Solving Momentum & Energy for Particles: Masses m & 3m

  • Thread starter Thread starter Maybe_Memorie
  • Start date Start date
  • Tags Tags
    Energy Momentum
Click For Summary

Homework Help Overview

The problem involves two particles of different masses, m and 3m, connected by a rigid rod on a smooth horizontal table. The heavier mass is projected with an initial velocity, and the task is to determine the velocities of both particles when the rod is oriented north-south.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation laws, questioning the inclusion of potential energy terms in the energy conservation equation. There is also a debate about the direction of motion and the implications of the rod's orientation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and questioning the assumptions made about the motion of the masses. Some guidance has been offered regarding the conservation equations, but no consensus has been reached.

Contextual Notes

Participants note that the system is constrained to horizontal motion, raising questions about the vertical height gain of the heavier mass and the implications for energy conservation. The initial conditions and the final orientation of the rod are also under scrutiny.

Maybe_Memorie
Messages
346
Reaction score
0

Homework Statement


Two particles of masses m and 3m are connected by a light rigid rod. The system rests on a smooth horizontal table, the heavier mass due east of the lighter mass. The heavier mass is projected with initial velocity uj. Find the velocities of the particles when the rod runs north-south.


Homework Equations


Conservation of momentum, conservation of energy.


The Attempt at a Solution


Applying conservation of momentum in the j direction, 3m(u) + m(0) = 3mv + mw.
3u = 3v + w.

Conservation of energy, 1/2(3m)u2 = 3mgh +1/2(3m)v2
u2 = 2gh + v2

I've no idea where to go from here.
This isn't homework, just a question I came across while revising for exams.
 
Last edited:
Physics news on Phys.org
Hi Maybe_Memorie! :smile:

(try using the X2 tag just above the Reply box :wink:)
Maybe_Memorie said:
… The system rests on a smooth horizontal table

Conservation of energy, 1/2(3m)u^2 = 3mgh +1/2(3m)v^2
u^2 = 2gh + v^2

erm :redface: … leave out the mgh ! :biggrin:
 
tiny-tim said:
erm :redface: … leave out the mgh ! :biggrin:

Why? The 3m mass gains vertical height.
 
Maybe_Memorie said:
Why? The 3m mass gains vertical height.

No …
Maybe_Memorie said:
Two particles of masses m and 3m are connected by a light rigid rod. The system rests on a smooth horizontal table, the heavier mass due east of the lighter mass. The heavier mass is projected with initial velocity uj. Find the velocities of the particles when the rod runs north-south.
… everything's horizontal.
 
tiny-tim said:
No …

… everything's horizontal.

The 3m mass is projected with a velocity uj, i.e. a velocity of u in the j-> direction, upwards.
 
Maybe_Memorie said:
The 3m mass is projected with a velocity uj, i.e. a velocity of u in the j-> direction, upwards.

Where I live, that's the k direction. :wink:

Anyway, if the rod starts east-west, and ends up north-south (as in the question), how can it ever move upwards??
 
tiny-tim said:
Where I live, that's the k direction. :wink:

Anyway, if the rod starts east-west, and ends up north-south (as in the question), how can it ever move upwards??

It's like the large mass moves in a circle around the small mass. So the 3m mass gains height in rotating.

It may be possible to treat it as a compound pendulum.
 
Any suggestions?
 

Similar threads

Replies
23
Views
2K
Replies
21
Views
3K
Replies
10
Views
3K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
55
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
10
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K