JEE Advance - Physics (2024 - Paper 2 Online - No. 17)

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 $\omega$.

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 $\omega$.

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 ______.
Answer
4.25

Explanation

$$\mathrm{t}_0=\frac{\pi}{2 \omega}=\frac{\mathrm{T}}{4}$$

Particles are at extreme position

JEE Advanced 2024 Paper 2 Online Physics - Simple Harmonic Motion Question 1 English Explanation 1

After collision

JEE Advanced 2024 Paper 2 Online Physics - Simple Harmonic Motion Question 1 English Explanation 2

in C-frame

JEE Advanced 2024 Paper 2 Online Physics - Simple Harmonic Motion Question 1 English Explanation 3

JEE Advanced 2024 Paper 2 Online Physics - Simple Harmonic Motion Question 1 English Explanation 4

using WET,

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

Comments (0)

Advertisement