Ice Skater Momentum problem

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  • #1
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I'm kinda having trouble with this problem. can n e one help me out?

Two ice skaters undergo a collision, after which their arms
are intertwined and they have a common velocity of 0.85 m/s
[27° S of E]. Before the collision, one skater of mass 71 kg
had a velocity of 2.3 m/s [12° N of E], while the other skater
had a velocity of 1.9 m/s [52° S of W]. What is the mass of
the second skater?

I tried to use the law of conservation of momentum and maybe draw a vector diagram, but i only have one momentum that i can actually calculate now. How can i do this (preferably without using components)?
 

Answers and Replies

  • #2
HallsofIvy
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On the contrary, you have THREE momentum vectors! Yes, two of them depend on the unknown mass but surely that wont bother you. Call the unknown mass "m".

"one skater of mass 71 kg had a velocity of 2.3 m/s [12° N of E]"
Okay, that skater's momentum vector is <71*2.3*cos(12),71*2.3*sin(12)> where I am taking the positive x-axis East and the positive y-axis North.

"the other skater had a velocity of 1.9 m/s [52° S of W]". This is the one with mass m so the momentum vector is <m*1.9*cos(52), m*1.9*sin(52)>

After the collision, "they have a common velocity of 0.85 m/s [27° S of E]" and their total mass is 71+m kg so their common momentum vector is <(71+m)*.85*cos(27), (71+m)*.85*sin(27)>.

"Conservation of momentum" means that the sum of the first two must equal the third:
<71*2.3*cos(12)+m*1.9*cos(52),71*2.3*sin(12)+m*1.9*sin(52)>= <(71+m)*.85*cos(27), (71+m)*.85*sin(27)>.

That really gives two equations for the single unknown, m:
71*2.3*cos(12)+m*1.9*cos(52)= (71+m)*.85*cos(27)
71*2.3*sin(12)+m*1.9*sin(52)= (71+m)*.85*sin(27)
so the problem is really "over-determined". You should be able to solve either of these equations for m and then check in the other one. If a value of m satisfies one equation but not the other, this is a BAD problem- it's physically impossible.
 
  • #3
arildno
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"If a value of m satisfies one equation but not the other, this is a BAD problem- it's physically impossible."
I would say that from a pedagogical side, this is a BAD problem, even if it happens to work out.
I do not think it is a good idea to make beginning students work on possibly overdetermined systems.
 
  • #4
ehild
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HallsofIvy said:
"one skater of mass 71 kg had a velocity of 2.3 m/s [12° N of E]"
Okay, that skater's momentum vector is <71*2.3*cos(12),71*2.3*sin(12)> where I am taking the positive x-axis East and the positive y-axis North.

"the other skater had a velocity of 1.9 m/s [52° S of W]". This is the one with mass m so the momentum vector is <m*1.9*cos(52), m*1.9*sin(52)>
Sorry, I do not know what [52° S of W] means. Does it mean that the velocity makes the angle of 52° with the negative x axis in the terms that the positive x-axis points to East and the positive y-axis points to North? If so, than both px and py should be negative, shouldn't they?


After the collision, "they have a common velocity of 0.85 m/s [27° S of E]" and their total mass is 71+m kg so their common momentum vector is <(71+m)*.85*cos(27), (71+m)*.85*sin(27)>.
Not <(71+m)*.85*cos(27), - (71+m)*.85*sin(27)> instead?

With these changes, the numbers are about all right, but I agree that such a problem is a bad problem for the students.

ehild
 
  • #5
53
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Thanx a lot. It's still kinda cloudy, but u cleared it up a lot
 

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