# Off center elastic collision

1. Nov 1, 2008

### jromeo

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

The problem statement, all variables and given/known data
The mass m1 has the velocity (v1i)\hat{i} and makes an off-center collision with m2=2m1. The final velocities are v1f=a1\hat{i}+b1\hat{j}, and v2f=a2\hat{i}+b2\hat{j}. Assuming elastic collision and v2i=0m/s, obtain the values of a1, a2, and b2 for the given value of b1. Also obtain the angles \theta1 and \theta2 of v1f and v2f with the x-axis. Retain the solutions for a1>0.

m1 = 3.20kg
v1i = 11.2m/s
b1 = 4.12m/s

2. Relevant equations

3. The attempt at a solution

First I broke down the conservation of momentum equation into it's vector components.
m1v1i = m1a1 + m2a2 and 0 = m1b1 + m2b2. I then solved for b2 by setting b2 equal to
-m1b1/m2. Then I attempted to solve for a1 and a2 by using KEi=KEf because energy is conserved in elastic collisions. KEf = 1/2m1v1i2 = 1/2m1(a12 + b12) + 1/2m2(a22 + b22)

I keep getting the wrong units when I solve for a1 or a2 in one equation though. I can't figure out what i'm missing about this problem. Thanks in advance for any help!

2. Nov 1, 2008

### djeitnstine

Ok your first step is correct however to simplify matters, you can use V1 - V2 = -(V1 - V2) (x and y components of course)

Then solve for either V1 or V2 and plug it into your momentum equations.

Also note that the final angle adds up to 90 degrees