Collision of Hockey Pucks: Solving for Final Speed and Angle

[ChrisBrandsborg]

Homework Statement

A hockey puck of mass M hits two other, identical pucks of mass m. The two pucks fly off with the same speed vf at angles of ±θ relative to the direction the original puck was traveling (see figure). The original puck had initial speed vi, and the two other pucks were initially at rest. We will assume that the pucks are sliding frictionlessly on the ice.

a) If the collision is elastic and the first puck ends up at rest after the collision, what is the final speed vf of the two other pucks and the angle θ?

b) What relation must one require between m and M in order for the scenario of a) to be possible? What happens if this requirement on m and M is not met?

For each question, provide an algebraic expression in terms of (some or all of) M, m, vi.

Homework Equations

E_k: 1/2mv_i² = 1/2mv_f²

The Attempt at a Solution

a) E_k conserved:

1/2Mv_i² = 2(1/2mv_f²)

v_f = √(Mv_i)²/2m

How do I find the angle?

Px = 0 -> 0 = 2mv_f⋅sinθ
Py = 0 -> Mv_i = 2mv_f⋅cosθ

I know the angle should be 45°, but can't figure out exactly how to get there.

b) 1/2Mv_i = 2(1/2mv_f)

Make an equation for M:

M = 2m (v_f²/v_i²)

Which means that if this requirement isn't met, than its not elastic and K is not conserved.

 

[PeroK]

Homework Statement

A hockey puck of mass M hits two other, identical pucks of mass m. The two pucks fly off with the same speed vf at angles of ±θ relative to the direction the original puck was traveling (see figure). The original puck had initial speed vi, and the two other pucks were initially at rest. We will assume that the pucks are sliding frictionlessly on the ice.

a) If the collision is elastic and the first puck ends up at rest after the collision, what is the final speed vf of the two other pucks and the angle θ?

b) What relation must one require between m and M in order for the scenario of a) to be possible? What happens if this requirement on m and M is not met?

For each question, provide an algebraic expression in terms of (some or all of) M, m, vi.

Homework Equations

E_k: ½mv_i = ½mv_f

The Attempt at a Solution

a) E_k conserved:

½Mv_i = 2(½mv_f)

v_f = √(Mv_i)²/2m

This is correct. Almost: I didn't notice those brackets.

How do I find the angle?

Px = 0 -> 0 = 2mv_f⋅sinθ
Py = 0 -> Mv_i = 2mv_f⋅cosθ

I know the angle should be 45°, but can't figure out exactly how to get there.

b) ½Mv_i = 2(½mv_f)

Make an equation for M:

M = 2m (v_f²/v_i²)

Which means that if this requirement isn't met, than its not elastic and K is not conserved.

This last equation is not what they are looking for. This is simply energy conservation. Instead, you need to combine this with an equation you get from conservation of momentum.

 

[ChrisBrandsborg]

This is correct.

This last equation is not what they are looking for. This is simply energy conservation. Instead, you need to combine this with an equation you get from conservation of momentum.

Mv_i = 2 (mv_f) ? That is conservation of momentum?

 

[PeroK]

Mv_i = 2 (mv_f) ? That is conservation of momentum?

No. Momentum is a vector. It has $x$ and $y$ components. As you had in your first post.

 

[NascentOxygen]

Homework Equations

Ek: ½mvi = ½mvf

K.E. = ½mv² [emoji1429]

 

[PeroK]

K.E. = ½mv² [emoji1429]

Yes, I didn't look closely enough. I've edited my original response.

 

[ChrisBrandsborg]

Yes, I didn't look closely enough. I've edited my original response.

Why isn't the first one correct?

 

[PeroK]

Why isn't the first one correct?

You seem to be struggling with latex a bit. Look at post #5. You've defined KE with $v$ rather than $v^2$. Then you get $(Mv)^2$ rather than $Mv^2$ and finally you seem to get the right answer for $M$, which all suggests you may just be mistyping.

 

[ChrisBrandsborg]

You seem to be struggling with latex a bit. Look at post #5. You've defined KE with $v$ rather than $v^2$. Then you get $(Mv)^2$ rather than $Mv^2$ and finally you seem to get the right answer for $M$, which all suggests you may just be mistyping.

Oh yeah, I see that I´ve been mistyping.. I meant v²

 

[PeroK]

Oh yeah, I see that I´ve been mistyping.. I meant v²

Okay, just to take stock. You have, from conservation of energy:

$v_f^2 = \frac{Mv_i^2}{2m}$

Now, conservation of momentum?

 

[ChrisBrandsborg]

Okay, just to take stock. You have, from conservation of energy:

$v_f^2 = \frac{Mv_i^2}{2m}$

Now, conservation of momentum?

Yes, and from conservation of momentum:

Px: 0 = 2mv_f⋅sinθ
Py: Mv_i = 2mv_f⋅cosθ

 

[PeroK]

Yes, and from conservation of momentum:

Px: 0 = 2mv_f⋅sinθ
Py: Mv_i = 2mv_f⋅cosθ

The first equation is clearly not correct. That implies $\theta = 0$. Remember that momentum is a vector so you must respect the direction the object is moving. The first equation is not needed for this question anyway. But, you need to understand why it's wrong.

What can you do with the second?

 

[ChrisBrandsborg]

Do I put my equation for v_f into my two equations for momentum?

 

[PeroK]

Do I put my equation for v_f into my two equations for momentum?

The first equation is of little interest in this problem. Focus on the second.

 

[ChrisBrandsborg]

The first equation is clearly not correct. That implies $\theta = 0$. Remember that momentum is a vector so you must respect the direction the object is moving. The first equation is not needed for this question anyway. But, you need to understand why it's wrong.

What can you do with the second?

Oh, I thought since they are moving only in y-direction, the x equation has to be = 0 (but I guess it´s just = 0), not 0 = 2mv_fsinθ ?
Hmm.. what I can I do with the second one.. Hmm. Can I put $v_f = √\frac{Mv_i^2}{2m}$ in for v_f ?

 

[ChrisBrandsborg]

The first equation is of little interest in this problem. Focus on the second.

Oh okay.. hmm

 

[ChrisBrandsborg]

Do I need something else? Or can I just divide by 2mv_f to get cosθ alone?

 

[PeroK]

Oh, I thought since they are moving only in y-direction, the x equation has to be = 0 (but I guess it´s just = 0), not 0 = 2mv_fsinθ ?
Hmm.. what I can I do with the second one.. Hmm. Can I put $v_f = √\frac{Mv_i^2}{2m}$ in for v_f ?

Sounds like a plan!

 

[PeroK]

Do I need something else? Or can I just divide by 2mv_f to get cosθ alone?

Why not just do it and see?

 

[ChrisBrandsborg]

Sounds like a plan!

I thought you told me that the first equation wasn't relevant for finding the angle?

 

[ChrisBrandsborg]

Why not just do it and see?

If so, I get:

cosθ = Mv_i/2mv_f
θ = cos^-1(Mv_i/2mv_f)

 

[PeroK]

If so, I get:

cosθ = Mv_i/2mv_f
θ = cos^-1(Mv_i/2mv_f)

Look, the whole point of having two equations for $v_i$ and $v_f$ is so that you can eliminate one or both. That's what you are trying to do here! If you put the equations side by side:

$v_f = v_i \sqrt{ \frac{M}{2m}}$

$v_f(2m \cos \theta) = Mv_i$

You should be able to see that you can eliminate both $v_i$ and $v_f$ here.

 

[ChrisBrandsborg]

Look, the whole point of having two equations for $v_i$ and $v_f$ is so that you can eliminate one or both. That's what you are trying to do here! If you put the equations side by side:

$v_f = v_i \sqrt{ \frac{M}{2m}}$

$v_f(2m \cos \theta) = Mv_i$

You should be able to see that you can eliminate both $v_i$ and $v_f$ here.

If I put v_f into the other one, I get:

cosθ = M/(2m⋅√M/2m) ?

 

[PeroK]

If I put v_f into the other one, I get:

cosθ = M/(2m⋅√M/2m) ?

Okay, but that can be simplified surely?

 

[ChrisBrandsborg]

Okay, but that can be simplified surely?

Yeah, probably.. Cos θ = M/√2Mm, isn't that right?

 

[PeroK]

Yeah, probably.. Cos θ = M/√2Mm, isn't that right?

What about $\cos \theta = \sqrt{\frac{M}{2m}}$?

More important: what does that tell you?

 

[ChrisBrandsborg]

What about $\cos \theta = \sqrt{\frac{M}{2m}}$?

More important: what does that tell you?

I first thought that meant that if M = 2m, then cos θ = 1, which mean that cos^-1(1) = 45°, but then I remembered that, that was for tan^-1(1)
For cos: if M = 2m, then the angle is 0° (but is that correct?), shouldn't it always be 45°?

 

[ChrisBrandsborg]

But it does tell you that the angles depends on the masses and not the velocity:)

 

[PeroK]

I first thought that meant that if M = 2m, then cos θ = 1, which mean that cos^-1(1) = 45°, but then I remembered that, that was for tan^-1(1)
For cos: if M = 2m, then the angle is 0° (but is that correct?), shouldn't it always be 45°?

No it isn't always 45°, which bis what that equation is telling you.

What happens if $M > 2m$?

What (physically) happens if $M = 2m$.

What happens as $M$ gets smaller?

 

[ChrisBrandsborg]

No it isn't always 45°, which bis what that equation is telling you.

What happens if $M > 2m$?

What (physically) happens if $M = 2m$.

What happens as $M$ gets smaller?

If M > 2m, then I´m not sure.. I guess he will just crash through them (and won't stop at rest at the collision))

if M = 2, then v_f = v_i, so M stops at rest, and the 2m´s continue with the same speed as M had (but the angle is 0° ? )

if M is smaller, than the angle goes towards 90° (the closer it goes to 0), but then v_f is slower than v_i

 

[PeroK]

If M > 2m, then I´m not sure.. I guess he will just crash through them (and won't stop at rest at the collision))

if M = 2, then v_f = v_i, so M stops at rest, and the 2m´s continue with the same speed as M had (but the angle is 0° ? )

if M is smaller, than the angle goes towards 90° (the closer it goes to 0), but then v_f is slower than v_i

More or less. If $M > 2m$ then there is no solution where $M$ stops completely. (Note that $cos \theta \le 1$.) The two smaller pucks cannot take all the energy/momentum of the larger puck.

$M = 2m$ just means that the two pucks move in the original direction (think of square pucks perhaps).

And, yes, as $M$ gets smaller, the two pucks it hits go up and down at a larger angle.

That's about as far as you can take it. You have $v_f$ in terms of $v_i$ and $\theta$ in terms of the ratio of the masses. There are no more equations to squeeze out of the problem.

 

[ChrisBrandsborg]

More or less. If $M > 2m$ then there is no solution where $M$ stops completely. (Note that $cos \theta \le 1$.) The two smaller pucks cannot take all the energy/momentum of the larger puck.

$M = 2m$ just means that the two pucks move in the original direction (think of square pucks perhaps).

And, yes, as $M$ gets smaller, the two pucks it hits go up and down at a larger angle.

That's about as far as you can take it. You have $v_f$ in terms of $v_i$ and $\theta$ in terms of the ratio of the masses. There are no more equations to squeeze out of the problem.

But what you said about M = 2m.. What is their original direction? Because if he hit them both (on the sides), they both won't go straight in the same direction? Mathematical the answer is 0°, but I don't understand that..

 

[PeroK]

But what you said about M = 2m.. What is their original direction? Because if he hit them both (on the sides), they both won't go straight in the same direction? Mathematical the answer is 0°, but I don't understand that..

The mathematics provides possible solutions. There may be other physical constraints that prevent some or all of these solutions. That's why I said think of square pucks. The solution is possible with square pucks, but not round pucks. $M$ must hit both $m$'s full on - square pucks could do that.

Notice that the question assumed that $M$ stopped after the collision. We can see that, with round pucks, this may be hard to achieve. For example, if the angle of impact and trajectory is determined by the shape of the pucks (let's say it is 45°), then we see that

$\frac{M}{2m} = \cos^2 (45°) = \frac12$

So, we see that we must have $M = m$ in order for $M$ to stop. Any other mass and we will not be able to get $M$ to stop.

Also, with square pucks, we may have a constraint that $\theta = 0$, so we must have $M = 2m$ or $M$ will not stop.

In reality, the problem would probably be the other way round: the angle $\theta$ may be observed and we might know $m$ and if we observe the change in velocity of $M$ we could determine its mass, say.

This problem could be seen as a special case of that. $M$ stopped and from the measurement of $\theta$ alone, then the mass $M$ could be determined - assuming the mass $m$ we already knew

 

[ChrisBrandsborg]

The mathematics provides possible solutions. There may be other physical constraints that prevent some or all of these solutions. That's why I said think of square pucks. The solution is possible with square pucks, but not round pucks. $M$ must hit both $m$'s full on - square pucks could do that.

Notice that the question assumed that $M$ stopped after the collision. We can see that, with round pucks, this may be hard to achieve. For example, if the angle of impact and trajectory is determined by the shape of the pucks (let's say it is 45°), then we see that

$\frac{M}{2m} = \cos^2 (45°) = \frac12$

So, we see that we must have $M = m$ in order for $M$ to stop. Any other mass and we will not be able to get $M$ to stop.

Also, with square pucks, we may have a constraint that $\theta = 0$, so we must have $M = 2m$ or $M$ will not stop.

In reality, the problem would probably be the other way round: the angle $\theta$ may be observed and we might know $m$ and if we observe the change in velocity of $M$ we could determine its mass, say.

This problem could be seen as a special case of that. $M$ stopped and from the measurement of $\theta$ alone, then the mass $M$ could be determined - assuming the mass $m$ we already knew

So the requirement I´m looking for in b is that M = 2m (but doesn't M stop when M < 2m ?)

 

[PeroK]

So the requirement I´m looking for in b is that M = 2m (but doesn't M stop when M < 2m ?)

Let's go back to the problem. You found that:

$\cos \theta = \sqrt{\frac{M}{2m}}$

And, I told you something you hopefully already knew, that;

$\cos \theta \le 1$

In order for this scenario to be possible, you must have $M \le 2m$. If $M > 2m$ it is impossible. With $M \le 2m$ it is possible - that's all we are saying.