# Homework Help: Which speed is greater? Different question, Is this possible?

1. Nov 20, 2012

### bored2death97

This isn't a homework question, just did a midterm and I am curious about a question. it had this diagram. Same initial speed.

and it asked which statement is most correct about the speed at height, h (the line going across the graph)

a) It depends on their masses
b) Ball 3 has smallest velocity
c)All have same speed

There were a couple other options, but I forget them, and these were the ones I was contemplating. I put that ball 3 has a lower speed because when it is at the height indicated the speed goes to zero.

I also have another.
Two carts sit on horiz. frictionless track. Equal masses. Cart A is moving towards cart B. Cart B is at rest. Is it ever possible for Cart A to move in reverse after the collision.

a) depends on conditions of collision
b) No, never
c) Yes

I put no, although I was thinking about that it depends on the conditions. I was thinking about if it was glues to the track, then it would go in reverse. But from what I know, the 2 cars either join together, one transfers all of its momentum to the 2nd, or they go off at different angles (since this is a horizontal track that wouldnt be possible).

Last edited: Nov 20, 2012
2. Nov 20, 2012

### K^2

The rules of this board require you to demonstrate your own attempt or considerations for solving any homework/test problem. Yes, that includes the past ones. Don't worry if you can't formalize it too well. Do the best you can.

There are many reasons for that. One is that it honestly helps us explain the problem to you when we see what it is that you are having problems with.

3. Nov 20, 2012

### bored2death97

Fixed it, thanks for letting me know.

4. Nov 20, 2012

### Vorde

Right about the second, wrong about the first. The speed at that given height isn't zero for the 3rd ball because it is still moving towards the right. Because the acceleration due to gravity is the same for all three balls, the vertical speed of all three will be the same, but the horizontal speed will be different. You could calculate the total speed using Pythagoras's theorem, but just realize that if the vertical speed is the same, then the ball with the highest horizontal speed will have the highest total speed, and vice versa.

So because the ball on the right has the highest velocity to the right, it will have the greatest total velocity, even if it's vertical speed is zero.

5. Nov 20, 2012

### JustinRyan

The speed of ball 3 is not zero at h.

And I think that "frictionless" means no gluing of the cart to the track is allowed :)

6. Nov 20, 2012

### K^2

You are over-thinking and making a logical mess that leads you to a wrongful conclusion.

What does conservation of energy has to say about all that?

It doesn't have to transfer all of its momentum. It can transfer just a fraction. Or it can even transfer more momentum than it has!

Picture it this way. Instead of cart B, a railroad cart on an actual railroad. Instead of cart A, you'll throw a ball at it. What's the outcome going to be?

To work it out more systematically, you can use the fact that center of mass will continue to move at the same velocity before and after collision. Try using three test cases mA = mB, mA = 2*mB, and mA = (1/2) * mB

Last edited: Nov 20, 2012
7. Nov 20, 2012

### JustinRyan

8. Nov 20, 2012

### Vorde

Ah crap, nevermind, I have to think more before I post.

9. Nov 20, 2012

### K^2

10. Nov 20, 2012

### JustinRyan

Sorry, I was replying to this.

11. Nov 20, 2012

### Vorde

Ignore that, when I was doing the math for myself I was doing something different from the original question, K^2 is right.

12. Nov 20, 2012

### jimmyly

Someone please correct me if i am wrong, i just started my first physics class this year.

so for questions 1
All 3 balls have different horizontal components but horizontal components and vertical components are independent of each other therefore we can neglect it.

the way i look at it is. assuming initial velocity(vertically) is equal among all 3 balls, they will all have the same vertical component, acceleration due to gravity,and time. thus, when all 3 balls reach the same height, they have equal speed(vertically not horizontally) which would be zero seeing that the 3rd ball is a projectile reaching its maximum height.

again if someone can tell me if i'm on the right track that would be great. I feel i learn better when i try to explain things to someone else

Last edited: Nov 20, 2012
13. Nov 21, 2012

### JustinRyan

You are correct but you initially said they begin with the same speed. That isn't the same thing. Speed is the magnitude of velocity regardless of the direction.

If they begin with the same speed then ball 1 has a greater vertical velocity because it has zero horizontal velocity. So it will go higher than h.

Maybe you could look up the question again and list all the options.

EDIT: Sorry I just realised you won't be able to do that.

Have you figured it out yet?

Last edited: Nov 21, 2012
14. Nov 21, 2012

### Staff: Mentor

The problem statement defines mA = mB. It cannot transfer more than its own momentum in this setup.

This assumption is wrong.
Energy conservation is the best way to solve problem 1, as K^2 already posted.

15. Nov 21, 2012

### K^2

Yeah. Missed that bit. OP's right about that one, then.

16. Nov 21, 2012

### jimmyly

I havent started conservation of energy yet so im a bit confused abot that part.

also why is this assumption wrong?

17. Nov 21, 2012

### Staff: Mentor

The speed (the magnitude of the velocity vector) is the same for all 3 balls, but the direction is different. You can write the vertical component as $v \sin(\alpha)$ where $\alpha$ is the angle between initial direction and the floor. As this is variable, the vertical component of the velocity is different, too.

18. Nov 21, 2012

### bored2death97

I asked my prof about the cart question, and he said the answer was no, which is what I put.
Didn't ask him about the other question, but a lot of people here are saying that it would have the same speed.