Metal ball sticks to putty. Where does the energy go?

Click For Summary

Homework Help Overview

The problem involves a collision between a metal ball and a ball of putty, focusing on the energy transformations and conservation laws during the event. The subject area includes concepts from mechanics, specifically momentum and energy conservation in collisions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the energy lost by the metal ball and gained by the putty during the collision, with attempts to calculate these values. Questions arise regarding the implications of energy conservation versus momentum conservation, particularly in relation to heat and other forms of energy conversion.

Discussion Status

Participants are exploring various interpretations of energy loss and conservation principles. Some have provided calculations for energy changes, while others question the phrasing of the problem and the validity of the teacher's explanation regarding energy conservation.

Contextual Notes

There is a noted ambiguity in the question regarding the terminology used, specifically whether "energy" should be understood as "kinetic energy" or "mechanical energy." This has led to differing interpretations among participants.

Subrosian
Messages
4
Reaction score
0
A 6.0kg metal ball moving at 4.0m/s hits a 6.0 kg ball of putty at rest and sticks to it. The two go on at 2.0 m/s.
a) How much energy does the metal ball lose in the collision?
b) How much energy does the putty ball gain in the collision?
c) What happened to the rest of the energy?

For a):
E_{klose} = E_{k} - E_{k'}
E_{k} = 1/2mv^2<br /> = 1/2(6.0kg)(4.0m/s)^2 - 1/2(6.0kg)(2.0m/s)^2<br /> = 36J<br />

Another method I tried was:

<br /> W = F \Delta d<br /> =m( \Delta v/ \Delta t)*(v_{f}+v_{i})/2* \Delta t<br /> =6.0kg(-2m/s)*(3m/s)<br /> =-36J<br />

b) is simple:
<br /> E_{k} = 1/2mv^2<br /> = 1/2(6.0kg)(2.0m/s)^2<br /> = 12J<br />

The only problem I have with this question is the last part, C. The thing is, the momentum was conserved completely, so the energy could not be lost from friction or sound. Here is an example of trying to find the velocity of the ball and putty combined if no energy was lost to heat/sound/friction/etc:

1/2(6.0kg)(v)^2 + 1/2(6.0kg)(v)^2 = 1/2(6.0kg)(4.0m/s)^2
2v^2 = 48J*2/6.0kg
v = \sqrt{8}

Which is impossible because of
mv = m^{&#039;}v^{&#039;}
6.0kg*4.0m/s = 12.0kg*\sqrt{8}m/s
Which is clearly unequal.

I asked this question to my teacher and he said that the conservation of energy does not always work. That is a ridiculous answer :rolleyes:
 
Last edited:
Physics news on Phys.org
I'm not sure why you say no energy will be lost to heat. The deformation of the putty will warm it up.
 
If energy was lost to heat, then why is the momentum the same?
 
Conservation of momentum and conservation of total energy are independent laws of physics.

Energy isn't lost, it's just converted into other forms of energy, such as heat.

What is "lost" in this particular problem is the kinetic energy associated with the center-of-mass motions of the two balls.

The center-of-mass momentum is, by definition, always equal to the total momentum of the system. But the center-of-mass energy is not the same as the total energy.
 
So, if the center of mass energy is not the same as the total energy, does that mean the answers to a) and b) are incorrect as well? They are also in the answer key.
 
The question is phrased wrongly: you answered correctly, but "energy" should be replaced by "kinetic energy", or "mechanical energy".

Given the way the question is phrased, and given your teacher's answer, I doubt he knows a lot of physics.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
27
Views
1K
Replies
4
Views
16K
Replies
8
Views
4K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 13 ·
Replies
13
Views
8K