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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Advanced 2024 · Paper 2 · Physics · Center of Mass, Momentum and Collision

Paragraph II: Two particles, 1 and 2 , each of mass m , are...

Paragraph II: Two particles, 1 and 2 , each of mass mm, 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 x0x_0, are oscillating with amplitude aa and angular frequency ω\omega. Thus, their positions at time tt are given by x1(t)=(x0+d)+asin⁡ωtx_1(t)=\left(x_0+d\right)+a \sin \omega t and x2(t)=(x0−d)−asin⁡ωtx_2(t)=\left(x_0-d\right)-a \sin \omega t, respectively, where d>2ad>2 a. Particle 3 of mass mm moves towards this system with speed u0=aω/2u_0=a \omega / 2, and undergoes instantaneous elastic collision with particle 2 , at time t0t_0. Finally, particles 1 and 2 acquire a center of mass speed vcmv_{\mathrm{cm}} and oscillate with amplitude bb and the same angular frequency ω\omega. Paragraph II: Two particles, 1 and 2 , each of mass m , are connected by a massless spring, and are on a horizontal If the collision occurs at time t0=0t_0=0, the value of vcm/(aω)v_{\mathrm{cm}} /(a \omega) will be ________ .

Correct answer

0

Step-by-step solution

At t0=0\mathrm{t}_0=0 Before collision Paragraph II: Two particles, 1 and 2 , each of mass m , are connected by a massless spring, and are on a horizontal After collision Paragraph II: Two particles, 1 and 2 , each of mass m , are connected by a massless spring, and are on a horizontal vCM=m⋅aω2+m⋅aωm+mvCM=3aω4VCMaω=34VCMaω=0.75\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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