Momentum Conservation in Collisions: Equal Mass Carts

AI Thread Summary
Momentum is conserved at every instant during a collision involving equal mass carts, including before, during, and after the event. During the collision, the forces between the carts are equal and opposite, resulting in equal and opposite changes in momentum, thus conserving total momentum. If cart 1 is moving and cart 2 is at rest, after the collision, cart 1 may not necessarily take cart 2's velocity unless the collision is perfectly elastic. The masses of the carts do not affect the conservation of momentum, but they do influence the outcome of their velocities post-collision. Understanding these principles clarifies the behavior of colliding carts in physics.
nietzsche
Messages
185
Reaction score
0

Homework Statement



When two carts collide, is momentum conserved at each instant before, during and after the collision? If the carts are of equal mass what happens to their individual momenta during and after the collision?

Homework Equations


The Attempt at a Solution



I'm actually unsure of what the question means. "If the carts are of equal mass what happens to their individual momenta during and after the collision?" During the collision? I'm confused because we normally just consider before the collision and after the collision, leaving out the complex in-between stuff.

As for after the collision, I think the answer is this. If cart 1 is moving at a velocity v relative to cart 2, and cart 2 is at rest, then after the collision, cart 1 will have 0 final momentum relative to cart 2, and cart 2 will have the same initial momentum that cart 1 had. Please correct me if I'm wrong.
 
Physics news on Phys.org
Momentum is conserved at each instant before, during and after a collision. This is why,

1. Before and after a collision Newton's 1st Law holds, "An object will maintain its state of motion unless an unbalanced force acts on it." Each cart experiences no unbalanced force.

2. During the collision momentum is conserved because of Newton's 2nd and 3rd Laws. Cart 1 experiences a force due to Cart 2 and exerts an equal and opposite force on Cart 2 (Newton's 3rd Law.) Now from Newton's 2nd Law, momentum change (impulse) is force times time interval over which the force acts. Since the forces that the carts exert on one another are equal and opposite and act for the same time (as long as the carts are in contact), then the changes in momentum are equal and opposite. This means that their sum is zero, i.e. momentum is conserved. The masses of the carts are irrelevant.
 
Thank you.

So does this necessarily mean that cart 1 will end up with cart 2's initial velocity, and cart 2 will end up with cart 1's initial velocity?
 
No it does not necessarily mean that. The two carts will exchange velocities if (a) the masses are equal and (b) the collision is perfectly elastic.
 
Ah thank you. Makes sense.
 
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 '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