2 boxes attached to sprint. concept questions.

  • Thread starter Thread starter reaperkid
  • Start date Start date
  • Tags Tags
    Concept
Click For Summary

Homework Help Overview

The problem involves two masses connected by a spring on a frictionless surface, with the goal of analyzing their speeds, heights, forces, and kinetic energies after being released. The context is rooted in concepts of momentum, energy conservation, and the effects of mass on motion.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the relationships between speed, kinetic energy, and momentum for the two masses, questioning the implications of mass differences on these quantities. They explore how the force duration from the spring affects each mass differently.

Discussion Status

There is an ongoing exploration of the relationships between the various statements, with participants questioning their own reasoning and the implications of their assumptions. Some participants suggest that the kinetic energy of the smaller mass may be greater than that of the larger mass, while others are uncertain about the duration of force application.

Contextual Notes

Participants express uncertainty regarding the timing of force application and the conditions under which the forces are considered equal. There is a lack of numerical data, which complicates the analysis of the problem.

reaperkid
Messages
14
Reaction score
0

Homework Statement



A spring of negligible mass is compressed between two masses (M1 is less massive than M2) on a frictionless table with sloping ramps at each end. The masses are released simultaneously. Select the appropriate symbol for each statement: G (Greater than), L (Less than), or E (Equal to).

A) The speed of M1 is ... the speed of M2 once they both lose contact with the spring.
B) The final height up the ramp reached by M2 is ... the height reached by M1.
C) The duration of the force exerted by the spring on M1 is ... the time the force acts on M2.
D) The kinetic energy of M2 is ... the kinetic energy of M1 once they both lose contact with the spring.
E) The magnitude of the force exerted by the spring on M2 is ... that it exerts on M1.
F) The magnitude of M2's momentum is ... that of M1 before and just after loss of contact with the spring.


Homework Equations



Well, there are no numbers so I don't really need to use equations but:

p=mv
f=ma

The Attempt at a Solution



This is what I put, but at least one of these are wrong.

A) Greater because the same force is being applied to a smaller mass.
B) Less because the smaller mass would have a greater velocity
C) duration of m1 is greater because it would travel farther.
D) the KE is equal because they are pushed at the same force and even though one is going faster, it's a lesser mass, so it balances out.
E) the force exerted is equal because they are being pushed by the same force.
F) the momentum is the same because the force is the same. The lower mass balances with the higher velocity.

Any help would be appreciated, thanks!
 
Physics news on Phys.org
How do you reconcile C) and E)?
 
Interesting.

For (D), we have (1) 1/2.m1.v1^2 = 1/2.m2.v2^2 ie kinetic energies equal

For (F) we have momentum is equal - which must be the case because momentum is conserved - which is m1v1 = m2v2

From (F) we can derive that if m1 is smaller than m2, v1 is bigger than v2. If we take (F) to be true, then it follows that KE of smaller mass is greater than that of the larger mass. So I think (D) is less than

(C) is tricky, but I don't think it's right. Because there's less resistance from the smaller mass, the half of the spring pushing that mass would fully extend faster. The half of the spring pushing against the larger mass would extend slower, so I'm inclined to say Less for that.

My 2 cents - I'm not positive.
 
LowlyPion said:
How do you reconcile C) and E)?

Well, I wasn't too sure on C because I don't fully understand when the force stops acting on it I guess. My reasoning was because M1 would be in motion longer than M2.

For E, I've been assuming that since it's getting pushed by the same spring that the force exerted on M1 and M2 are equal.
 
So in E) the force is the same on both.

But in C) it acts for a longer time on one and not the other?

Doesn't that seem like a contradiction?
 

Similar threads

Replies
20
Views
3K
Replies
23
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 19 ·
Replies
19
Views
12K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K