# Elastic collision

1. May 15, 2014

### Karol

1. The problem statement, all variables and given/known data
A body of mass 3 kg slides on a friction less surface with velocity 4 m/s and collides elastically with a resting mass of 2 kg. calculate the final velocities.

2. Relevant equations
Conservation of momentum: $m_1v_1+m_2v_2=m_1u_1+m_2u_2$
Conservation of energy: $\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2$

3. The attempt at a solution
I solved those two equations and got a correct answer. but my question is as follows:
If we solve, as a general case, the two above equations we get a third equation:
$$v_1-v_2=-(u_1-u_2)$$
Which states that the relative velocities before and after the collision remain the same.
When i solved this third equation together with the first equation, that of the conservation of momentum i got, of course, a wrong answer.
Can you explain? both ways seem correct.

2. May 15, 2014

### Fightfish

It might help to show your work. I don't see why the latter approach should give you a wrong answer.

3. May 15, 2014

### Karol

My work

When i use the first method, the long way by solving fully the 2 equations, right from the start there is a difference: from the Conservation of momentum:
$$3\cdot 4=3u_1+2u_2 \Rightarrow u_1=4-\frac{2}{3}u_2$$
The second equation in both methods is the conservation of energy which remains the same.
In the second method, from the equation:
$$v_1-v_2=-(u_1-u_2) \Rightarrow 3=-(u_1-u_2) \Rightarrow u_1=u_2-3$$
The two u1 differ.

4. May 15, 2014

### Fightfish

Based on sign convention, if positive is towards right, then it should be
$$u_{1} - u_{2} = 4$$
and that gives same answer as the previous approach since $u_{2}$ is zero.
You might have mixed up some numbers in your attempt (eg. the 3 you have here instead of 4).

5. May 15, 2014

### Karol

Thanks, i wouldn't have found it, but i think you have a mistake, if i take it as you wrote:
$$u_1=u_2+4$$
the answer would be wrong. and why should i change signs? i shouldn't take into consideration the positive direction since the equation:
$$v_1-v_2=-(u_1-u_2)$$
Is general, it doesn't take in consideration directions, no?
So, if i use:
$$u_1=u_2-4$$
then it comes out right

6. May 15, 2014

### Fightfish

I think we might have been using different letters for initial and final. I assumed that u referred to the initial velocity. If so, then u1 = 4 is the correct answer since u2 = 0 and the direction of the initial block is taken to be positive.

7. May 15, 2014

### Fightfish

Yup, seems that you mean final velocities by u. Then the two equations that you mentioned earlier:
$$u_1=4-\frac{2}{3}u_2$$
$$u_1=u_2-4$$
are correct.

There is no disagreement between them; they form a pair of simultaneous equations that allow you to determine both $u_{1}$ and $u_{2}$

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