# 2D Elastic Collision

1. Dec 11, 2015

### Valenti

1. The problem statement, all variables and given/known data
A red and a blue rubber puck are free to slide along a frictionless air table. Each has a mass of 40 grams. They collide in an elastic collision. Initially the red one is at rest and the blue one is traveling in the x direction with a speed of 4 m/s. After the collision the blue one is traveling in the direction +30 degrees, with the red one traveling in the direction ‐55 degrees. Using conservation of energy and momentum find the speed of each puck after the collision.

2. Relevant equations
m1v1i+m2v2i = m1v1f + m2v2f

3. The attempt at a solution

Solve for the momentum in each direction
m1v1i+m2v2i = m1v1f + m2v2f
X Direction
m1v1i+m2v2i = m1v1f + m2v2f
0.04kg(4m/s) + 0.04kg(0m/s) = 0.04 (v1f cos30) + 0.04(v2f cos-55)
0.16kgm/s = 0.03v1f + 0.02v2f

Y Direction
m1v1i+m2v2i = m1v1f + m2v2f
0.04kg(0) + 0.04kg(0m/s) = 0.04kg(v1f sin30) + 0.04kg (v2f sin-55)
0 = 0.02v1f - 0.03 v2f

Solve for one of the Velocities
0 = 0.02v1f - 0.03 v2f
0.03v2f = 0.02v1f
v2f = 2/3 v1f

Replace with new velocity
0.16kgm/s = 0.03v1f + 0.02v2f
0.16kgm/s = 0.03v1f + 0.02(2/3)v1f
0.16kgm/s = 0.03v1f + 1/75 v1f
0.16kgm/s = 13/300 v1f
3.69m/s = v1f

Solve for v2f
v2f = 2/3 v1f
v2f = 2/3 (3.69 m/s)
v2f = 2.46 m/s

2. Dec 12, 2015

### haruspex

Rounding to only 1 significant digit at intermediate steps might be making your final answer very inaccurate.
Best is to avoid plugging in numbers until the final step. Work entirely symbolically, even providing symbolic variables to represent data given as numbers. It has numerous advantages.
When you do plug in the numbers, keep 3 or 4 significant digits in order to quote an answer of 2 or 3 digits. Even then, watch out if at some point you find you take the difference between two numbers that are close in value.
I reconstructed the initial velocity from your answers and the x momentum equation and got 4.6 m/s.

3. Dec 12, 2015

### Valenti

Ah alright thanks I'll be sure to do that for upcoming questions. As for this question not counting my errors in significant digits was it wrong, or did i have the right procedure?

4. Dec 12, 2015

### haruspex

The method was fine.

5. Dec 12, 2015

### ehild

You round off too much during the calculation. If you keep one significant digit only, the result will be less accurate than one digit. So you can write 4m/s = v1f and v2f = 2m/s. The data were given with two digits. You need keep at last 3 digits during the calculation to reach the result with accuracy of one digit.
By the way, the problem gives both angles, so you can solve for the velocities using conservation of momentum alone. According to the results, the collision is not elastic, energy is not conserved. Your result gives more energy after collision it was before, so it is impossible. Calculating with enough accuracy, the final energy is a bit less than the initial one, a Physically realistic result.