Conservation of linear momentum

In summary, a question was asked about the conservation of linear momentum in a problem involving two blocks with springs attached. The law of conservation of linear momentum still holds despite the external forces acting on the blocks. This is because the interaction between the blocks happens so quickly that the springs are unchanged during the collision. However, if the blocks are made of a material that can deform, this assumption may not hold. In the case where there is only one spring attached horizontally to one of the blocks, linear momentum would not be conserved. The pool ball example mentioned also demonstrates how momentum can be conserved in a collision with external forces. Ultimately, the time interval during which the blocks are in contact is negligible and the collision can be considered as a system
  • #1
peeks
3
0
Hi everyone,

I had a doubt about conservation of linear momentum while doing the following question(I managed to get the right answer but am not sure why my method worked). A bmp file of the problem has been attached.

Both blocks shown are confined to move in the horizontal slot. Block B has 2 springs attached to it. Initially, block A has a velocity V and block B is at rest. Also, each spring has a stiffness constant 'k' and an initial extension of dx. Coefficient of restitution, 'e' is 0.7. Friction is negligible.

I applied conservation of linear momentum to the collision between block A and B in order to find the velocity of B after the impact. What I don't understand is how the law still holds despite the external force acting on block B? I know the net external impulse on the system must be zero in order for the initial and final momentum to be equal but the presence of the spring forces has confused me a bit.

I would greatly appreciate some clarification regarding this.


Thanks!


pk
 

Attachments

  • problem.bmp
    26.4 KB · Views: 480
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  • #2
You make the assumption that the interaction of the two blocks happens so quickly (over such a short distance) that the springs are unchanged during the collision. It is a very good assumption if the blocks are made of hard material, less good but still not bad if they are able to deform like lumps of clay.
 
  • #3
ah! So we assume that the springs remain vertical during the collision and the spring forces cancel out? The law only specifies that the net external impulsive force must be zero. Suppose, instead of two springs attached perpendicular to the block B, there was only one spring attached horizontally to it(refer attached pic). Linear momentum wouldn't be conserved then, right?
 

Attachments

  • problem_mod.bmp
    26.4 KB · Views: 468
  • #4
You would make the same assumption. The collision happens so fast the spring is not a factor in the collision process. The forces involved in the collision are enormous, and are gone in an instant.
 
  • #5
peeks,

Linear momentum is always conserved!

Older Dan is mistaken; you can't assume that the interaction time between the two blocks is very short when B is attached to a spring (or to two springs, as in your original problem).

To figure out what's going on here, think about how momentum can be conserved when a pool ball hits a rail perpendicularly and bounces back. The ball's momentum has gone from mv to -mv, a change of -2mv. Where is the +2mv that cancels this out to conserve momentum?
 
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  • #6
hmmm ok I think I understand now. Thanks for the help guys!
 
  • #7
jdavel said:
peeks,

Linear momentum is always conserved!

Older Dan is mistaken; you can't assume that the interaction time between the two blocks is very short when B is attached to a spring (or to two springs, as in your original problem).

To figure out what's going on here, think about how momentum can be conserved when a pool ball hits a rail perpendicularly and bounces back. The ball's momentum has gone from mv to -mv, a change of -2mv. Where is the +2mv that cancels this out to conserve momentum?

Sorry, but you are quite wrong. Linear momentum is only conserved in a system in the absence of external forces. Your pool ball problem can be considered from two different points of view. On the one hand, the table and the Earth to which it is connected are the system and the forces acting on each are an internal action-reaction pair that does not alter the momentum of the system. The 2mv is the recoil momentum of the table that is imperceptible because of the huge mass. Actually, since the force is not directed through the center of mass of the earth, there is a torque and a change in angular momentum that is imperceptible, but there is no need to explore that complication.

From another point of view, the pool ball is subjected to an external force provided by the table. Its change in linear momentum is the result of an implulse delived to it by the rail of the table. We don't know the details of the force and the time involved but we do know that the integral of the force over the contact time (the impulse) is the change in momentum of the ball. The pool table/earth experiences an equal impulse in the opposite direction.
 
  • #8
OlderDan,

I apologize for saying you were mistaken! I should have said that I didn't understand what you meant.

It seemed to me you were saying that if the collision is elastic then you can assume the two masses are only in contact for an instant. Is that what you meant? I don't think that's true.
 
  • #9
One more try...

The time interval during which A and B are in contact is only negligible if the mass of A is less than or equal to that of B.
 
  • #10
jdavel said:
OlderDan,

It seemed to me you were saying that if the collision is elastic then you can assume the two masses are only in contact for an instant. Is that what you meant? I don't think that's true.

The collision is not elastic in the original problem. It is stated that the coefficient of restitution is 0.7. Nevertheless, the two masses are only in contact for an instant. The important thing is that to a very good approximation there is no translation to stretch the springs (or compress the one spring in the second variation) while the collision is taking place. The second block is on its way with the momentum it acquired from the collision before there is any stretching of the springs that needs to be considered. The collision of the two blocks may be considered as a system of two masses with no external forces acting, so linear momentum will be conserved.

After the collision the first mass will recoil if it is lighter than the second, and the stretching of the springs will provide a force that will eventually stop the translation of the second block and bring it back to its original position. In the absence of friction, the second block will be in periodic motion (not harmonic if the displacement is large) about its original position.

If the two blocks have the same mass and if the collision were elastic, the first block would stop dead until the second block comes back and hits it

If the first block is heavier than the second block, it will keep moving toward the second block after the intial impact and eventually catch it again. There will be several impacts before the springs finally turn everything around. In the secondary impacts there will be an unbalanced force due to the springs, but the collision time involved is so short and the contact forces so high that the spring forces are of little consequence until the relative velocities of the two blocks gets small and the implulse forces get weak.
 
  • #11
jdavel said:
One more try...

The time interval during which A and B are in contact is only negligible if the mass of A is less than or equal to that of B.

Not true. The contact time is a function of the elasticity of the materials. There will however be a point in time where both blocks come to rest with the springs strecthed. After several impacts the springs will not be negligable.
 
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  • #12
OlderDan,

You are absolutely right! And nicely explained. In the future, I'll think more carefully before questioning your posts.

jdl
 
  • #13
jdavel said:
OlderDan,

You are absolutely right! And nicely explained. In the future, I'll think more carefully before questioning your posts.

jdl

Don't stop questioning! Sometimes our intuition leads us down the wrong path and analysis is needed to bring us back. Your intuition was not all bad on this. As the springs gets stretched if both blocks are moving in the same direction eventually the springs become important.

My statement about both masses eventually being at rest was not totally accurate, but at the time I needed to get away. Depending on the ratio of the masses and the coefficient of restitution and all the details, there could be a secondary collision after the second block turns around that reverses the direction of the first block. You would have to work out all the details to see if the spring force ever became an issue. It's more likely that it never will be because all the collisions will be at high enough relative velocity to ignore them.
 

Related to Conservation of linear momentum

1. What is the law of conservation of linear momentum?

The law of conservation of linear momentum states that the total momentum of a closed system remains constant, unless acted upon by an external force. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction.

2. How is linear momentum calculated?

Linear momentum is calculated by multiplying an object's mass by its velocity. Mathematically, it can be represented as p = mv, where p is momentum, m is mass, and v is velocity.

3. What is an example of conservation of linear momentum in action?

An example of conservation of linear momentum is a game of pool. When the cue ball strikes the other balls, the momentum is transferred from the cue ball to the other balls, causing them to move. The total momentum of the system remains constant, even though individual ball velocities may change.

4. How does conservation of linear momentum relate to Newton's Third Law?

Conservation of linear momentum is closely related to Newton's Third Law, which states that for every action, there is an equal and opposite reaction. In the case of linear momentum, when two objects interact, they exert equal and opposite forces on each other, resulting in a conservation of momentum.

5. Can linear momentum be lost or gained?

No, according to the law of conservation of linear momentum, the total momentum of a closed system remains constant. Momentum can be transferred from one object to another, but it cannot be created or destroyed.

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