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Question

In the figure shown, a plate of mass 60 grams is at rest and in equilibrium. A particle of mass m = 30 grams is released from height 4.5mgk from the plate. The particle sticks to the plate. Neglecting the duration of collision, the time from the collision of the particle and the plate to the moment when the spring has maximum compression is in the form of Xπ ms (millisecond) where X is (Spring has force constant of 100 N/m)

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Solution

Velocity of the particle just before collision

u=√2g×4.5 mgk⇒u=3g√mk

Now it collides with the plate.

Now just after collision velocity of system of plate + particle

mu=3mV⇒V=u3=g√mk

Now system perform's SHM with time period T=2π√3mk and mean position as mgk distance below the point of collision.

Let the equation of motion y=sin(ωt+ϕ)

for t=0, y=mgk

mgk=Asinϕ (1)Now for amplitude V=ω√A2−y2g√mk=√k3m√A2−m2g2k2⇒A=2mgk (2)By (1) and (2)

y=A2 to y=0 ⇒t=T12y=0 to y=A ⇒t=T4

Total time t=T12+T4=2π3√3mk

Using values, t=20π ms

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