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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

Two particles, 1 and 2 , each of mass m , are connected by a...

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. Two particles, 1 and 2 , each of mass m , are connected by a massless spring, and are on a horizontal frictionless If the collision occurs at time t0=π/(2ω)t_0=\pi /(2 \omega), then the value of 4b2/a24 b^2 / a^2 will be ________ .

Correct answer

0

Step-by-step solution

t0=π2ω=T4t_0=\frac{\pi}{2 \omega}=\frac{T}{4} Particles are at extreme position After collision in C-frame using WET, Wspring =ΔK12k(2b)2−12k(2a)2=2×12m×(aω4)2(k= spring constant )4kb2−4ka2=2×m×a216×2km4b2=174a24b2a2=4.25\begin{aligned} & \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 \end{aligned}
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