Momentum problem!

  • Thread starter BlasterV
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  • #1
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One sport at the Winter Olympics is short-track speedskating. The relay event features teams of four skaters in a long race. Each team member typically skates one or two laps, then passes the race to another member. Instead of handing the teammate a baton, however, a skater must simply touch the teammate. Skaters are trained to collide with the next skater, in order to give him or her a big push.

Consider this year's US Speedskating Women's relay team. Caroline Hallisey is just finishing her lap, moving at 11.17 m/s, and runs into Alyson Dudek, who is at rest. If Caroline stops dead, how fast will Alyson move forward?

14.4167594178 m/s - Correct.

(Yes, you need to know their masses. Go look them up. If you can't find them after 10 minutes of searching, let me know. You should find that Caroline's mass is larger than Alyson's mass).
Caroline is 60.3kg - Correct.
Alyson is 46.72kg - Correct.

Suppose that Caroline is again moving at 11.17 m/s, and again preparing to push Alyson. This time, however, Caroline pushes Alyson forward as hard as she can during their brief encounter: Caroline exerts an average force 1200 N on Alyson over a duration of .75 seconds.

Now how fast does Alyson leave the starting line after the collision?

22.8264024752 m/s - Correct.

What is Caroline's velocity after the collision?

Here is where it goes to hell, I done this over 6 different ways and got it wrong, and because of that I really can't show work because it wouldn't really be going anywhere (I am at the point where I have no idea whats wrong). I tried to use the Conservation of momentum equations and it didn't work for me.

Can someone show me how to do this part step by step and give an answer (so after I solve I know if I'm agreeing with the right answer)

Thanks alot.
 

Answers and Replies

  • #2
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Hi BlasterV,

Are you sure that your answer of 22.83m/s for Alyson's velocity is correct? After the following calculation, I got a different answer:

Impulse = Change in Momentum = Force * Time

I = Ft

I = 1200 * 0.75 = 900Ns

This is Alyson's momentum after being pushed by Caroline, i.e.

mv = 900Ns

v = 900 / 46.72 = 19.26m/s

Then by conservation of momentum;

m1u1 = m1v1 + m2v2

where m1 is Caroline's mass, u1 is Caroline's initial velocity, v1 is her final velocity, m2 is Alyson's mass and v2 is Alyson's final velocity

60.3 * 11.17 = 60.3 * v1 + 900

60.3 * v1 = -226.449

v1 = -3.755m/s

i.e. Caroline moves at 3.755m/s in the opposite direction to Alyson

An alternative way of considering this second part is to use Newton's third law (every action has an equal and opposite reaction). Therefore the force acting on Caroline is also 1200N for 0.75s and the change in her momentum is -900Ns (negative since it is in the opposite direction). Then her final momentum is:

11.17 * 60.3 - 900 = -226.449

mv = -226.449

v = -3.755m/s

Hope that helped (and also that I haven't made some mistake :smile: )

Chris
 
  • #3
NateTG
Science Advisor
Homework Helper
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Answer 1: There is not enough information. Caroline will be pushing off on the ground while pushing on Alyson.

Answer 2:
Conservation of momentum:
[tex]m_{a}v_{a0}+m_{c}v_{c0}=m_{a}v_{a1}+m_{c}v_{c1}[/tex]
Fill in the known quantities:
[tex]m_{a},m_{c},v_{a0},v_{c0},v_{c1}[/tex]
Using rough values:
[tex]40kg(0\frac{m}{s})+60kg(10\frac{m}{s})=40kg(20\frac{m}{s})+60kg(v_{c1})[/tex]
simplify and solve
[tex]600\frac{mkg}{s}=800\frac{mkg}{s}+60kg v_{c1}[/tex]
[tex]-\frac{200\frac{mkg}{s}}{60kg}=v_{c1}[/tex]
[tex]v_{c1} \approx -3 \frac{m}{s}[/tex]
 
  • #4
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Chris, I submit homework to an online thing that says if its right or not, 22.8 etc is definitley right, but I'll try using my number with what you did
 
  • #5
Doc Al
Mentor
44,945
1,216
BlasterV said:
Suppose that Caroline is again moving at 11.17 m/s, and again preparing to push Alyson. This time, however, Caroline pushes Alyson forward as hard as she can during their brief encounter: Caroline exerts an average force 1200 N on Alyson over a duration of .75 seconds.

Now how fast does Alyson leave the starting line after the collision?

22.8264024752 m/s - Correct.
No. How did you come up with this answer?

Chris's answers are correct.

Making the assumption that during the collision of the two skaters the only forces that you need consider are what they exert on each other, then you can find the change in velocity by considering the impulse:
Impulse on Alyson: [itex]F \Delta T = m_A \Delta v_A[/itex], solve for [itex]\Delta v_A[/itex].
What is Caroline's velocity after the collision?
The impulse on Caroline is equal and opposite to the impulse on Alyson:
[itex]-F \Delta T = m_C \Delta v_C[/itex], solve for [itex]\Delta v_C[/itex].
 

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