# Work and Energy (Kinetic Energy/Grav. Potential Energy)

1. Apr 22, 2013

### kitty075

NOTE: I know this seems like quite a bit, but it's really not! Most of the questions involve simple 1-line equations, so please don't be discouraged by the amount of text - any assistance is greatly appreciated :)

1. The problem statement, all variables and given/known data

A 6kg steel ball moving at 4m/s hits a 6kg putty ball, initially at rest, and sticks to it. After the collision the two move on (stuck together) at 2m/s.

a) What is the kinetic energy of the metal ball before the collision?
b) What is the kinetic energy of the metal ball after the collision?
c) What is the kinetic energy of the putty ball after the collision?
d) How much energy does the metal ball 'lose' as a result of the collision? (fixed)
e) How much kinetic energy is transferred from the metal ball to the putty ball?
f) What happened to the rest of the energy? Explain.

Also, there's a second segment which reads: A coconut falls out of a tree 8m above the ground and hits a bystander, 1.75m tall, in top of his head. It bounces up 0.5m before falling back towards the ground. If the mass of the coconut is 2kg, calculate the potential energy of the coconut relative to the ground at each of the following times:
a) While it's still in the tree
b) When it hits the bystander's head
c) When it bounces to its maximum height

2. Relevant equations
Ek=0.5mv^2
Eg=mgh

3. The attempt at a solution

1. a) For part 1, I figured out that kinetic energy of the metal ball (prior to collision) was 48J. I'm fairly confident in this answer. :)

1. b) and c) After the collision, however, I calculated the kinetic energy of the metal ball and the putty ball to both be 12J ( 0.5*6kg*2m/s^2). Can they actually both be the same?

1. d) If that's the case, then would the metal ball's 'lost' energy be equal to 48J-12J=36J?

1. e) If the putty ball has a kinetic energy of 12J, then I'm fairly confident that the metal ball transfers 12J of kinetic energy.

1. f) Not sure about this one. Er, I know that some of the energy will be transformed into heat (by friction) and sound, but our teacher was hinting at something involving the change in shape of the putty. Can someone expand on this please? :)

PART 2

2. a) Eg=mgh, so Eg=(2)(9.81)(8). I'm fairly confident with this :)

2. b) When it hits the bystander on the head would I simply change the height to 1.75m? So Eg=(2)(9.81)(1.75)?

2. c) Not sure about this one. Would I just add 0.5m to the height of 1.75m? Eg=(2)(9.81)(2.25)?

Thank you!

Last edited: Apr 22, 2013
2. Apr 22, 2013

### cepheid

Staff Emeritus
Re: 1b and 1c

If they are stuck together, they have the same speed. If they have the same speed, they must have the same KE, because they also have the same mass, and KE depends only on speed and mass.

Your answer for 1d is wrong. If you think about it, the total kinetic energy of the system before was 48 J, and the total kinetic energy after was 24 J (12 + 12). So half the kinetic energy is lost from the system. This is because the metal ball keeps "12 J for itself, and transfers 12 J to the putty" (although really they are a single object). So you didn't account for this 12 J transferred in your tally. Of course the metal ball alone has 36 J fewer than it started with, but arguably the metal ball alone doesn't exist anymore, it's a metal + putty ball with 24 J of KE.

For 1f, work needs to be done to deform the putty when it collides.

3. Apr 22, 2013

### kitty075

Thanks, I found your post to be very helpful :) Is 2 done correctly though?

Oh and also, I think I wrote out the question incorrectly; 1.d) should read: " How much energy does the metal ball lose as a result of the collision?"

At first I thought it would be 24J like you were saying, but is that the case if it's only pertaining to the metal ball's kinetic energy?

4. Apr 22, 2013

### cepheid

Staff Emeritus
Yes, you have done number 2 correctly (EDIT: the physics and equations and heights are right. I didn't bother checking your multiplication). It's a really really simple question. PE = mgh, and it asks you compute it at three different heights. That's it.

Last edited: Apr 22, 2013
5. Apr 22, 2013

### cepheid

Staff Emeritus
Yes the kinetic energy of "just the metal ball" after the collision could be stated to be 12 J. However, it's the part in quotes that I take issue with. There is no "just the metal ball" after the collision: it's mass has been combined with that of something else, and it really only makes sense to talk about the kinetic energy of the overall object. So I don't think the question is very well posed. Still, I suppose 12 J is the best answer to the question "what the KE of the metal ball after the collision."

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