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JEE Advanced · Physics · 7. COM & Collisions

Paragraph II: 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 \(x_0\), are oscillating with amplitude \(a\) and angular frequency \(\omega\). Thus, their positions at time \(t\) are given by \(x_1(t)=\left(x_0+d\right)+a \sin \omega t\) and \(x_2(t)=\left(x_0-d\right)-a \sin \omega t\), respectively, where \(d>2 a\). Particle 3 of mass \(m\) moves towards this system with speed \(u_0=a \omega / 2\), and undergoes instantaneous elastic collision with particle 2 , at time \(t_0\). Finally, particles 1 and 2 acquire a center of mass speed \(v_{\mathrm{cm}}\) and oscillate with amplitude \(b\) and the same angular frequency \(\omega\).

If the collision occurs at time \(t_0=0\), the value of \(v_{\mathrm{cm}} /(a \omega)\) will be ________ .

  1. A 0.75
  2. B 0.74
  3. C 0.73
  4. D 0.72
Verified Solution

Answer & Solution

Correct Answer

(A) 0.75

Step-by-step Solution

Detailed explanation

At \(\mathrm{t}_0=0\)
Before collision

After collision

\(\begin{aligned} \mathrm{v}_{\mathrm{CM}} & =\frac{\mathrm{m} \cdot \frac{\mathrm{a} \omega}{2}+\mathrm{m} \cdot \mathrm{a} \omega}{\mathrm{m}+\mathrm{m}} \\ \mathrm{v}_{\mathrm{CM}} & =\frac{3 \mathrm{a} \omega}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =\frac{3}{4} \\ \frac{\mathrm{V}_{\mathrm{CM}}}{\mathrm{a} \omega} & =0.75\end{aligned}\)
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