Inelastic collision in opposite directions: KE lost & angular momentum

Click For Summary

Homework Help Overview

The problem involves an inelastic collision between two boys sliding toward each other on a frictionless surface, focusing on kinetic energy loss and angular momentum after they lock arms and rotate about their common center of mass.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of kinetic energy before and after the collision, questioning the discrepancy between their results and the expected answer. They also explore the calculation of angular speed and the rotational inertia involved in the scenario.

Discussion Status

Some participants have provided arithmetic checks and clarification on the application of formulas, while others are seeking further understanding of the concepts involved, particularly regarding the rotational inertia and the common center of mass. There is an acknowledgment of the correct value for velocity derived from momentum conservation, but no consensus on the overall approach has been reached.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The original poster expresses a need for clear, step-by-step guidance to understand the problem better.

vaxxa
Messages
1
Reaction score
0
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.
 
Last edited:
Physics news on Phys.org
vaxxa said:
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?
 
Where did v3 = 0.361 m/s come from?
 
rude man said:
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.
 
haruspex said:
The working was not shown, but I confirm it is correct from momentum conservation in part a.

I should have realized that. Thanks.
 

Similar threads

Replies
9
Views
3K
Replies
26
Views
2K
Replies
335
Views
17K
Replies
10
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
14
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K