An oscillator of mass M is at rest in its equilibrium position in…

MCQ

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
$\frac {1}{2}$
✓
$\frac {1}{\sqrt 3}$
C
$\frac {2}{3}$
D
$\sqrt {\frac {3}{5}}$

✓

Answer

Correct option: B.

$\frac {1}{\sqrt 3}$

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

On $13^{\text {th }}$ collision,

$\mathrm{m} \rightarrow {\mathrm{M}+12} ; \quad \mathrm{M}+13 \mathrm{m} \rightarrow \mathrm{V}$

$\mathrm{mu}=(\mathrm{M}+13 \mathrm{m}) \mathrm{v} \Rightarrow \mathrm{v}=\frac{\mathrm{mu}}{\mathrm{M}+13 \mathrm{m}}=\frac{\mathrm{u}}{15}$

$v=\omega A \Rightarrow \frac{u}{15}=\sqrt{\frac{K}{M-13 m}} \times A$

Putting value of $M, m, u$ and $K$ we get amplitude

$A=\frac{1}{15} \sqrt{\frac{75}{1}}=\frac{1}{\sqrt{3}}$

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