Solving Physics Puzzle: Hockey Puck Collision

Click For Summary
SUMMARY

The discussion focuses on solving a physics problem involving the collision of two hockey pucks, where puck A, traveling at 40.0 m/s, strikes puck B at rest. After the collision, puck A is deflected at 30 degrees above the x-axis, while puck B moves at 45 degrees below the x-axis. The conservation of momentum equations are established for both x and y directions, leading to two equations with two unknowns, which can be solved to find the speeds of both pucks and the fraction of kinetic energy dissipated during the collision.

PREREQUISITES
  • Understanding of conservation of momentum in two dimensions
  • Knowledge of trigonometric functions (sine and cosine) in physics
  • Familiarity with kinetic energy calculations
  • Basic algebra for solving simultaneous equations
NEXT STEPS
  • Study the principles of conservation of momentum in elastic and inelastic collisions
  • Learn how to apply trigonometric functions to resolve vector components in physics problems
  • Explore kinetic energy conservation and dissipation in collisions
  • Practice solving simultaneous equations involving trigonometric functions
USEFUL FOR

Students in introductory physics courses, educators teaching physics concepts, and anyone interested in understanding momentum and energy conservation in collisions.

RussG
Messages
3
Reaction score
0
Ok, this is probably the simplest problem I was assigned out of an intro physics course and simply cannot figure out how to go about it.

Hockey Puck B at rest is struck by puck A, which was traveling at 40.0m/s in the +x direction. After they hit, A is deflected 30 degrees above the x axis, and Puck B acquires a velocity at 45 degrees below the x axis. The pucks have the same mass and friction forces aren't considered.

a) The speed of each puck after collision
b) what fraction of the original kinetic energy of puck A dissipates during the collision

I have the solutions, but can't figure out how to solve for them. Yes, I suck at physics :( Any help would be appreciated.

Since the message says solutions can't be solved for us, just a hint on where to start would be useful. I'm just ending up with ridiculous equations with sin/cos trying to account for the conservation of momentum.
 
Last edited:
Physics news on Phys.org
Those equations might not be so ridicuous after all. The total momentum is 40*m in the positive x-direction. Keep that in mind.

So the total momentum in the y-direction is 0. The pucks do get a speed in the y-direction though, so these must be equal and opposite to make the total momentum 0:
Va*sin(30 degrees) *m = - Vb*sin(45 degrees)*m (the positive y direction is above the x-axis). So here's equation [1]. (Just forget about the mass btw. It can be annihilated from the equation)

Now in the x-direction there is also conservation of momentum, meaning it has to be 40 m/s * m
Now we achieve:
Va*cos(30 degrees)*m + Vb*cos(45 degrees)*m = 40 *m (Just take out the m again.)
This leaves you with 2 equations and 2 unknowns. Do some basic maths and you'll be able to solve it. Good luck!
 
Hey - thanks a lot for the help. Those are the same equations as what I was getting albeit one difference, and looking at the problem I thought "wow, this one's easy." yet I couldn't get any solutions. I see what I was doing wrong now. Instead of the velocities, Va and Vb, I was using their component velocities with the trig functions, like Vaysin30 Vbysin45 etc. Needless to say that made things complicated when trying to equate the two. Looking back, I have no clue why I was doing that, maybe some inner desire to make things harder than they really are. ;) Thanks for the clarification.
 

Similar threads

Replies
4
Views
3K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K