Elastic Collision and Momentum of Ice Skaters

[lpettigrew]

Homework Statement

Hello, I have come across the question below and despite having rather a firm grasp of the topic I am a little confused and would appreciate any help.

There are two figure skaters who are moving together at 10 m/s in a straight line. The first skater pushes the other skater who continues on at 12 m/s.
The mass of skater 1 is 80kg, the mass of skater 2 is 60kg.
1. What is the final speed of the first skater?
2. Is this an elastic or inelastic collision? If the collison is not elastic, where has the additional energy come from or been lost to?

I would be very grateful if anyone could check my calculations and suggest any improvements to my method. I think it may be clear that I am a little uncertain in myself here.

Relevant Equations

p=mv
pbefore=pafter

1. Hello, so the difficulty I am having with this problem is that is seems relatively straightforward. I have tried to solving it by assuming that this is a collision in which momentum is conserved. Therefore, I found the total momentum before the collision and used this to resolve it must be the same as the total momentum after the collision i.e. momentum before= momentum after.
Thus, by rearranging the equation I was able to isolate the final velocity of the first skater as below:

p total before= (m1*u1)+(m2*u2)
p total before =(80*10)+(60*10)
p total before= 800 kg ms^-1 +600kgms^-1
p total before = 1400 kgms^-1

Thus, 1400 kgms^-1 before = 1400kgms^-1 after
Since skater 1 is traveling in the opposite direction to skater 2 after they separate, the negative value for the momentum and final velocity of skater 1 shows that they are moving in the opposite direction. Since both momentum and velocity are vector quantities it is important to consider magnitude and direction.
p total after=(m1*v1)+(m2*v2)
p total after=(80*v1)+(60*12)
p total after=(80*v1)+(720kgms^-1)
1400kgms^-1=(80*v1)+(720kgms^-1)
-680kgms^-1=80*v1
v1=-680/80=-8.5ms^-1 in the opposite direction to skater 2. 2. This is where I am having most difficulty in conclusively identifying whether the collision is elastic or inelastic.
Would I evaluate whether the kinetic energy in the collision is conserved to solve this?
For example if the collison is elastic then KE before=KE after
Before the collision;
KE1=1/2m1u1^2=1/2*80*10^2=4000J
KE2=1/2m2u2^2=1/2*60*10^2=3000J
KE total before=(1/2m1u1^2)+(1/2m2u2^2)=7000J

After collision;
KE1=1/2m1v1^2=1/2*80*-8.5^2=2890J
KE2=1/2m2v2^2=1/2*60*12^2=4320J
KE total after=(1/2m1v1^2)+(1/2m2v2^2)=7210J

Since 7000J does not equal 7210J it is clear (providing my calculations are correct) that 210J of kinetic energy have been gained, therefore, this is an inelastic collision. But where would the energy have been gained from? Stored at elastic potential energy and released when the first skater pushes the second?

I am greatly appreciate anyone who replies

 

[kuruman]

I would say the energy is stored as biochemical energy in the muscles of the pushing skater who probably had a good breakfast before hitting the ice rink. Muscles generate power on demand.

Check your algebra. The final velocity of the first skater has the incorrect sign. Everything else is OK.

 

[lpettigrew]

I would say the energy is stored as biochemical energy in the muscles of the pushing skater who probably had a good breakfast before hitting the ice rink. Muscles generate power on demand.

Check your algebra. The final velocity of the first skater has the incorrect sign. Everything else is OK.

Thank you very much for your reply. Ah yes biochemical energy sounds far more senible, thank you for the advice. Should the final velocity of the first skater be positive and of the second skater be negative?

 

[kuruman]

Thank you very much for your reply. Ah yes biochemical energy sounds far more senible, thank you for the advice. Should the final velocity of the first skater be positive and of the second skater be negative?

You have defined the positive direction as that of the initial velocity 10 m/s. The problem tells you that skater 2 moves on at 12 m/s. This means the velocity of skater i1 s whatever is necessary to satisfy the momentum conservation equation. If it has a positive sign, skater 1 also moves on in the same direction; if it has a negative sign skater 1 moves in the opposite direction. You cannot predict the sign before substituting the numbers.

 

[lpettigrew]

You have defined the positive direction as that of the initial velocity 10 m/s. The problem tells you that skater 2 moves on at 12 m/s. This means the velocity of skater i1 s whatever is necessary to satisfy the momentum conservation equation. If it has a positive sign, skater 1 also moves on in the same direction; if it has a negative sign skater 1 moves in the opposite direction. You cannot predict the sign before substituting the numbers.

Oh right, thank you for your reply. Your statement has been tremendously helpful and I appreciate your explanation. So since skater 2 continues in the original direction (which we take to be positive) and hence the direction skater 1 travels in, opposite skater 2, would be negative?
So, should I have written;
p total after=(m1*v1)+(m2*v2)
p total after=-(80*v1)+(60*12)
p total after=-(80*v1)+(720kgms^-1)
1400kgms^-1=-(80*v1)+(720kgms^-1)
680kgms^-1=-80*v1
-v1=680/80=-8.5ms^-1 in the opposite direction to skater 2.
Sorry I have confused myself a little

 

[PeroK]

I'm struggling to make sense of this. $10m/s$ is pretty fast ($36 kmph$). That's a helluva collision between two people. Not only that, one of the skaters comes out of the collision at $12 m/s$, which is $43.2 kmph$.

There's no way you could project someone back at those speeds. It can't be a head-on collision.

Instead, I suggest that skater 1 pushes skater 2 in the back as they pass, increasing skater 2's speed from $10m/s$ to $12m/s$ in the same direction. Meanwhile, skater 1 also accelerates (in their original direction). The question says:

The first skater pushes the other skater who continues on at 12 m/s.

It doesn't say that skater 1 pushes skater 2 back in the opposite direction. That would be absurd.

 

[kuruman]

Oh right, thank you for your reply. Your statement has been tremendously helpful and I appreciate your explanation. So since skater 2 continues in the original direction (which we take to be positive) and hence the direction skater 1 travels in, opposite skater 2, would be negative?
So, should I have written;
p total after=(m1*v1)+(m2*v2)
p total after=-(80*v1)+(60*12)

Why are you replacing the mass of skater 1 with -80 kg? What does negative mass even mean?

 

[lpettigrew]

Why are you replacing the mass of skater 1 with -80 kg? What does negative mass even mean?

Sorry right, mass is scalar and can only have magnitude. I think I was just trying to satisfy the momentum conservation equation in a way I could understand. My mistake.

 

[lpettigrew]

I'm struggling to make sense of this. $10m/s$ is pretty fast ($36 kmph$). That's a helluva collision between two people. Not only that, one of the skaters comes out of the collision at $12 m/s$, which is $43.2 kmph$.

There's no way you could project someone back at those speeds. It can't be a head-on collision.

Instead, I suggest that skater 1 pushes skater 2 in the back as they pass, increasing skater 2's speed from $10m/s$ to $12m/s$ in the same direction. Meanwhile, skater 1 also accelerates (in their original direction). The question says:
It doesn't say that skater 1 pushes skater 2 back in the opposite direction. That would be absurd.

Thank you for your reply. No, I had taken skater 1 and 2 to intially be traveling in the same direction together and then for skater 2 to be pushed by skater 1 and continue in this direction. I thought upon speration this would cause skater 1 to move in the opposite direction, would I be wrong?

 

[kuruman]

I'm struggling to make sense of this. $10m/s$ is pretty fast ($36 kmph$). That's a helluva collision between two people. Not only that, one of the skaters comes out of the collision at $12 m/s$, which is $43.2 kmph$.

There's no way you could project someone back at those speeds. It can't be a head-on collision.

Instead, I suggest that skater 1 pushes skater 2 in the back as they pass, increasing skater 2's speed from $10m/s$ to $12m/s$ in the same direction. Meanwhile, skater 1 also accelerates (in their original direction). The question says:
It doesn't say that skater 1 pushes skater 2 back in the opposite direction. That would be absurd.

The problem states that "There are two figure skaters who are moving together at 10 m/s in a straight line." They could be holding hands as they do so. I agree that 10 m/s is a bit fast, but if you think that the world record for the 100 m dash is now a bit under 10 s for an average speed of about 10 m/s, that figure is not unreasonably excessive.

 

[kuruman]

I thought upon speration this would cause skater 1 to move in the opposite direction, would I be wrong?

That's exactly where you are wrong. As I said earlier, conserve momentum and whatever comes out of the equation is it. You cannot insert bias into the problem this way.

 

[PeroK]

Thank you for your reply. No, I had taken skater 1 and 2 to intially be traveling in the same direction together and then for skater 2 to be pushed by skater 1 and continue in this direction. I thought upon speration this would cause skater 1 to move in the opposite direction, would I be wrong?

Ah, okay, I took "moving together" to be mean on a collision course. That makes sense now.

 

[PeroK]

The problem states that "There are two figure skaters who are moving together at 10 m/s in a straight line." They could be holding hands as they do so. I agree that 10 m/s is a bit fast, but if you think that the world record for the 100 m dash is now a bit under 10 s for an average speed of about 10 m/s, that figure is not unreasonably excessive.

Yes, moving as a pair! It needed a diagram. Since it was a collision, I assumed they were on a crash course.

 

[kuruman]

Ah, okay, I took "moving together" to be mean on a collision course. That makes sense now.

Now I understand your post #6. I prefer "moving as one" to describe the same situation.

 

[lpettigrew]

@kuruman and @PeroK thank you for both of your replies. Ah, I see would my error be assuming that after separation they continue in different directions? So in actuality if they just separate then the momentums of skater 1 and 2 are both positive and this removes my misplaced negative sign.

So; question 1:
p total after=(m1*v1)+(m2*v2)
p total after=(80*v1)+(60*12)
p total after=(80*v1)+(720kgms^-1)
1400kgms^-1=(80*v1)+(720kgms^-1)
680kgms^-1=80*v1
v1=680/80=8.5ms^-1

 

[kuruman]

That's the correct answer. It is entirely possible for two objects to be moving in the same direction, collide and keep moving in the same direction after the collision. In that case the trailing object must be moving faster than the leading object before the collision otherwise there will be no collision. After the collision, the leading object will have picked up forward velocity while the trailing object will have lost some.

 

[PeroK]

In addition, it's (always) worth looking at this from the centre of momentum frame. In that frame the skaters are initially at rest, then skater 1 pushes skater 2 away at $2m/s$ and recoils at $1.5m/s$.