Elastic collision conceptual problem

AI Thread Summary
In an elastic collision involving two equal masses, where one is initially at rest, the first mass comes to a complete stop while the second mass moves forward with the initial speed of the first mass. This outcome occurs because both momentum and kinetic energy are conserved during the collision. The mathematical explanation shows that when analyzing relative velocities, the two masses effectively swap their velocities. If the masses are unequal, the behavior changes: a lighter mass will rebound, while a heavier mass will result in both moving forward post-collision. Understanding these principles clarifies the mechanics behind elastic collisions.
alingy2
Messages
16
Reaction score
0
Hello,

My textbook says that if two equal masses, with the second one of them at rest initially, collide in an elastic collision. The first mass will stop and the second one will have the initial speed of the first mass.

WHY? This makes no sense to me. Elastic collision only means momentum + energy is conserved.

Why can't the first ball rebound?
 
Physics news on Phys.org
alingy2 said:
WHY? This makes no sense to me. Elastic collision only means momentum + energy is conserved.

Why can't the first ball rebound?

Because if you work out the math, conserving energy and momentum, the answer is that first ball doesn't rebound. If you don't believe it, play with one of these toys: http://en.wikipedia.org/wiki/Newton's_cradle.

If you know about relative velocities, the math is very easy. Suppose the velocities before the collision are v and 0. Now, imagine you are moving with velocity v2. You see the balls approaching each other with velocities +v/2 and -v/2. When they bounce, conserving energy and momentum, they have velocities -v/2 and +v/2. In other words, the two velocities "swap over". Relative to the ground, the first ball stops and the second one moves with velocity v.

This is only true if the two masses are equal. If the moving ball is lighter than the other one, it will rebound back. If it is heavier, both balls will be moving forwards after the collision.
 
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