Elastic collision of neon atom problem

Click For Summary
SUMMARY

The discussion revolves around solving an elastic collision problem involving a neon atom with a mass of 20.0 u colliding with an unknown atom at rest. After the collision, the neon atom moves at a 57.9° angle, while the unknown atom moves at a -45.4° angle. The conservation of momentum and energy equations are established, but the user encounters difficulties due to having more unknowns than equations. The law of sines is suggested as a method to relate the velocities and angles, leading to a solution for the mass of the unknown atom.

PREREQUISITES
  • Understanding of elastic collisions and conservation of momentum
  • Familiarity with the law of sines in trigonometry
  • Basic algebra skills for solving equations
  • Knowledge of vector components in physics
NEXT STEPS
  • Study the application of the law of sines in collision problems
  • Review conservation of momentum in two dimensions
  • Learn how to isolate variables in algebraic equations
  • Explore examples of elastic collision problems involving angles
USEFUL FOR

Physics students, educators, and anyone interested in understanding elastic collisions and momentum conservation in two-dimensional scenarios.

mithanon
Messages
1
Reaction score
0

Homework Statement


A neon atom (m = 20.0 u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 57.9° angle from its original direction and the unknown atom travels away at a -45.4° angle. What is the mass (in u) of the unknown atom? [Hint: You can use the law of sines.]

variables
m = neon = 20 u
M = mass of unknown
\theta = 57.9°
\phi = 45.4°
v = initial velocity of neon
v' = final velocity of neon
V = final velocity of unknown


Homework Equations


I've set up conservation of momentum equations for both directions, and the energy equation, but I'm stuck with more unknowns than equations


The Attempt at a Solution


mv2=mv'2+MV2
mv = mv'cos\theta + MVcos\phi
0 = mv'sin\theta - MVsin\phi

using a momentum vector sum diagram and law of sines? I got
mv/sin(180-\theta-\phi)=mv'/sin(\phi)

not sure how that helps but I guess I can then express v' in terms of v, eliminating one variable

no idea where to go from here
 
Physics news on Phys.org
Everything looks OK so far I think. You can also use the law of sines to get the other "side" of the triangle. The problem is now just one of algebra. I would pick one equation (maybe the energy one), and substitute in for the variables you don't want and then isolate M.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
21
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K