# Elastic collision of a ball on a wire

1. Feb 22, 2010

### jahrollins

1. The problem statement, all variables and given/known data
A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.60 kg and 2.40 kg, and the length of the wire is 1.20 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

2. Relevant equations
.5m1v2 = mgh
m1v1 + m2v2 = m1vf1 + m2vf2

3. The attempt at a solution
So I've got part a ~
.5m1v2 = mgh
v2 = 2gh

and I'm lost on part b:
m1v1 + m2v2 = m1vf1 + m2vf2
since the final velocities of the ball and box are unknown how do I solve?

2. Feb 22, 2010

### rl.bhat

After collision, total momentum and the KE is conserved.
Write down equations for these two conservation laws and solve for the final velocities.

3. Feb 22, 2010

### E=mc^84

Hi, for a) The ball starts from rest and has potential energy. The potential energy is converted to kinetic energy at the bottom of the string, therefore: mgh = 0.5mv^2, you shud get v = sqrt2gh, where h = the wire legnth = 1.2m.

For b) since the ball collides with the block, initially at rest, you have to use the law of conservation of momentum and adding the block mass and the ball mass and using the speed in the above question as the initial velocity of only the ball: mava = v(mball +mblock), where mava = the speed and mass of the ball respectively from the above question. Therefore, Solve for v.

Hope this helps:)

4. Feb 22, 2010

### jahrollins

In mava = v(mball +mblock)
why would the final velocity be the same for both the ball and the block?

5. Feb 22, 2010

### ehild

The velocities will not be the same after collision.

The collision is elastic, both the momentum and the energy is conserved during the collision.

$$m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}$$

$$\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{i2}^2=\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2$$

Just before collision, the block is in rest, vi1=0, and the velocity of the ball is obtained from the equation vi2=sqrt(gL) where the length of wire is L=1.2 m. Calculate vi2 and plug in the values for the initial velocities in both equations.

ehild

6. Feb 22, 2010

### E=mc^84

Because they are in contact with each other during the collision for a reallt small amount of time. That velocity cud = the ball or the block since they are both moving at such initial speeds. But, since you are concerned with only the the velocity of the ball, you use only that.

7. Feb 23, 2010

### ehild

You are wrong. We do not care what happens during that very short time while the colliding bodies are in contact: it is beyond point mechanics. Only the result counts. If the collision is elastic, the kinetic energy is conserved, it is the same before and after collision.

The velocities will be the same if the collision is totally inelastic.

ehild

8. Feb 23, 2010

### E=mc^84

You are talking about what you "care about", this is what happens in the problem. Work out the problem and show me that i am wrong :)

9. Feb 23, 2010

### ehild

Well, do you know what is elastic collision?

ehild

10. Feb 23, 2010

### jahrollins

From those two, I came out with vi2 = vf2?

11. Feb 23, 2010

### ehild

No. Plug in the numbers and solve the system of equations.

ehild