    Question

# An oscillator of mass M is at rest in its equilibrium position in a potential V=12k(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
12
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B
13
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C
23
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D
35
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Solution

## The correct option is C 1√3Initial momentum of mass 'm' = mu =5Final momentum of system= (M+m)v=mu = 5For second collision, mass (m=5, u = 1) coming from right strikes with system of mass 15, both momentum have opposite direction.∴ net momentum = zeroSimilarly for 12th collision momentum is zero.For 13th collision, total mass = 10 +12 × 5 = 70Using conservation of momentum70×0+5×1=(70+5)v′v′=115Total mass =10+13×5=75 Finald KE of system =12mv2=12×75× 12k A2=12 75×115×115 =17×(1)A2=1275×115×115 A2=13 A=1√3  Suggest Corrections  0      Similar questions  Related Videos   Aftermath of SHM
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