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In the figure shown, a plate of mass 60 gm is at rest and in equilibrium. A particle of mass m=30 gm is released from height 4.5mgK above the plate. The particle hits the plate and sticks to it. If the time interval between the collision of the particle and the time at which it comes to momentary rest for the first time is given by πx, then find x.
Spring has force constant 1 N/m. Neglect the duration of collision


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Solution


Velocity of the particle just before collision
u=2g×4.5mgK=3gmK
Just after collision, let velocity of the system be (V)
Then, from momentum conservation,
mu=3mV
V=u3=gmK
Now the system performs SHM with time period
T=2π3mK
And mean position will be mgK distance below the point of collision, corresponding to balancing of weight of m.
Let the equation of motion be
y=Asin(ωt+ϕ) ..... (i)
v=dydt=Aωcos(ωt+ϕ) ......(ii)
At t=0,y=mgK and v=gmK
From Eqs. (i) and (ii), mgK=Asinϕ ...(iii)
and 3mgK=Acosϕ ... (iv)
[because ω=K3m]
Squaring and adding (iii) and (iv),
A=2mgK ....(v)
From Eqs. (iii) and (v), tanϕ=13 or ϕ=5π6
Therefore, y=2mgK sin(K3mt+5π6)
The plate will be at rest again when y=A [maximum compression]
i.e A=2mgK=2mgK sin(K3mt+5π6)
K3mt+5π6=3π2
t=2π33mK
Using values K=1 N/m,m=30 gm
t=π5s
Thus, x=5 is the answer.

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