# Particle collision

1. Mar 12, 2009

### twentysix26

So there is a particle with m1 that hits a particle m2 at rest, they bounce off at angles theta1 and theta2 from the horizontal. The original problem proposes that you can find the mass of the second particle from knowing the first particle, and the angles that they both make from the horizontal after the elastic collision. So i solved the equation when m1 = 20 and theta1 = 55.6 degrees and theta2 = 50 degrees, and i found out that m2 = 40.

But what im trying to find is an easier way to solve, so a formula to discover the ratio of the two masses from the total angle after the elastic collision

So conservation of momentum in x-direction:
m1v (before)= m1v1cos(theta1) + m2v2cos(theta2) (after)

y-direction:
m1v1sin(theta1) - m2v2sin(theta2) = 0

conservation of KE:
1/2(m1)(v^2) = 1/2(m1)(v1^2) + 1/2(m2)(v2)^2

2. Mar 12, 2009

### berryland1jmm

Here's the equation that my friend and I found...

We can find the ratio based off of the observed angles. We want to know how to find the angles (or spread) that result for any given ratio of masses.

#### Attached Files:

• ###### Physics.doc
File size:
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3. Mar 12, 2009

### berryland1jmm

Py is wrong
------
y-direction:
m1v1sin(theta1) - m2v2sin(theta2) = 0
------

Negative angles correlate to negative sin..
it should be: m1v1sin(theta1) + m2v2sin(theta2) = 0

With that modification the equation becomes what I have attached.

What puzzles me is how a span of 105.6 does not always correlate to 2m1=m2...
A span of 90 always correlates to m1=m2...

File size:
18 KB
Views:
113