Spring, blocks and conservation of energy

[zorro]

Homework Statement

A mass M is in static equilibrium on a massless vertical spring as shown in the figure. A ball of mass m dropped from certain height sticks to the mass M after colliding with it. The oscillations they perform reach to height 'a' above the original level of scales & depth 'b' below it. What is the height above the initial level from which the mass m was dropped? (Given that M = m; m = 1 kg , b=3mg/k, k = 10 N/m, g =10m/s²)

The Attempt at a Solution

By conservation of energy (with 0 potential energy level the along the dotted line),

mgh + 0.5*k*(mg/k)² = -(2mgb) + 0.5*k(b+mg/k)²

On solving, I got h as 1.5m which is incorrect.
Can somebody point out my mistake?

 

[Sourabh N]

Because energy is not conserved

Remember - "..sticks to the mass.." => Inelastic collision => Momentum conserved, energy not conserved.. so on.

 

[zorro]

Because energy is not conserved

kinetic energy. Total energy is always conserved

Remember - "..sticks to the mass.." => Inelastic collision => Momentum conserved, energy not conserved.. so on.

How is the momentum conserved? There is an external gravitational force on the system.

 

[Sourabh N]

kinetic energy. Total energy is always conserved

Oops sorry, I meant total mechanical energy is not conserved. It being an inelastic collision, some energy (mechanical energy) is converted to vibrational energy to shake the molecules.

How is the momentum conserved? There is an external gravitational force on the system.

Just before the collision and just after the collision, momentum is conserved. So the way I did it, is to conserve energy till the moment before collision to find the momentum before collision, use conservation of momentum and the fact that they stick together to find *new* mechanical energy, conserve energy again.

Tell me if this makes things clearer :P

 

[zorro]

Tell me if this makes things clearer :P

Actually that confused me more

Just before the collision and just after the collision, momentum is conserved.

momentum of what?

 

[Sourabh N]

Okay, haha let's go slow.

First, draw images of the system just before the ball hits the pad and just after. (Make the pictures like a time series, so in the initial figure you posted, these two new images will be in between your two images.) Label the images 1, 2, 3 and 4.

Ask yourself, what happens from 1-> 2, from 2 -> 3 and from 3-> 4. Among energy and momentum, what is conserved and what is not, for each case. Good luck

 

[zorro]

I understand what you are talking about.
If you consider the system of the two blocks just before and after the collision, there is a vertical gravitational force (weight) acting on the system. How can we conserve the momentum?. Your point will be right if the process takes place on a horizontal plane, where weight is perpendicular to the motion.

 

[Sourabh N]

I understand what you are talking about.
If you consider the system of the two blocks just before and after the collision, there is a vertical gravitational force (weight) acting on the system. How can we conserve the momentum?. Your point will be right if the process takes place on a horizontal plane, where weight is perpendicular to the motion.

Hence the "just". For the brief moment when the collision happens, one can assume that the change in potential energy due to gravitational force is negligible.

 

[zorro]

Hmm...neglecting it, I got h=3m by applying W-E theorem. Thanks for your help!