Question

a block of mass 200g is suspended througha vertical spring. The spring is stretched by 1cm when the block is in equilibrium. A particle of mass 120g is dropped on the block from a height of 45cm . The particle sticks to the block after the impact . Find the maximum extension of the spring .(Take g=10m/s^2)

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Answer 1

Answer:

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Explanation:

Mass of block , M = 200 g

Block of particle = m = 120 gm = 0.12 kg

In equlibrium condition ,

Spring is stretched by a distance , x = 1.00 cm = 0.01 m

0.2 x g = K x

2 = K x 0.01

K = 200 N/m

Velocity with which particle m will strike M is given by u = √( 2 x 10 x 0.45)

= √9 = 3 m/s

After collision , the velocity of particle and block is

V = 0.12 x 3 / 0.32 = 9/8 m/s

Let the spring be stretched through an extra deflection of δ

0 - (1/2) x 0.32 x (81/64) = 0.32 x 10 x δ - (1/2 x 200 x (δ + 0.1)² -(1/2)x200x(0.01)²

After solving above equation we get ,

δ = 0.045 m = 4.5 cm