Newton's Laws of Motion (Is my theory correct?)

AI Thread Summary
When two cars of equal mass, each 2.0 kg, are pushed apart by a spring with a total force of 14 N, the acceleration of each cart can be calculated using Newton's second law, F=ma. Each cart experiences the full force of 14 N, leading to an acceleration of 7 m/s² for each cart when considering the individual forces acting on them. The discussion highlights the importance of recognizing that the forces are equal and opposite, resulting in a net force of zero when considering the system as a whole. It is also noted that acceleration is a vector quantity, requiring direction specification. The conclusion emphasizes that the initial assumption of distributing the force was incorrect; each cart indeed experiences the full force of 14 N.
math989
Messages
5
Reaction score
0

Homework Statement


If two cars of equal mass are pushed apart by a spring mounted on cart #1 with a force of 14 N what will be the resulting acceleration of the carts.

Mass of carts 1 and 2= 2.0Kg
f=14N
A?

Homework Equations


F=ma

The Attempt at a Solution


My theory is that the Newtons will be distributed between the two carts so that each cart will receive 7N of force. Then i can apply f=ma to solve for the A so 7/2= 3.5m/s^2
 
Physics news on Phys.org
I think you've discovered the center of mass coordinate system for this problem.

Since both carts have the same mass and the spring is pushing the, apart then you
Insight is correct.

Remember that acceleration is a vector so to be absolutely correct you must specify a direction too for each carts acceleration.
 
"Pushed apart with a force of 14 N" means that each cart experiences a force of 14 N.
If they were pushed each by 7 N, there would be no reason to add the 7 by 7 and give a value of 14. The two forces are opposite so their sum is zero and not 14.
 
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 hanged 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