The feedback from my proffessor is here: Now, on the CX, and specifically Equati

Question

The feedback from my proffessor is here:

Now, on the CX, and specifically Equation 2: I don't know where it came from or how you arrived at it. The one-dimensional two-body expression of the momentum conservation principle on the equation sheet is: m1v1i + m2v2i = m1v1f + m2v2f

If you're going to assume the collision is elastic, I'd support that - make it explicit, and then write an equation expressing that assumption, namely Kei = Kef. Both expressions will be somewhat simplified by the fact that v1i = -v2i, so you're down to two unknowns (v1f and v2f), which is solvable with two equations.

Now, while in general "outside" equations are forbidden in CX assignments, the textbook does show how you can use those two equations to derive another relationship that applies only to elastic collisions. With suitable introduction, I'd be ok with you using that.

But pulling half of an equation (iirc, there isn't even an equals sign, which is why I asked what the expression is equal to) from somewhere with no explanation is of course going to leave me puzzled about what is going on. I'm sure it makes total sense to you, and it may be correct. But I need to be able to follow the train of thought, and much of it appears to have taken place somewhere off the page.

Explanation / Answer

We assume the collision is elastic and no energy is lost.

wemconsider the motion in COM co-rodinates.

The total PE (M+m)gh is converted to KE of the COmas it bounces back.

m is the mass of the tennis bal and

M that of basket ball

velocity of the COM is sqrt(2gh)

if d is the seperation between the two balls then distance of the two balls from COM is given by

basket ball = md(M+m)

and

tennis ball distance from COM   = Md/(m+M),

heavier the basket ball farther the tennsi ball from the COM and will fly hiegher

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