A ball of mass m is dropped from height 'h' as shown. The collision is perfectly inelastic. What is the speed of combined body and amplitude of oscillations of system. Actually the figure is a spring of spring constant k that is vertically attached to wall and and a mass M is attached to it and ball is falling on M

Question

Answer

Momentum conservation in y direction which gives the velocity of the combined mass , then find angular frequency using w=√k/√(m+M) and thus find amplitude by v=Aw find A

Answer

The Setup:
A ball of mass m is dropped from a height h onto a mass M, which is attached to a vertical spring with a spring constant k. The collision between the ball and the mass M is perfectly inelastic, meaning that they stick together after the collision.

Conservation of Momentum:
Before the collision, the ball is in free fall and has a velocity v given by v = sqrt(2gh), where g is the acceleration due to gravity. The mass M is initially at rest.

After the Collision:
Since the collision is perfectly inelastic, the ball and the mass M stick together and move as a combined body. The mass of the combined body is m + M.

Conservation of Energy:
The energy of the ball-M system is conserved. Initially, the ball has potential energy of mgh and no kinetic energy. After the collision, the combined body has potential energy of (m + M)gh and kinetic energy.

Deriving the Speed:
Using the conservation of energy, we can equate the initial potential energy to the final kinetic energy:

mgh = (m + M)(v')^2/2

where v' is the speed of the combined body after the collision. Solving for v', we get:

v' = sqrt(2mgh / (m + M))

where M is the mass of the object attached to the spring.

Amplitude of Oscillations:
The amplitude of the oscillations of the system can be determined by analyzing the potential energy of the system when it reaches its maximum displacement from the equilibrium position.

At maximum displacement, the spring is stretched or compressed by a certain amount, which we'll call x. The potential energy stored in the spring is given by:

Potential Energy = (1/2)kx^2

Since the potential energy of the system is (m + M)gh at its maximum displacement, we can equate the two expressions:

(m + M)gh = (1/2)kx^2

Solving for x, we get:

x = sqrt((2(m + M)gh) / k)

Summary:
The speed of the combined body after the collision is given by v' = sqrt(2mgh / (m + M)). The amplitude of the oscillations of the system is given by x = sqrt((2(m + M)gh) / k).

Similar JEE Doubts

There are two identical particles A and B. One is projected vertically upward with speed √2gh from ground and other is dropped from height h along the same vertical line. Collision between them is perfectly inelastic. Find time taken by them to reach the ground after collision in terms of.

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