Compressed Spring Between Two Boxes

  • Thread starter Thread starter Soniteflash
  • Start date Start date
  • Tags Tags
    Compressed Spring
Click For Summary
In a scenario with two blocks of masses M and 2M on a frictionless surface, a compressed spring between them is released, causing the block of mass 2M to move with velocity v. The discussion centers on determining the velocity of the center of mass of the system after the release. It is concluded that the center of mass remains stationary, resulting in a velocity of zero. This is supported by the principle of conservation of linear momentum, which indicates that the total momentum before and after the release remains constant. Understanding the relationship between the blocks' movements and the center of mass is crucial for solving the problem accurately.
Soniteflash
Messages
36
Reaction score
1

Homework Statement


Two blocks of masses M and 2M are on a frictionless horizontal surface and are held in place with a compressed spring of negligible mass between them. If the blocks are then released and the block of mass 2M leaves the spring with a velocity v, the velocity of the center of mass of the blocks is

A zero
B -(v/2)
C -(2v/3)
D -(3v/2)
E -2v

Homework Equations


n/a ?

The Attempt at a Solution


I do not understand how the center of mass relates to the movement of the blocks. I remember from class that if an object exploded into thousands of pieces the center of mass would still be at the original position.
So I assume it is A. Zero.
 
Physics news on Phys.org
Soniteflash said:
I do not understand how the center of mass relates to the movement of the blocks. I remember from class that if an object exploded into thousands of pieces the center of mass would still be at the original position.
So I assume it is A. Zero.

Can you explain why it would be zero? I mean, you are right. But it might be beneficial to examine the problem a little more deeply.
 
AlephNumbers said:
Can you explain why it would be zero? I mean, you are right. But it might be beneficial to examine the problem a little more deeply.
I am not quite sure but I thought it is similar or identical sort of to an explosion.
 
It is very similar to an explosion. Try using conservation of linear momentum and the equation for the velocity of the center of mass of a system of particles (or blocks, in this case) to show why your answer is correct.
 
For linear momentum :
Pi=Pf
0 = m(-2v) + (2mv)
0 = -2mv + 2mv
0 = 0
Linear momentum is conserved.

Hmm, I don't really know an equation for the velocity of the center of mass of a system of particles. I know that the velocities of the boxes are both in opposite directions.
 
It isn't too difficult to prove.

Start with the equation for the center of mass of a system of particles. If you don't know that either, you can find it here http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html.

Any ideas on what you can do to this equation to create an equation for the velocity of the center of mass?
It involves calculus.
 
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?

Similar threads

Replies
17
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
29
Views
3K
Replies
8
Views
6K
Replies
30
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K