JEE Challenger
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PhyHardAdv 2024 • Paper 2

Center of Mass Speed After Elastic Collision of Oscillating Spring Mass System

Comprehension Passage

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) = (x_0 + d) + a \sin \omega t and x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega t, respectively, where d>2ad > 2a. 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_{\text{cm}} and oscillate with amplitude bb and the same angular frequency ω\omega.

If the collision occurs at time t0=0t_0 = 0, the value of vcm/(aω)v_{\text{cm}}/(a\omega) will be ________.

Question Diagram 1
Official Numerical Answer0.75

Topics & Concepts

OscillationsSHMCenter of MassCollision

Step-by-Step Solution

To find the center of mass speed vcmv_{\text{cm}} of the two-particle system (particles 1 and 2) after the collision at t0=0t_0 = 0, we follow these steps:

1. Initial Velocities of the Particles (at t=0−t = 0^-)

The position equations for particles 1 and 2 prior to the collision are: x1(t)=(x0+d)+asin⁡ωtx_1(t) = (x_0 + d) + a \sin \omega t x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega t

Differentiating these with respect to time tt gives their respective velocities: v1(t)=dx1dt=aωcos⁡ωtv_1(t) = \frac{dx_1}{dt} = a \omega \cos \omega t v2(t)=dx2dt=−aωcos⁡ωtv_2(t) = \frac{dx_2}{dt} = -a \omega \cos \omega t

At time t0=0t_0 = 0: v1(0−)=aωcos⁡(0)=aωv_1(0^-) = a\omega \cos(0) = a\omega v2(0−)=−aωcos⁡(0)=−aωv_2(0^-) = -a\omega \cos(0) = -a\omega

Particle 3 moves to the right towards particle 2 with speed u0=aω2u_0 = \frac{a\omega}{2}, so its velocity just before collision is: v3(0−)=u0=aω2v_3(0^-) = u_0 = \frac{a\omega}{2}

2. Velocities Immediately After Collision (at t=0+t = 0^+)

Particle 3 undergoes an instantaneous head-on elastic collision with particle 2. Since both particle 2 and particle 3 have equal mass mm:

  • The velocities of particle 2 and particle 3 are exchanged elastically.
  • The spring force is non-impulsive during the instantaneous collision, so the velocity of particle 1 remains unchanged.

Thus, immediately after the collision: v1(0+)=v1(0−)=aωv_1(0^+) = v_1(0^-) = a\omega v2(0+)=v3(0−)=aω2v_2(0^+) = v_3(0^-) = \frac{a\omega}{2}

3. Center of Mass Velocity of the System

The velocity of the center of mass of particles 1 and 2 immediately after the collision is given by: vcm=mv1(0+)+mv2(0+)m+m=v1(0+)+v2(0+)2v_{\text{cm}} = \frac{m v_1(0^+) + m v_2(0^+)}{m + m} = \frac{v_1(0^+) + v_2(0^+)}{2}

Substituting the values of v1(0+)v_1(0^+) and v2(0+)v_2(0^+): vcm=aω+aω22=32aω2=34aω=0.75aωv_{\text{cm}} = \frac{a\omega + \frac{a\omega}{2}}{2} = \frac{\frac{3}{2}a\omega}{2} = \frac{3}{4}a\omega = 0.75 a\omega

Therefore, the required ratio is: vcmaω=34=0.75\frac{v_{\text{cm}}}{a\omega} = \frac{3}{4} = 0.75

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Center of Mass Speed After Elastic Collision of Oscillating Spring Mass System | Physics PYQ Solution - JEE Challenger