A marble with a mass of 2 grams moves to the left with a velocity of 2 m/s when it collides with a 3 gram marble moving in the opposite direction with a velocity of 2 m/s; if the first marble has a velocity of 1.5 m/s to the right after the collision, determine the velocity of the second marble after the collision. I thought it could be solved using the conservation of momentum: momentum before = momentum after p.moving right + p.moving left = p.moving left + p.moving right 3(2) + 2(-2) = 3(v1') + 2(+1.5) solving for v1' gives -0.3333... I think and ignoring sig.figs. But then I checked using the conservation of energy. It didn't check. So I guess one of the velocities is wrong. Let's say the velocity of the first marble after the collision is unknown. Then what would v1' and v2' be? Someone else on that website said v2' would be 2.333...., but then that doesn't seem right. Why would the larger mass have more energy aft.than before? I don't have my books at home for the summer, and I forgot the formulas. I tried deriving, but I am not sure how to deal with the signs after equating the momentum factors in the difference of two squares. In fact I don't think I have seen this derivation for probably ten years - I couldn't even find it on the web. Finally, could it be that the collision is not totally elastic? How would that work? I guess then the velocity of the 3 gram after could be 1.5, but not sure. Idk, brain fried. I stayed awake trying to figure it out last night. I guess what I really want to know is what do you get for velocities when it is a totally elastic collision?
The data in this problem appears to be incorrect. If you think about this logically, is it possible for the smaller marble to actually slow down after being hit by the larger marble? If you could simulate this, you can see that the smaller marble would go faster. Try it with this site: https://www.msu.edu/~brechtjo/physics/airTrack/airTrack.html
This is correct. That's because mechanical energy isn't conserved in this problem, and you probably just added the kinetic energies of the two balls before and after the collision to check. Some of the kinetic energy becomes work to change the velocities (speed and direction) of the two balls. Remember that mechanical energy of an object is only conserved when only INTERNAL forces (gravity and spring) work on the object. When an external force is applied (such as running ino something!), you have to account for the work applied or performed by an object.
Thanks everyone, but I don't think anyone answered my last question. What if the 1.5 was not given? That is, what would the two velocities be after the collision if it was totally elastic?
I guess since the OP didn't mention that the marbles collided perfectly elastically, this would apply. I just assumed that marbles collide elastically, in which case Kinetic Energy is conserved as well as momentum. If the collision were inelastic, then kinetic energy would be lost as you say.
Using the simulation site i mentioned earlier, the 3.0 gram marble would be travelling at 1.2 m/s and the 2.0 g marble 2.8 m/s, in the opposite directions from their initial velocities.
Momentum is always conserved, KE is not. If it were relevant, the problem would have probably stated it was elastic and then you'd have the joy of solving simultaneous equations for momentum and KE. I was just too lazy to do that last night. m1v1a + m2v2a = m1v1b + m2v2b .5 m1(v1a)^2 + .5 m2(v2a)^2 = .5 m1(v1b)^2 + .5 m2(v2b)^2 If you substitute what you know and solve, you get the two solutions of v1=2 m/s, v2 = -2 m/s (original conditions) and v1= -1.2 m/s, v2= 2.8 m/s
I don't understand why it wouldn't be an elastic collision. Since when did marbles behave differently? It's just a matter of whether the collision is perfectly elastic or not.
