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# A block of mass 200 g is suspended through a vertical spring.The spring is stretched by 1.0 cm when the block is in equilibrium.A particle of mass 120 g is dropped on the block from a height of 45 cm.The particle sticks to the block after the impact.Find the maximum extension of the spring.Take g=10 m/s2. Open in App
Solution

## Mass of block M=200 g=0.20 kg Mass of the particle,m=120gm=0.12 kg In the equilibrium condition,the spring is stretched by a distance, x = 1.00 cm = 0.01 mBy balancing the forces in the equilibrium position we have,Mg=kx ⇒0.2×g=k×x ⇒2=k×0.01 ⇒k=200 N/m The velocity with which the particle m will strike M is given by, u=√2gh=√2×10×0.45 =√9=3 m/sec So,after the collision, the velocity of the particle and the block is v=0.12×30.32=98 m/sec Let the spring be stretched through an extra deflection of δ. 0−(12)×0.32×(8164) =0.32×10×δ−(12)×200×(δ+0.1)62−(12)×200×(0.01)2 Solving the above equation,we can get δ=0.045mso xmax=1cm+δxmax=1+4.5 = 5.5 cm  Suggest Corrections  10      Similar questions  Explore more