Question

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

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=3g√mK

Just after collision, let velocity of the system be (V)

Then, from momentum conservation,

mu=3mV

⇒V=u3=g√mK

Now the system performs SHM with time period

T=2π√3mK

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=g√mK

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=1√3 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=2π3√3mK

Using values K=1 N/m,m=30 gm

⇒t=π5s

Thus, x=5 is the answer.

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