JEE Main On 16 April 2018 Question 15
Question: 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 comes from right with speed u and collides completely in elastically with and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after collisions is: $ (M=10,m=5,u=1,k=1). $ [JEE Main 16-4-2018]
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{\sqrt{3}} $
C) $ \frac{2}{3} $
D) $ \sqrt{\frac{3}{5}} $
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Answer:
Correct Answer: B
Solution:
Initial momentum of mass ’m’ = mu =5
Final momentum of system $ =(M+m)v=mu=5 $
For second collision, mass (m=5, u = 1)
coming from right strikes with system of mass 15,
both momentum have opposite direction.
$ \therefore $ net momentum = zero
Similarly for12th collision momentum is zero.
For 13th collision, total mass $ =10+12\times 5=70 $
Using conservation of momentum $ 70\times 0+5\times 1=(70+5)v’ $
$ v’=\frac{1}{5} $ Total mass $ =10+13\times 5=75 $
Finally KE of system $ =\frac{1}{2}mv^{2}=\frac{1}{2}\times 75\times [ \frac{1}{15} ][ \frac{1}{15} ] $ $ \frac{1}{2}k,A^{2}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15} $
$ =\frac{1}{7}\times (1)A^{2}=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15} $ $ A^{2}=\frac{1}{3} $ $ A=\frac{1}{\sqrt{3}} $
