If the collision occurs at time 𝑡₀ = 0, the value of 𝑣_cm/(𝑎𝜔) will be __________.

The correct answer is 0.75.

To find the value of the ratio $v_{\text{cm}}/\left(a\omega \right)$ when the collision occurs at time $t_0=0$, we analyze the motion of the particles step-by-step:

Step 1: Determine the velocities of the particles just before the collision
Before the collision, particles 1 and 2 (each of mass $m$) are connected by a spring and undergo simple harmonic motion. At $t=0$, their velocities are maximum in magnitude and opposite in direction:
$v_1=a\omega$
$v_2=-a\omega$
Here, $a$ is the amplitude and $\omega$ is the angular frequency of the oscillation.

Step 2: Apply the conditions of the elastic collision
A third particle of mass $m$ moving with an initial speed $u_0=\frac{a\omega }{2}$ undergoes a head-on elastic collision with particle 2 at $t_0=0$.
Since the colliding particles (particle 3 and particle 2) have equal mass $m$ and the collision is perfectly elastic, they completely exchange their velocities.
Therefore, immediately after the collision, the new velocity of particle 2 ($v_2^{\prime }$) becomes equal to the initial velocity of particle 3:
$v_2^{\prime }=u_0=0.5a\omega$
The velocity of particle 1 ($v_1^{\prime }$) remains unaffected during the instantaneous collision:
$v_1^{\prime }=v_1=a\omega$

Step 3: Calculate the velocity of the center of mass
The velocity of the center of mass $v_{\text{cm}}$ of the system consisting of particles 1 and 2 immediately after the collision is given by:
$v_{\text{cm}}=\frac{m{v}_1^{\prime }+m{v}_2^{\prime }}{m+m}=\frac{v_1^{\prime }+v_2^{\prime }}{2}$
Substituting the post-collision velocity values:
$v_{\text{cm}}=\frac{a\omega +0.5a\omega }{2}=\frac{1.5a\omega }{2}=0.75a\omega$

Step 4: Find the required ratio
Dividing both sides by $a\omega$ yields:
$\frac{v_{\text{cm}}}{a\omega }=0.75$