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Question: An oscillator of mass M is at rest in its equilibrium position in a potential V =\(\dfrac{1}{2}\) k(...

An oscillator of mass M is at rest in its equilibrium position in a potential V =12\dfrac{1}{2} k(x – X)2^2. 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\dfrac{1}{2}
B 13\dfrac{1}{{\sqrt 3 }}
C 23\dfrac{2}{3}
D 35\sqrt {\dfrac{3}{5}}

Explanation

Solution

An inelastic collision is one in which objects stick together after impact, and kinetic energy is not conserved. In an inelastic collision the momentum of the system is conserved therefore we can write mu+M×0=(m+M)v. Where, v is the final velocity.
Using this conservation method we get the velocity after 13 collisions. The total potential is equal to the total kinetic energy of the oscillator which provides the energy to the oscillator to perform oscillation K.E(13)K.E_(13)=12\dfrac{1}{2}kA2^2.

Complete step by step answer:
Given, m=5, M=10, u=1, k=1.
By conservation of momentum we can write,
Initial momentum =final momentum
For first collision equation one becomes, mu+M×0=(m+M)v
Therefore final momentum after 1st collision is (m+M)v=mu=5×1=5
For a second collision, mass (m=5, u=1) coming from right strikes with the system of mass (m+M=15), both momentum have opposite directions. Net momentum= zero.
Similarly for 12th collision, total mass=M+12×m=10+12×5
=10+60=70
Using conservation of momentum for 13th collision,
70×0+5×1=(70+5)v’
5=75v’
v’=575=115\dfrac{5}{{75}} = \dfrac{1}{{15}}
Final KE=12\dfrac{1}{2}mv’2^2=12×75×(115)2\dfrac{1}{2} \times 75 \times {\left( {\dfrac{1}{{15}}} \right)^2}
This kinetic energy is equal to the potential of the oscillator to perform oscillation.
Potential of the system= final KE
12kA2=12×75×1225\dfrac{1}{2}k{A^2} = \dfrac{1}{2} \times 75 \times \dfrac{1}{{225}}
1×A2=131 \times {A^2} = \dfrac{1}{3}
A=13\sqrt {\dfrac{1}{3}} .
The amplitude after 13 collision=13\dfrac{1}{{\sqrt 3 }}

So, the correct answer is “Option B”.

Additional Information:
Collisions are of three types:
1. perfectly elastic collision.
2. inelastic collision.
3. perfectly inelastic collision

Note:
In all types of collision momentum is conserved.
In inelastic collision kinetic energy is not conserved whereas inelastic collision kinetic energy is also conserved.