JEE Advanced · Physics · 7. COM & Collisions
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 \(t_0=\pi /(2 \omega)\), then the value of \(4 b^2 / a^2\) will be ________ .
- A 4.3
- B 4.25
- C 4.35
- D 4.44
Answer & Solution
Correct Answer
(B) 4.25
Step-by-step Solution
Detailed explanation
\(t_0=\frac{\pi}{2 \omega}=\frac{T}{4}\)
Particles are at extreme position

After collision

in C-frame


using WET,
\(\mathrm{W}_{\text {spring }}=\Delta \mathrm{K} \)
\( \frac{1}{2} \mathrm{k}(2 \mathrm{~b})^2-\frac{1}{2} \mathrm{k}(2 \mathrm{a})^2=2 \times \frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{a} \omega}{4}\right)^2 \quad\)\((\mathrm{k}=\text { spring constant }) \)
\( 4 \mathrm{~kb}{ }^2-4 \mathrm{ka}^2=2 \times \mathrm{m} \times \frac{\mathrm{a}^2}{16} \times \frac{2 \mathrm{k}}{\mathrm{m}} \)
\( 4 \mathrm{~b}^2=\frac{17}{4} \mathrm{a}^2 \)
\( \frac{4 \mathrm{~b}^2}{\mathrm{a}^2}=4.25\)
Particles are at extreme position

After collision

in C-frame


using WET,
\(\mathrm{W}_{\text {spring }}=\Delta \mathrm{K} \)
\( \frac{1}{2} \mathrm{k}(2 \mathrm{~b})^2-\frac{1}{2} \mathrm{k}(2 \mathrm{a})^2=2 \times \frac{1}{2} \mathrm{~m} \times\left(\frac{\mathrm{a} \omega}{4}\right)^2 \quad\)\((\mathrm{k}=\text { spring constant }) \)
\( 4 \mathrm{~kb}{ }^2-4 \mathrm{ka}^2=2 \times \mathrm{m} \times \frac{\mathrm{a}^2}{16} \times \frac{2 \mathrm{k}}{\mathrm{m}} \)
\( 4 \mathrm{~b}^2=\frac{17}{4} \mathrm{a}^2 \)
\( \frac{4 \mathrm{~b}^2}{\mathrm{a}^2}=4.25\)
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