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 ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Conservation of Momentum

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

ω

. Thus, their positions at time

t

are given by

x_{1} (t) = (x_{0} + d) + a sin ω t

and

x_{2} (t) = (x_{0} - d) - a sin ω t

, respectively, where

d > 2 a

. Particle 3 of mass

m

moves towards this system with speed

u_{0} = aω /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_{cm}

and oscillate with amplitude

b

and the same angular frequency

ω

.

If the collision occurs at time

t_{0} = π / (2 ω)

, then the value of

4 b^{2} / a^{2}

will be ____.

Answer: 4.25

Solution

t_{0} = π / (2 ω) = T /4

, both particles are at their extreme positions with zero velocity.

Particle 2 is at rest when particle 3 (moving with speed

aω /2

) collides with it. After elastic collision between equal masses, they exchange velocities:

Particle 2 acquires velocity $aω /2$
Particle 3 comes to rest

The center of mass velocity of particles 1 and 2:

v_{cm} = \frac{0 + m ( aω /2 )}{2 m} = \frac{aω}{4}

In the CM frame, the initial spring extension is

2 a

(particles at extremes). After collision, particle 2 has velocity

aω /4

relative to CM.

Using energy conservation in the CM frame:

\frac{1}{2} k (2 b)^{2} - \frac{1}{2} k (2 a)^{2} = 2 \times \frac{1}{2} m (\frac{aω}{4})^{2}

Since

ω^{2} = k / (2 m)

for the two-particle system:

4 k b^{2} - 4 k a^{2} = m \times \frac{a ^{2} ω ^{2}}{8} = m \times \frac{a ^{2} k}{16 m} = \frac{k a ^{2}}{16}

4 b^{2} = 4 a^{2} + \frac{a ^{2}}{4} = \frac{17 a ^{2}}{4}

\frac{4 b ^{2}}{a ^{2}} = 4.25

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Numerical answer

Practise more: all JEE Advanced PYQs on Conservation of Momentum →

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Conservation of Momentum

Solution

More JEE Advanced questions on Conservation of Momentum