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.
Mass of block M = 200g = 0.2 kg,
Mass of particle is m = 120g = 0.12 kg,
Acceleration due to gravity g = 10 m/s,
Particle is dropped on the block from a height of 45cm =0.45m,
The string is stretched by a distance x =1.00 cm= 0.01m.
In the equilibrium condition
The velocity with which the particle m will strike M is given by
So, after the collision the velocity of the particle and the block is 
Let the string be stretched through an extra deflection of δ,
According to law of conservation of energy
On solving above equation we get
