Inelastic collision in opposite directions: KE lost & angular momentum

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1. The problem statement.
Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at 7.98 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.7 m/s along the same line. When they meet, they grab each other and hang on.
a.) What fraction of their original kinetic energy is still mechanical energy after their collision?

That was so much fun that the boys repeat the collision with the same original velocities, this time moving along parallel lines 1.10 m apart. At closest approach, they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch.
b.) What is their angular speed?

2. Homework Equations .
a.) KEinitial = 1/2m1v1^2 + 1/2m2v2^2
KEfinal = 1/2(m1+m2)v3^2
b.) w = L/I
L =mvr
I =mr^2

3. Attempt at solution.
a.)1/2(45)(7.98)^2 + 1/2(31)(10.7)^2=3,207.41J
1/2(45+31)(0.361)^2=0.901J
0.901J/3,207.41J = 2.81 x 10^-4, but the given answer is 0.154. I've absolutely no clue how they arrived at that. I need to know how to solve it for my quiz.

b.)
I know w=L/I, but I don't know how that will help. I'm confused as to how to find the rotational inertia (I), among other things. I believe it's mr^2. Using L=mvr and I=mr^2, I get:
L=(76kg)(0.361m/s)(.55m)^2=15.09
I=76kg(.55m)^2=23
so when 15.09/23=0.66, but the given answer is 17! How in the world do I find that? Clearly written, step-by-step directions would be greatly appreciated. Thank you.
 
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Answers and Replies

  • #2
haruspex
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1/2(45+31)(0.361)^2=0.901J
Check the arithmetic
L=(76kg)(0.361m/s)(.55m)^2=15.09
The 0.361m/s is for part a. It does not apply to part b.
For part b, consider the line taken by their common mass centre before they meet. How far is each boy from that? What is their combined moment about a point on that line?
 
  • #3
rude man
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Where did v3 = 0.361 m/s come from?
 
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haruspex
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Where did v3 = 0.361 m/s come from?
The working was not shown, but I confirm it is correct from momentum conservation in part a.
 
  • #5
rude man
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The working was not shown, but I confirm it is correct from momentum conservation in part a.

I should have realized that. Thanks.
 

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