Question 48 of 194

An oscillator of mass $M$ is at rest in its equilibrium position in a potential, $V=\dfrac{1}{2}k{\left(x–X\right)}^{2}$ . A particle of mass $m$ comes from the right with speed $u$ and collides completely inelastic 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: $\left(M=10,m=5,u=1,k=1\right)$

A$\dfrac{2}{3}$B$\dfrac{1}{\sqrt{3}}$C$\sqrt{\dfrac{3}{5}}$D$\dfrac{1}{2}$

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Answer: (B) $\dfrac{1}{\sqrt{3}}$

In the first collision $\text{mu}$ momentum will be imparted to the system. In the second collision when the momentum of $(M+m)$ is in the opposite direction mu momentum of the particle will make its momentum zero. On ${13}^{th}$ collision,


Applying momentum conservation
$mu=(M+13m)v$

$\Rightarrow v=\dfrac{mu}{M+13m}=\dfrac{u}{15}$ $\Rightarrow M=2m$
$\Rightarrow v=\omega A$

$\Rightarrow \dfrac{u}{15}=\sqrt{\dfrac{K}{M+13m}}\times A$

$\Rightarrow A=\dfrac{1}{15}\sqrt{\dfrac{75}{1}}=\dfrac{1}{\sqrt{3}}$