Two particles, 1 and  2, each of mass m , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x₀ , are oscillating with amplitude a and angular frequency ω . Thus, their positions at time t are given by x₁(t) = (x₀ + d) + a sin ωt and x₂(t) = (x₀ − d) − a sin ωt , respectively, where d > 2a . Particle 3 of mass m moves towards this system with speed u₀ = aω/2 , and 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 .

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

Correct Answer: The value of $\frac{v_{\text{cm}}}{a\omega }$ is 0.75.

Step-by-Step Explanation:

1. Understanding the Initial Motion of Particles 1 and 2:

As shown in the figure, particles 1 and 2 (each of mass $m$) are connected by a spring on a frictionless horizontal plane, with their center of mass at position $x_0$.

The positions of particles 1 and 2 as a function of time $t$ before collision are given by:

$x_1\left(t\right)=(x_0+d)+a\sin \omega t$

$x_2\left(t\right)=(x_0-d)-a\sin \omega t$

Differentiating these equations with respect to time $t$, we obtain their velocities before collision:

$v_1\left(t\right)=\frac{d{x}_1}{dt}=a\omega \cos \omega t$

$v_2\left(t\right)=\frac{d{x}_2}{dt}=-a\omega \cos \omega t$

At time $t_0=0$:

$v_1\left(0\right)=a\omega \cos \left(0\right)=a\omega$

$v_2\left(0\right)=-a\omega \cos \left(0\right)=-a\omega$

2. Elastic Collision of Particle 3 with Particle 2:

Particle 3 has mass $m$ and moves to the right with initial velocity $u_0=\frac{a\omega }{2}$.

At time $t_0=0$, particle 3 undergoes an instantaneous head-on elastic collision with particle 2.

Since both particles 2 and 3 have identical mass $m$, an elastic collision between them causes an exact interchange of their velocities:

$v_2^{\prime }=u_0=\frac{a\omega }{2}$

Particle 1 is not directly affected by the instantaneous collision, so its velocity immediately after collision remains:

$v_1^{\prime }=v_1\left(0\right)=a\omega$

3. Calculating the Center of Mass Speed $v_{\text{cm}}$ of Particles 1 and 2:

After the collision, the combined system of particles 1 and 2 (total mass $2m$) moves with a center of mass velocity given by:

$v_{\text{cm}}=\frac{m{v}_1^{\prime }+m{v}_2^{\prime }}{m+m}=\frac{v_1^{\prime }+v_2^{\prime }}{2}$

Substitute the values of $v_1^{\prime }$ and $v_2^{\prime }$ into the equation:

$v_{\text{cm}}=\frac{a\omega +\frac{a\omega }{2}}{2}=\frac{\frac{3}{2}a\omega }{2}=\frac{3}{4}a\omega =0.75a\omega$

Thus, the required ratio is:

$\frac{v_{\text{cm}}}{a\omega }=0.75$