Conservation of Momentum and Energy

Click For Summary
The discussion focuses on the conservation of linear and angular momentum in a physics problem involving a block and clay. The key equations for momentum conservation are presented, indicating that the total momentum before and after the impact should be equal. The participant expresses uncertainty about whether their understanding of momentum conservation is correct, particularly regarding the potential energy change as the system moves. However, it is clarified that potential energy considerations are separate from momentum conservation. The conversation emphasizes the importance of distinguishing between momentum and energy in analyzing the system's dynamics.
ryankunzzz
Messages
21
Reaction score
0

Homework Statement


media%2F655%2F6552a508-d11e-4c8c-9c6e-5e12276b9d7f%2Fphp4Rr6Au.png


Homework Equations


Conservation of liner and angular momentum
H1=H2
r x M1V1= r x M2V2

G1=G2
M1V1=M2V2

The Attempt at a Solution


I've gotten an equation for the velocity of the block and clay after the impact, and I also know that when the block reaches the highest point of its arc, its linear velocity is going to be the same as the velocity of the cart. Is the momentum going to be conserved and I can say MbV1=Vtot(M1+M2+M3)
 
Physics news on Phys.org
ryankunzzz said:
Is the momentum going to be conserved and I can say MbV1=Vtot(M1+M2+M3)
It seems like that would be true. On the other hand, I'm thinking, "It seems too easy. Am I missing something?" But I can't think of what that might be.
 
ryankunzzz said:
Is the momentum going to be conserved and I can say MbV1=Vtot(M1+M2+M3)
Yes, and I believe that is where the hint is directed.
 
  • Like
Likes TomHart
But what about the about the change in potential energy as the arm swings upward?
 
ryankunzzz said:
But what about the about the change in potential energy as the arm swings upward?
What about it? That's energy, not momentum.
 

Thread 'Magnitude of buoyant force in fluids of different densities'

The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?
View full post »

Similar threads

Inelastic collisions -- how is momentum conserved but not energy?
  • · Replies 5 ·
Replies
5
Views
3K
Velocity from spring pushing on 2 masses
  • · Replies 3 ·
Replies
3
Views
1K
Finding the Line of Motion of Two Particles
  • · Replies 16 ·
Replies
16
Views
1K
Momentum conservation: block-wedge problem
  • · Replies 4 ·
Replies
4
Views
2K
How do I solve for an elastic collision going up a ramp?
  • · Replies 10 ·
Replies
10
Views
3K
Using conservation of energy and momentum in projectile motion
  • · Replies 18 ·
Replies
18
Views
2K
Conservation of Mechanical Energy - Which Equation To Use?
  • · Replies 9 ·
Replies
9
Views
1K
Confusion with "explosive" part, Conservation of Momentum
  • · Replies 6 ·
Replies
6
Views
2K
Help with Final Velocity Collision Problem
  • · Replies 1 ·
Replies
1
Views
2K
Elastic collision of block on block
  • · Replies 4 ·
Replies
4
Views
2K