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Question

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.

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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 m

By 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=0.045m

so xmax=1cm+δ

xmax=1+4.5 = 5.5 cm


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