The term mutual velocity, can be described as?

  • Thread starter Thread starter JohnnyB212
  • Start date Start date
  • Tags Tags
    Term Velocity
Click For Summary

Homework Help Overview

The discussion revolves around the concept of "mutual velocity" in the context of a collision between two players in a physics problem. The original poster presents a scenario involving a tackler and a halfback, providing their masses and initial velocities, and seeks to understand the mutual velocity after their collision.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of mutual velocity and its implications in a collision scenario. The original poster questions the use of certain velocities in their calculations and expresses confusion about determining a common velocity after the collision. Others suggest re-evaluating calculations and applying conservation of momentum to understand the situation better.

Discussion Status

The discussion is active, with participants providing guidance on the concept of mutual velocity and the nature of the collision. There is an ongoing exploration of the correct equations and methods to apply, particularly regarding the assumption that the two players move together after the collision.

Contextual Notes

Participants note that the collision is perfectly inelastic, which influences the approach to calculating the mutual velocity. There is also mention of directionality being important in the context of the problem.

JohnnyB212
Messages
14
Reaction score
0
The term "mutual" velocity, can be described as?

Homework Statement



A 110 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 95 kg halfback moving at 7.5 m/s. What will be their mutual velocity in meters/second immediately after the collision?

Homework Equations



m1 = 110kg
V01 = 3.0 m/s
m2 = 95 kg
V02 = 7.5 m/s
V1f = Final Velocity of the 110 kg tackler
V2f = Final Velocity of halfback

V1f = ((m1-m2)/(m1+m2))V01

V2f = ((2m1)/(m1+m2))V01

The Attempt at a Solution



V1f = 1.68
V2f = 24.68Am I correct for both? How would I determine the mutual velocity? I noticed V02 wasnt used, so I'm a little confused, any help would be appreciated, thanks.
 
Physics news on Phys.org


"Mutual" in the sense of "shared in common". The two players stick together after the collision and thus have the same velocity.

Redo your calculations with that in mind.
 


Doc Al said:
"Mutual" in the sense of "shared in common". The two players stick together after the collision and thus have the same velocity.

Redo your calculations with that in mind.

Was the equation correct?
 


Oh! I caught myself on the V1f, which is actually .5897

Still, confused about the mutual portion. How are you to determine the common similarities of two completely different numbers?
 


JohnnyB212 said:
Was the equation correct?
No.

JohnnyB212 said:
Oh! I caught myself on the V1f, which is actually .5897

Still, confused about the mutual portion. How are you to determine the common similarities of two completely different numbers?
Instead of plugging into some formula (which doesn't apply to this situation), why not just apply conservation of momentum?

The collision is perfectly inelastic (they collide and move together). Hint: Direction matters--they are headed toward each other.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
5K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K