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: $(M = 10,\, m = 5,\, u = 1,\, k = 1 ).$
JEE MAIN 2018, Diffcult
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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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