1. Apr 3, 2015

gracy

My teacher taught us conservation of momentum.She gave us an example (or way to think )as if one of the blocks has spring attached to it.As given in the image.As u1 >u2 there will be collision ,though surface is not shown it is assumed to be smooth so that friction does not act,this is necessary condition for external force to be zero and in turn for conservation of momentum.The block one will collide with block 2.So,in this process spring is compressed so in order to restore it's position it will apply force on both the blocks,force applied by spring on the left block would be in left direction and force applied by spring on the right block would be in right direction.So,left block which was having greater velocity will decelerate after collision and m2 will accelerate after collision.What I don't understand is why spring applies force on both the blocks to restore its position?Shouldn't it apply force only on m1 in left direction so that it comes to it's original position?

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2. Apr 3, 2015

PWiz

No, the spring will apply a force on both of them. A body that undergoes elastic compression pushes on all objects attached to it along the axis of compression.While I don't want you to do what's not recommended by your teacher, I would suggest that you don't take this "spring" idea too far in more complex problems. Your intuition might work against you (as was in this case).

3. Apr 3, 2015

4. Apr 3, 2015

PWiz

A force is exerted on the wall as well.

5. Apr 3, 2015

6. Apr 3, 2015

PWiz

The compressed string pushes both the masses.

7. Apr 3, 2015

gracy

My teacher also says that there is maximum compression when u1=u2 .I really did not understand.Because when u1=u2 there is no head on collision so there should be no compression.

8. Apr 3, 2015

PWiz

That does not make much sense to me either. Perhaps your teacher meant it in a different context? And for the compression thing, try this:
Use a small string and place it on your thumb. Press it in with your index finger. You'll notice that you feel the pressure on both your index finger as well as your thumb, even though it is your index finger that's pushing the string in.

9. Apr 3, 2015

jbriggs444

During a single collision the speeds of the two masses will be changing as the spring is compressing and then expanding. Before the moment of maximum compression, the rear mass will still be catching up to the front mass. After the moment of maximum compression, the rear mass will be falling behind. At the moment of maximum compression, the two masses will necessarily be moving at the same speed.

10. Apr 3, 2015

gracy

Did you mean m1 in my case?

11. Apr 3, 2015

gracy

What is wrong in it?

12. Apr 3, 2015

jbriggs444

Yes.

13. Apr 3, 2015

jbriggs444

The problem is the interpretation of u1 and u2. Your teacher is taking u1 and u2 to be the current instantaneous velocities of m1 and m2 during a single collision. You are taking them to be the velocities of m1 and m2 before a collision.

14. Apr 3, 2015

gracy

Ok,even if I take u1 and u2 as instantaneous velocities of m1 and m2 why there should be maximum compression when these instantaneous velocities are equal?

15. Apr 3, 2015

PWiz

16. Apr 3, 2015

gracy

Do you want me to tie the string around my thumb?

17. Apr 3, 2015

jbriggs444

If u1 is greater than u2 then the masses are getting closer together. Compression is not yet maximum because it is still increasing.
If u1 is less than u2 then the masses are getting farther apart. Compression is not maximum because it was greater in the past.
The remaining possibility is that u1 = u2.

18. Apr 3, 2015

PWiz

No, just hold it between your thumb and finger and then push it in with the help of your finger only.

19. Apr 3, 2015

gracy

That's what I want to understand why such obligation?What's the reason behind it?

20. Apr 3, 2015

jbriggs444

What is the first derivative of a continuous function at a position where it has a local maximum?