Use Conservation of Momentum or Conservation of KE for spring problem

AI Thread Summary
In the spring problem involving two masses, the key focus is on understanding the conservation laws applicable to the system. The masses have different magnitudes, with one being three times greater than the other. It is clarified that the question pertains to the kinetic energy and momentum of each mass individually, rather than their combined totals. The discussion emphasizes that since the system is closed and perfectly elastic, both kinetic energy and momentum should remain conserved. However, participants note that the masses have not yet collided, which affects the application of these conservation principles.
brainyasian
Messages
13
Reaction score
0

Homework Statement



The system shown above has a spring attached to two masses. Mass I is 3 times greater in mass than Mass II. The spring is initially stretched and let go to allow the two masses to approach each other. Which of the following is the same for both masses?
a. speed
b. velocity.
c. acceleration
d. kinetic energy
e. magnitude of momentum

Homework Equations


Conservation of momentum
conservation of energy


The Attempt at a Solution


I know it's not the first three, but I'm not sure if you use the conservation of momentum or the conservation of energy for this problem
 
Physics news on Phys.org
Try proving each one and post what you get here.

What would cause momentum to be lost?
 
Screen said:
Try proving each one and post what you get here.

What would cause momentum to be lost?

But this is a closed system. Since no energy is lost or gained and this is a perfectly elastic collision, wouldn't both KE and momentum remain the same afterwards?
 
brainyasian said:
But this is a closed system. Since no energy is lost or gained and this is a perfectly elastic collision, wouldn't both KE and momentum remain the same afterwards?

Try writing it mathematically.

The problem is asking about the kinetic energy and momentum of each mass, not the combined total.
 
Screen said:
Try writing it mathematically.

The problem is asking about the kinetic energy and momentum of each mass, not the combined total.

but doesn't KE(M1) = KE(M2) because of conservation of mass and doesn't M(A)V(A) = M(B)V(B) make momentum equal?
 
brainyasian said:
but doesn't KE(M1) = KE(M2) because of conservation of mass and doesn't M(A)V(A) = M(B)V(B) make momentum equal?

Read the question carefully; they haven't collided yet.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top