# Momentum - Hit and stick problem solving

1. Jun 26, 2011

### AceInfinity

Momentum - "Hit and stick" problem solving

1. The problem statement, all variables and given/known data

6. A 75.0 kg hockey player moving at +10.0 m/s crashes into a second, stationary
hockey player. After the collision, the two skaters move as a unit at +4.50 m/s. In
the collision, the impulse received by the second hockey player was

A. +1.09 × 10^3 kg.m/s
B. +7.50 × 10^2 kg.m/s
C. +4.13 × 10^2 kg.m/s
D. +3.38 × 10^2 kg.m/s

2. Relevant equations

p = mv

3. The attempt at a solution

I had forgotten how to do this type of question, i'm reviewing over my previous school material.

Last edited: Jun 26, 2011
2. Jun 26, 2011

### cepheid

Staff Emeritus
Re: Momentum - "Hit and stick" problem solving

Let's start with this: what is the definition of impulse?

3. Jun 26, 2011

### fizika_kz

Re: Momentum - "Hit and stick" problem solving

Hello! It's quite simple question :D. We just need to use the conservation law of the momentmum:
the initial momentum of the system is only running player's - 75kg * 10m/s = 750 kg* m/s.
After the collision the total momentum is just the sum of momentums of two players.
So, the momentum of the running player after the collision is 75kg * 4.5 m/s = 337.5 kg * m/s.
And two find the impulse, received by the second player, you need to substract the above momentums: 750 kg*m/s - 337.5 kg*m/s = 412.5 kg * m/s
Answer: 412.5 kg* m/s

4. Jun 26, 2011

### cepheid

Staff Emeritus
Re: Momentum - "Hit and stick" problem solving

We don't give out complete solutions to homework problems. Please READ THE PF RULES before posting!!!

5. Jun 26, 2011

### AceInfinity

Re: Momentum - "Hit and stick" problem solving

This is not homework however. It's a practice Diploma i've found on the internet. I already have the answer notice, i'm only asking how you'd go about getting the answer through steps.

I'd find it a lot easier if I had the steps to follow as well, but impulse I know is the same as momentum.

Thanks though I now understand, so the 750 kg•m/s was the total, since the non-moving player had 0 momentum. (p = mv > p = m•0 > 0), and the conservation of momentum allowed for some of the 750 kg•m/s to be transferred over to the non-moving player (412.5kg•m/s) while the rest of the momentum stayed with the original player that caused the collision who now has 337.5kg•m/s momentum magnitude after the collision.

I'm not here to cheat homework, I'm here to re-learn what I already once knew so that I can prep for the big year end test this week. So I would hope people know that they can trust to give me answers. I can give you a link to the diploma i'm testing even if you don't believe me trustworthy of telling you the truth.

Unless i'm posting this in the wrong area, because I want to be able to learn, and hopefully get some help from people who are willing to help me out with the steps to complete a problem I'm struggling with. I just need a refresher

Last edited: Jun 26, 2011
6. Jun 26, 2011

### cepheid

Staff Emeritus
Re: Momentum - "Hit and stick" problem solving

You posted in homework help. Homework helpers are not supposed to give people complete solutions under any circumstances. I was admonishing fizika_kz, not you. It was not my intention to suggest that you did anything wrong.

The final answer is meaningless. The important part is the steps to arrive at it, i.e. the solution. On this site we are willing to help guide you in solving a problem, but we will not do it for you outright, because you would not learn anything if we did. I would say that that holds true even if this is not homework. You prepare for an exam by trying to learn physics, and you do that by solving problems, not by having others solve them for you.

Impulse is not quite "the same as" momentum. An impulse is equal to the change in momentum. (In other words, if you supply an impulse to an object, its momentum will change by that amount). The impulse is also equal to the product of the applied force and the time interval over which it is applied.

I don't understand what you are trying to say with the expression that I've bolded above. It doesn't make sense to me. But as for the rest of it, yes, your understanding of the problem is valid.

Again, this has nothing to do with you or your particular situation, so you shouldn't take my protest (which wasn't even directed at you) personally. We simply don't give people full solutions to problems in the homework help section of the site. Those are the rules, and the other homework helper who posted here violated them. The forum rules are intended to make sure that PF is a place where people can get help learning the material, not a place where they can avoid having to learn the material because others will do their work for them. Since your intention in coming here was to get help learning the material, everything is fine (again, you did nothing wrong). I'm just explaining to you why you shouldn't expect full solutions to be provided (and why you wouldn't want them to be). We're all more than happy to guide you through the steps of a problem on this site.

7. Jun 26, 2011

### AceInfinity

Re: Momentum - "Hit and stick" problem solving

Sorry, I understand. Change in momentum is indicated by the delta symbol in my equation, I forgot to note that.

For that, pretend the ">" signs are indicating a new step in the problem. I was solving/proving that the momentum of the motionless player is equal to zero because delta p = mv, and since his velocity is equal to 0, then his momentum must be 0 as well.

Thanks for the extra explanation and the effort taken to write all that detail out though in helping me confirm my understanding of the problem. Sorry for any misconceptions i've raised in what I said. I guess i'll have to take the long route in getting help from others.

Some of what I ask though just needs a quick refresher. I'll understand it all soon enough, to as best I can.

8. Jun 26, 2011

### cepheid

Staff Emeritus
Re: Momentum - "Hit and stick" problem solving

Ahh yes I see. I had interpreted them as "greater than" symbols.

No problem. The way I see it, there are sort of two related ways to solve this problem. The way I first approached it was as follows. Let J2 be the impulse that acted on the second (stationary) hockey player during the collision. I'm going to used the prime symbol (') for quantities after the collision. Now J2 is equal to the player's change in momentum:

J2 = p'2 - p2

= m2v' - 0

You already know v' (the velocity after the collision), so it remains to find m2. To do this, you can use conservation of momentum:

p1 + p2 = p', where p' is the momentum of the combined mass.

m1v + 0 = (m1 + m2)v'

(m1 + m2)/m1 = v/v'

1 + m2/m1 = v/v'

m2/m1 = v/v' - 1

Once you know this ratio, you can then compute m2 = (m2/m1)*m1

As you can see, my method is somewhat cumbersome. fizika_kz's method is simpler. Again, start with the conservation of momentum:

p1 + p2 = p' = p'1 + p'2

m1v + 0 = m1v' + m2v'

m2v' = J2 = m1(v-v')

Done.

Basically I did the extra step of computing m2 separately from m2v', which was unnecessary. If nothing else I guess it shows that if you were asked to find the mass of the second player, you would have been able to.

9. Jun 26, 2011

### AceInfinity

Re: Momentum - "Hit and stick" problem solving

This was the original way that we learned how to do it. Thanks for the second method :) I'll be able to do this kind of problem easily now, the key here is to know that it's the conservation of momentum.

10. Jun 26, 2011

### AceInfinity

Re: Momentum - "Hit and stick" problem solving

I was wondering, what about the problems that involve questions where 2 "objects" rebound off each other, or one stops and the other continues moving after collision. Would it be the same kind of "solution" to finding the momentum of one or the other?

11. Jun 26, 2011

### cepheid

Staff Emeritus
Re: Momentum - "Hit and stick" problem solving

Momentum is always conserved provided that no external forces act on the system. So, the conservation of momentum is the key principle in all such "collision" problems.

The "sticking together" problem is an example of an inelastic collision, so you know that kinetic energy is not conserved.

Sometimes you're given problems in which you're told the collision is perfectly elastic (i.e. kinetic energy is also conserved). So the conservation of momentum and the conservation of kinetic energy give you a set of two equations, and usually in such situations you need both in order to solve the problem (because you have two unknowns). You can actually solve this problem only once (for the 1D case) and from that point onwards just apply the result. But the derivation is kind of a pain (the algebra gets messy even in 1D). What exactly happens (rebound, or one stops and the other starts moving, etc.) depends on the mass ratio of the two objects.