Asked by ME
At the center of a 41.0 m diameter circular ice rink, a 61.0 kg skater traveling north at 1.05 m/s collides with and holds onto a 77.0 kg skater who had been heading west at 1.90 m/s. How long (in seconds) will it take them to reach the edge of the rink assuming negligible friction and drag?
Answers
Answered by
bobpursley
use conservation of momentum to find the new momentum.
New momentum: 61*1.05N+77*1.90W
absolute speed: sqrt [(61*1.05)^2 + (77*1.9)^2 ]
then time= distance/speed= 41/2 * 1/speed
New momentum: 61*1.05N+77*1.90W
absolute speed: sqrt [(61*1.05)^2 + (77*1.9)^2 ]
then time= distance/speed= 41/2 * 1/speed
Answered by
ME
I did this and it still says it is incorrect...
Answered by
bobpursley
I bet you made an error. Do you see anything wrong with the approach, or thinking? Have you checked your work? Post it here, and someone can critique it.
Answered by
ME
I used the equation m1vf1 + m2vf2 = m1vi1 + m2vi2 and got a vf of 1.52m/s
I then plugged that into the equation t=d/v => t=20.5/1.52 and got 13.5s
I then plugged that into the equation t=d/v => t=20.5/1.52 and got 13.5s
Answered by
bobpursley
Is there some reason you did not do what I told you? What you did is really very wrong. Please look at the solution I outlined. And I see I have an error, where I typed absolute speed, it should read absolute momentum, then a new line
absolute speed= absolute mometnum/total mass I get about 1.1 for the speed (not 1.1).
absolute speed= absolute mometnum/total mass I get about 1.1 for the speed (not 1.1).
Answered by
ME
Yeah I did do it the way you told me at first and got a number that made absolutely no sense but the way you just illustrated makes more sense....thank you
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