# Elastic collision problem

1. Mar 29, 2013

### Mangoes

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

A 10.0 g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.200 m/s.

Find the velocity of each marble after the collision.

3. The attempt at a solution

Since I am ignoring friction, the sum of external forces is zero and momentum is conserved.

$$m_1v_{A1} + m_2v_{B1} = m_1v_{A2} + m_2v_{B2}$$

$$(0.01 kg)(0.4 m/s) + (0.03 kg)(0.2 m/s) = (0.01 kg)v_A + (0.03 kg)v_B$$

I have two unknowns and one equation.

Since the collision is elastic, kinetic energy is also conserved.

$$(1/2)m_1{v_{A1}}^2 + (1/2)m_2{v_{B1}}^2 = (1/2)m_1{v_{A2}}^2 + (1/2)m_2{v_{B2}}^2$$

$$(1/2)(0.1kg)(0.4 m/s)^2 + (1/2)(0.3 kg)(0.2 m/s)^2 = (1/2)(0.1kg){v_{A2}}^2 + (1/2)(0.3kg){v_{B2}}^2$$

Simplifying the numbers a little and omitting units for simplicity gives the two equations:

$$0.01 = 0.01v_A + 0.03v_B$$
$$0.014 = 0.05{v_A}^2 + 0.15{v_B}^2$$

I didn't bother going through with the algebra, but plugging in the two equations into WA gives:
http://www.wolframalpha.com/input/?i=0.01+=+0.01x+++0.03y,+0.014+=+0.05x^2+++0.15y^2

Not only do I not have a way to discriminate between which pair of points I should use, but neither of the answers match up. The book tells me that the velocities are vA = 0.5 m/s and vB = -0.1 m/s.

Why isn't this working out?
This exact same thought process worked fine for the problem right before this one.

2. Mar 29, 2013

### Mangoes

I just realized I made an error in the first equation for the velocity of the marble. I forgot to include the negative to account for the fact that the marble is going towards the other marble.

After fixing that and getting the correct numbers, I got this result:

http://www.wolframalpha.com/input/?i=0.002+=+0.01x+++0.03y,+0.014+=+0.05x^2+++0.15y^2

One of the set of points in this equation is what the book is saying. But I still have one other problem... Why does the book omit the other point (-0.4, 0.2)?

EDIT: Nevermind I just realized why the answer doesn't make physical sense.

Last edited: Mar 29, 2013
3. Mar 30, 2013

### ehild

The equations for an elastic collision state conservation of momentum and conservation of energy. They are valid also if nothing happens, the particles do not collide, just avoid each other. You always get that solution in addition to the "correct" one.

ehild