Hockey puck collision with elastic constraint and rest condition

[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.

 

[ChrisBrandsborg]

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.

Yeah, that´s what I thought :-) Thanks for helping me!