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 $m=mu=m\times 5=5m$
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
$(M+m)=mu=5\times 1=5$
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
$70\times 0+5\times 1=(70+5)v^{\prime }$
$\Rightarrow v^{\prime }=\frac{5}{75}$
$\Rightarrow \frac{1}{15}$
Total mass after $13^{\text{th }}$ collision $=(10+13\times 5)=75$
Kinetic energy of system $=\frac{1}{2}m{v}^{\prime 2}$
$\Rightarrow KE=\frac{1}{2}\times 75\times \frac{1}{15}\times \frac{1}{15}$
$\Rightarrow \frac{1}{2}k{A}^2=\frac{1}{2}\times \frac{75}{225}=\frac{1}{6}$
$\Rightarrow \frac{1}{2}\times 1\times A^2=\frac{1}{6}$
$\Rightarrow A^2=\frac{1}{3}$
$\Rightarrow A=\frac{1}{\sqrt{3}}$