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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)

  • Option 1)

    \frac{1}{\sqrt 3}

  • Option 2)

    1/2

  • Option 3)

    2/3

  • Option 4)

    \sqrt(3/5)

 

Answers (1)

As we have learned

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

m_{1},m_{2}:masses

u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}

u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}

 

After the first collision , momentum of (m+M)  is mu at time of second collision momentum of (m+M) and momentum of new colliding bodies are equal and opposite hence p=0 hence momentum after every collision : mu (M+13m)v 

v=\frac{m}{M+13}u\rightarrow (1)

M = 2m, \Rightarrow v=\frac{u}{15}

v= wa \Rightarrow \frac{u}{15}=\sqrt{\left ( \frac{k}{M+13m} \right )}A\Rightarrow A=\frac{u}{15}\cdot \sqrt{\left ( \frac{15m}{k} \right )}

A=1/15\sqrt{\left ( \frac{15*5}{1} \right )}=1/ \sqrt 3

 

 

 

 

 

 

 

 


Option 1)

\frac{1}{\sqrt 3}

This is correct

Option 2)

1/2

This is incorrect

Option 3)

2/3

This is incorrect

Option 4)

\sqrt(3/5)

This is incorrect

Posted by

Vakul

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