Elastic Collision of hockey pucks

[americanforest]

Ok, I ran across this question as I was doing my homework and I thought it looked pretty easy. I worked on it for an hour and still had no luck with getting the correct answer. Here is the full text:

An elastic collision occurs between two air hockey pucks in which one puck is at rest and the other is moving with a speed of 0.1 m/s. After the collision, the puck initially in motion makes an angle of 35.00 deg with its original direction, and the struck puck moves at an angle of 55.00 deg on the other side of the original direction. What is the final speed of the first puck and second puck? (Also, mass is the same)

First thing I thought was that I could use conservation of momentum and conservation of energy in x or y direction. This would give me two equations, which would allow me to solve for final velocities. I'm not going to go through the tedium of solving all the systems of equations. I'll just tell you that in the end I got

V1=.0628 m/s (This is the initially stationary puck)
V2=.0897 m/s

Is the answer you guys get? It's not right according to my book.

from equations:

$$(.1cos(35))^2=(v_1cos(55))^2+(v_2cos(35))^2$$

$$0=v_2sin(35)-v_1sin(55)$$

I hate when seemingly simple problems stump me and this is a great example. I normally wouldn't be surprised if it was some math error, but I spent a large amount of time diligently checking and rechecking math. The answers look legitimate. Please help?

 

[OlderDan]

Your first equation should not have any trig ratios in it. It is the full velocity that contributes to the energy. You have left out some of the energy. Your second equation looks OK.

The angle between the two puck velocities should be 90 degrees, which is what they gave you.

 

[americanforest]

It was a stupid misunderstanding of the problem that got me. I didn't understand where the problem was telling me the angles were. Just one more example of the fact that physics questions like this cannot be considered valid without pictures. Turns out

$$V_1f=V_0cos (\theta)$$
$$V_2f=V_0sin (\theta)$$

$$\theta=35$$ degrees
$$V_0=.1 m/s$$

I didn't think of the angles being perpendicular until you brought it up. That's when I suspected I misunderstood and after that I had it right within 3 minutes. Thanks.

P.S. Looking forward to becoming a member in this community. Just changed to Physics major and really love the subject. I hope I can learn a lot here.

 

[OlderDan]

It was a stupid misunderstanding of the problem that got me. I didn't understand where the problem was telling me the angles were. Just one more example of the fact that physics questions like this cannot be considered valid without pictures. Turns out

$$V_1f=V_0cos (\theta)$$
$$V_2f=V_0sin (\theta)$$

$$\theta=35$$ degrees
$$V_0=.1 m/s$$

I didn't think of the angles being perpendicular until you brought it up. That's when I suspected I misunderstood and after that I had it right within 3 minutes. Thanks.

P.S. Looking forward to becoming a member in this community. Just changed to Physics major and really love the subject. I hope I can learn a lot here.

Welcome to the forum. It is a good place to visit often. There are several people here who are very helpful. You will soon get to know who they are.

Your problem was interesting in that it gave you more information than required by the situation. An elastic collision of two objects of equal mass, one being initially at rest, always results in a separtaion angle of 90 degrees. The problem could have been done even if they had given only one of the angles.