Question

An oscillator of mass $M$ is at rest in its equilibrium position in a potential $V = \frac{1}{2} k(x - X)^2$. A particle of mass $m$ comes from right with speed u and collides completely inelastically with $M$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after $13$ collisions is : $(M=10, m=5, u=1, k=1)$

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\sqrt{\frac{3}{5}}$

Answer 1

Answer: (A)

Potential of the given oscillator is $V=\frac{1}{2} k(x-k)^{2}$
Given: $M=10 ;\, m=5,\, u=1 ;\, k=1$
Initial momentum of the particle of mass $m=m u=m \times 5=5\, m$
Momentum of (oscillator + particle) after collision $=(M+ m)$
Velocity of oscillator after collision $=v$
So, momentum of system $=(M+ m) v$
From conservation of linear momentum, we have
$(M +m)=m u=5 \times 1=5$
For second collision, oscillator and particle have momentum in opposite direction.
Net or total momentum is zero.
Likewise after $4^{\text {th }}, 6^{\text {th }}, 8^{\text {th }}, 10^{\text {th }}, 12^{\text {th }}$ collision the momentum is zero.
After $12^{\text {th }}$ collision, Mass of oscillator and 12 particles will be $(10+12 \times 5)=70$
Now, from conservation of linear momentum, for $13^{\text {th }}$ collision, we have
$70 \times 0+5 \times 1=(70+5) v'$
$\Rightarrow v'=\frac{5}{75}$
$\Rightarrow \frac{1}{15}$
Total mass after $13^{\text {th }}$ collision $=(10+13 \times 5)=75$
Kinetic energy of system $=\frac{1}{2} m v'^{2}$
$\Rightarrow K E=\frac{1}{2} \times 75 \times \frac{1}{15} \times \frac{1}{15}$
$\Rightarrow \frac{1}{2} k A^{2}=\frac{1}{2} \times \frac{75}{225}=\frac{1}{6}$
$\Rightarrow \frac{1}{2} \times 1 \times A^{2}=\frac{1}{6}$
$\Rightarrow A^{2}=\frac{1}{3}$
$\Rightarrow A=\frac{1}{\sqrt{3}}$