rest.

2. Particle 3 (mass $m$) with initial speed $u_0$ collides elastically with particle 2. Since the collision is instantaneous and with identical masses, particle 3 comes to rest and particle 2 acquires the speed $u_0$.

3. Immediately after the collision, particle 1 is at rest and particle 2 moves with speed $u_0$. Thus, the center‐of‐mass velocity of particles 1 and 2 is

   $$v_{cm}=\frac{0+u_0}{2}=\frac{u_0}{2}.$$

4. In the centre-of-mass frame, the relative speed immediately after collision is $u_0$. For two masses $m$ connected by a spring, the reduced mass is $\mu=\frac{m}{2}$ and the energy in the relative motion is

   $$\frac{1}{2}\mu (u_0)^2=\frac{1}{2}\left(\frac{m}{2}\right)u_0^2=\frac{m\,u_0^2}{4}.$$

5. This energy completely converts into spring potential energy at maximum compression/extension:

   $$\frac{1}{2}k\,b^2,$$

   where $b$ is the amplitude and the effective spring constant if the oscillation frequency is $\omega$ is given by

   $$\omega=\sqrt{\frac{2k}{m}}\quad\Longrightarrow\quad k=\frac{m\omega^2}{2}.$$

6. Equate the energies:

   $$\frac{m\,u_0^2}{4}=\frac{1}{2}\left(\frac{m\omega^2}{2}\right)b^2 \quad\Longrightarrow\quad \frac{m\,u_0^2}{4}=\frac{m\,\omega^2\,b^2}{4}.$$

   Canceling common factors, we obtain:

   $$u_0^2=\omega^2\,b^2 \quad\Longrightarrow\quad b=\frac{u_0}{\omega}.$$

7. We are asked for $\frac{v_{cm}}{a\omega}$. Noting that the amplitude in the oscillation is given as $b$, we identify $a=b$. Thus:

   $$\frac{v_{cm}}{a\omega}=\frac{v_{cm}}{b\omega}=\frac{\frac{u_0}{2}}{\frac{u_0}{\omega}\,\omega}=\frac{\frac{u_0}{2}}{u_0}=\frac{1}{2}.$$

Explanation (minimal):

• Elastic collision transfers speed $u_0$ from particle 3 to particle 2.

• Centre-of-mass speed of particles 1 and 2 becomes $v_{cm}=\frac{u_0}{2}$.

• Energy conservation in the relative oscillation gives amplitude $b=\frac{u_0}{\omega}$.

• Hence, $\frac{v_{cm}}{b\omega}=\frac{1}{2}$.