Hockey puck velocity after impact

[nysnacc]

Homework Statement

Homework Equations

mv1 + mv2 = mv1' +mv2'

The Attempt at a Solution

(0.17) (10i - 4j) + (mass of stick) (v j) = (0.17) (sin 20 i + cos 20 j)

what do i need for the mass of stick?

 

[Simon Bridge]

Why do you think you need a mass for the stick?
Please show your reasoning.

 

[nysnacc]

mv1+mv2 =mv1' +mv2'

 

[haruspex]

mv1+mv2 =mv1' +mv2'

You are told to neglect the change in speed of the stick. I admit it might not be obvious how to deal with that. Here are two ways:
1. Let the stick have mass M and do not ignore its change in speed. You will get an answer that depends on M. Then let M tend to infinity and see what happens to the answer.
2. Work in the reference frame of the stick. That makes the stick head effectively a solid, immoveable floor. The puck is now like a ball bouncing on the ground.

 

[nysnacc]

SO for the puck,

mv1 = mv2 ??

 

[haruspex]

SO for the puck,

mv1 = mv2 ??

No, why?
By the way, this is wrong:

mv1 + mv2 = mv1' +mv2'

There are two different masses. m₁v₁+m₂v₂=m₁v₁'+m₂v₂'.

 

[nysnacc]

But setting m2(stick) as infinity?

 

[haruspex]

But setting m2(stick) as infinity?

If you do that straight away (option 2) you cannot use (and will not need) momentum conservation. The change in momentum of the stick becomes indeterminate (0 times infinity).
In option 1, letting M tend to infinity is the final step.

 

[nysnacc]

yes option one then

m1v1 + infinity v2 = m1v1' + infinity v2' ??

 

[haruspex]

yes option one then

m1v1 + infinity v2 = m1v1' + infinity v2' ??

As I explained, with option 1, setting M to infinity is the final step. You have to get a complete answer as a function of M first.

 

[nysnacc]

m1v1 + infinity v2 = m1v1' + infinity v2'
M (v2 - v2') = m1(v1' - v1)
M = m1 (v1' - v1) / (v2 - v2')

v1 is 0 and v2' is the desire velocity (dir) towards the goal??

 

[haruspex]

m1v1 + infinity v2 = m1v1' + infinity v2'

Please desist from posting this, it is unhelpful. We will let M go to infinity right at the end, not before.

M (v2 - v2') = m1(v1' - v1)

Right, but you need another equation. Use the given coefficient of restitution.

 

[nysnacc]

e (v2 - v1) = v2' - v1'

 

[haruspex]

e (v2 - v1) = v2' - v1'

Almost! You have a sign error there. Correct that and solve the pair of equations.

 

[nysnacc]

e (v2 - v1) = v1' - v2' (cuz v2' is 0?)

 

[haruspex]

e (v2 - v1) = v1' - v2'

That is the right restitution equation.

(cuz v2' is 0?)

As I wrote in post #4, using option 1, we do not set v₂' equal to v₂ (and certainly not to 0).
All the velocities we have referred to so far are in the y direction, right? At some point, we will have to consider the x direction in order to make use of the given desired angle, but that can wait. For now, we just have to assume we are going to be able to determine from that what v₁' needs to be. So the next step is to use the two equations you have (momentum and restitution) to find an expression for v₂. It can involve the given velocity v₁, the velocity we expect to be able to find, v₁', and the two masses. So what variable do we need to eliminate?