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# In the figure shown a plate 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 from the plate.The particle sticks to the plate. Neglecting the duration of collision find time from the collision of the particle and the plate to the moment when the spring has maximum compression. Spring has force constant 1 N/m. Calculate value of time in the form π/x and find the value of x. Open in App
Solution

## Velocity of the particle just before collision.u=√2g×4.5 mgK⇒u=3g√mKNow it collides with the plate.Now just after collision velocity (V) of the system of plate + particle.mu=3mV⇒V=u3=g√mKNow the system performs SHM with time period T=2π√3m/K(ω=√K3m) and mean position as mg/K distance below the point of collision.Let the equation of motion is y=Asin(ωt+ϕ)v=dydt=Aωcos(ωt+ϕ)At t=0,y=mgK and v=g√mKfrom Eqs (i) and (ii) mgK=AsinϕA=2mgKfrom Eqs (iii) and (iv) ⇒ϕ=5π6y=2mgKsin(√K3mt+5π6)Hence equation of SHM should bey=−A=−2mgK=2mgKsin(√K3mt+5π6)The plate will be at rest again wheny=−A=−2mgK=2mgKsin(√K3mt+5π6)⇒sin(√K3mt+5π6)=−1=sin3π2⇒√K3mt+5π6=3π2⇒t=2π3√3mKUsing value t=π/5 s.  Suggest Corrections  0      Related Videos   Relative
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