
#1
Sep2108, 02:23 PM

P: 72

1. The problem statement, all variables and given/known data
Puck A has a mass of 0.226 kg and is moving along the x axis with a velocity of 5.59 m/s. It makes a collision with puck B, which has a mass of 0.452 kg and is initially at rest. The collision is not headon. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of a) Puck A b) Puck B 2. Relevant equations No idea 3. The attempt at a solution No Idea  Where do I even start? 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Sep2108, 03:30 PM

HW Helper
P: 2,688

You have to show some work in order to get help here. Those are the rules. "No idea" doesn't cut it, sorry.
You have to have SOME thoughts on the problem. What have you tried? What concepts does the problem involve? 



#3
Sep2108, 04:10 PM

P: 72

I know that it has to do with momentum, and directions. I really just don't know where to start.. After I find the momentum, where would I go from there? 



#4
Sep2108, 04:37 PM

P: 218

Find speed of two objects after collision
Consider the relationship between the momenta before and after collision ...




#5
Sep2108, 04:59 PM

P: 72





#6
Sep2108, 05:37 PM

P: 218

That's what the law of conservation tells us ... and there are more than two (you wrote "both") momenta to consider.




#7
Sep2108, 05:41 PM

P: 72





#8
Sep2108, 05:50 PM

P: 218

p = mv
Both p (momentum) and v (velocity) are vectors so it's a vector equation. Start with the horizontal momenta, one for A before and one each for A and B after. After must be the same as before (conservation!). You don't know the magnitude of the velocities of A and B after so you'll have to call them Va and Vb (or some such) and find another equation to find their values. I've got to go now. 



#9
Sep2108, 06:52 PM

P: 72

So far I've come up with these equations: P=m*v = 0.226kg * 5.59m/s = 1.263314 kg*m/s Vay = Va sin 37  for A's Y direction Vax = Va cos 65  for A's X direction Vbx = Vb cos 37  for B's X direction Vby = Vb sin 65  For B's Y Direction So Va = Sqrt((Vax)^2+(Vay)^2) Just lost on how to find the velocity of Va and Vb... 



#10
Sep2108, 07:18 PM

P: 295

I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59




#11
Sep2108, 07:39 PM

P: 72





#12
Sep2208, 01:30 PM

P: 218

Let M stand for the mass of A, 0.226 kg. How many M is the mass of B? Slightly modifying what you wrote above, the total momentum in the x direction before the collision can be written P1x = (V1ax * m1) + (V1bx * mb) = (5.59 * M) + (0 * 2M) = 5.59M Use your expressions for velocities after the collision (V2ax etc.) in an expression for the total momentum in the x direction after the collision: P2x = ... What is the total momentum in the y direction before and after the collision? P1y = ... P2y = ... 



#13
Sep2208, 01:38 PM

P: 218




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