# Elastic Collision of hockey pucks

• americanforest
In summary: You would have just plugged in the other angle to find the final velocity. The final velocities are .0628 m/s for the initially stationary puck and .0897 m/s for the struck puck.
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?

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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.

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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.

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americanforest said:
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.

## 1. What is an elastic collision of hockey pucks?

An elastic collision of hockey pucks is a type of collision in which the total kinetic energy of the system is conserved. This means that the pucks do not stick together or lose any energy during the collision, but instead bounce off each other with the same speed and direction as before the collision.

## 2. How is the elasticity of a hockey puck collision determined?

The elasticity of a hockey puck collision is determined by the coefficient of restitution, which is a number between 0 and 1 that represents the ratio of the final relative speed of the pucks to the initial relative speed. A higher coefficient of restitution indicates a more elastic collision, while a lower coefficient indicates a less elastic one.

## 3. What factors affect the elasticity of a hockey puck collision?

The elasticity of a hockey puck collision can be affected by factors such as the material and surface properties of the pucks, the angle and speed at which they collide, and any external forces acting on the pucks during the collision.

## 4. How is momentum conserved in an elastic collision of hockey pucks?

In an elastic collision of hockey pucks, momentum is conserved because the total momentum of the system before and after the collision remains the same. This means that the sum of the individual momentums of the pucks before the collision is equal to the sum of their momentums after the collision.

## 5. Can an elastic collision of hockey pucks occur in real life?

Yes, an elastic collision of hockey pucks can occur in real life, although it is rare. In most cases, there will be some energy loss due to friction and other factors, resulting in a slightly less elastic collision. However, with careful control and measurement, it is possible to achieve a near-perfect elastic collision of hockey pucks.

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