# Elastic Collisions in one dimension

1. Oct 18, 2009

### patelkey

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

Particle 1 of mass m1 = 0.30 kg slides rightward along an x axis on a frictionless floor with a speed of 2.0 m/s. When it reaches x = 0, it undergoes a one-dimensional elastic collision with stationary particle 2 of mass m2 = 0.40 kg. When particle 2 then reaches a wall at xw = 70 cm, it bounces from it with no loss of speed. At what position on the x axis does particle 2 then collide with particle 1?
2. Relevant equations

1/2m1v1i^2 = 1/2m1v1f^2 + 1/2m2v2f^2
3. The attempt at a solution[/b
I found all the values in the equation however, I don't know how to apply that to figure out the position where they will collide.

2. Oct 18, 2009

### ordirules

that's good, but you need another conserved quantity to solve the equation. Have you tried using conservation of momentum?

$m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$

3. Oct 18, 2009

### patelkey

After using the equation you gave me I got v2f=3m/s and v1f=-2m/s
but what do I do with these two quantities to figure out the point of collision?

4. Oct 18, 2009

### ordirules

okay, so one is moving left at 2 m/s ($$v_{1f} = -2 \frac{m}{s}$$) and one is moving right at 3 m/s ($$v_{2f} = 3 \frac{m}{s}$$.

Another important fact is the particle loses no speed when it hits the wall, so you may assume it just changes direction with the same speed.
(side note: technically, to conserve momentum, you have to assume the wall it hits moves backwards as well, but you can say that the mass of the wall is so huge it barely moves, so $$m_1 v_{1i}= m_1 v_{1f} + m_{wall} v_{wall} \rightarrow \frac{m_1}{m_{wall}} (v_{1i} - v_{1f}) = v_{wall}$$, but $$m_{wall} >> m_1 so v_{wall}$$ is approx 0 )

You can start by calculating how far the one moving left at 2 m/s will get from the point of collision (say $$x_0 = 0$$) when the one moving right hits the wall, and then find how long it will take for them to meet with the simple equations:

$$x_1 (t) = v_{1f} t + x_1$$
$$x_2 (t) = v_{2f} t + x_2$$

where $$x_1, x_2$$ are the initial positions of the masses respectively from the point of collision at the time the second particle strikes the wall,

and find when $$x_1(t) = x_2 (t)$$, solve for that time and plug it in one of the equations to get the distance.

however, there is a slightly more clever way, using the same methods, where you can solve for it in one less step. In what way can you move the particle and slightly change the situation such that the time the particle takes to get to the wall and back in the original situation is the same but it involves no wall?

5. Oct 18, 2009

### patelkey

I got it
It turned out I was entering -0.28 cm but I forgot to convert it so it was actually -28 cm thanks for the help