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

The Correct Option is A

The correct option is(A): \(\frac{1}{\sqrt{3}}\)

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 \(=\frac{1}{2} m v'^{2}\)
\(\Rightarrow K E=\frac{1}{2} \times 75 \times \frac{1}{15} \times \frac{1}{15}\)
\(\Rightarrow \frac{1}{2} k A^{2}=\frac{1}{2} \times \frac{75}{225}=\frac{1}{6}\)
\(\Rightarrow \frac{1}{2} \times 1 \times A^{2}=\frac{1}{6}\)
\(\Rightarrow A^{2}=\frac{1}{3}\)
\(\Rightarrow A=\frac{1}{\sqrt{3}}\)