# Sanity check

1. Apr 20, 2006

### Jacques

Hi,

I have a physics problem I need to solve which I think is quite straight forward but would appreciate some input on whether I'm going down the right lines

Tow metal spheres hanging from the roof one of 10g (ball1) and one of 100g (ball2)

ball1 is falling from a height of 15cm accelerating at 9.81ms-2 and collides with sphere b which is stationary

After the collision sphere ball2 is moving at a velocity of 1ms

What is the velocity of ball1?

Now my logic is as follows and I would appreciate comments.

To get the velocity of ball1 we need to use the equation

v = (Uball1(Mball1- Mball2)+2 x Mball2 x Uball2) / Mball1+Mball2

Now we have everything we need apart from Uball1

Uball1 we need to calculate using

Uball1 = sqrt(2x9.81x0.15m)

Once I have Uball1 I can calculate the velocity of ball1 after the collision.

Is my thinking correct?

Thanks for your help.

2. Apr 20, 2006

### nazgjunk

At a quick glance, it looks fine to me. Except for one thing: how should ball 2 start moving when it is hanging on a rope or something and is hit on the top? Or even how can it be hit on the top, while it is obviously connected with the rope on top?

I guess this is a bad case of non-practical assignments.

3. Apr 20, 2006

### Jacques

Sorry, I was unclear on that

both balls are haning from ropes

ball1 is pulled up keeping the string tight to an elevation of 15 cm above ball2

Thanks

j

4. Apr 20, 2006

### Hootenanny

Staff Emeritus
I have it slightly different.

Conservation of momentum states;

$$m_{1}u_{1} = m_{1}v_{1} + m_{2}v_{2}$$

$$m_{1}u_{1} - m_{2}v_{2} = m_{1}v_{1}$$

$$v_{1} = \frac{m_{1}u_{1} - m_{2}v_{2}}{m_{1}}$$

I think this is different from yours?

~H

5. Apr 20, 2006

### Jacques

Thanks Hoot,

6. Apr 21, 2006

### Jacques

Me again.

In the question above I used a the equation v = sqrt(2as) to get the U of ball1

Is this correct? or should I be using some sort of shm calculation to get the U as the ball is hanging from a rope and thus acting as a pendulum?

thanks

j

7. Apr 21, 2006

### Hootenanny

Staff Emeritus
As you are not given the length of the rope or any time period of the oscillations, you cannot calculate the velocity if it is undergoing SHM, in addition you are not given a release point. Therefore, I think it as safe to assume that the ball is initially at rest, in which case your formula,$v = \sqrt{2as}$ is correct.

~H

8. Apr 21, 2006

### Jacques

THanks H

Just asking as I was discussing it with someone last night who reconed that its an shm calculation.

the question does however give that the ball1 is pulled up to 15 cm above the height of ballb, this does not give us the length of the rope but can't the 15cm somehow be used as an amplitude in shm?.
My gut feeling is no as we dont know what 15cm is of the total length of the rope.

one las thing: how do you get the mathematical symbols in your posts?

Thanks

j

9. Apr 21, 2006

### Hootenanny

Staff Emeritus
I think it is safe to assume that the ball was initially stationary. The mathematical symbols use a code called $\LaTeX$. If you click on any mathematical equations used in these forums a popup will display the code used. There is a tutorial on latex in this thread;