Tennis ball and basket ball falling, wad of sticky clay being thrown at block + move

  • Thread starter Roushrsh
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  • #1
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1) A tennis ball of mass mt is held just above a basketball of mass mb.
Their centers are vertically aligned. They fall through a height h, then there's an elastic collision with the ground which reverses the velcoity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling. Next, the two balls meet in an elastic collision.
A) to what height does the tennis ball rebound?
b) How do you account for the height in A being larger than h? Does it seem like a violation of conservation of energy

2) A wad of sticky clay of mass m is hurled horizontally at a wooden block of mass M on a horizontal surface. The clay sticks to the block. After impact the block slides a distance d before coming to rest. If the coefficient of friction between the block and the surface is u. What was the speed of the clay immediately before impact?

No relevent equations!

attempt for 1) Ok, so I've done this type of problem a bunch of times... but NORMALLY I HAVE THE MASSES of the balls and the height, making it waaaaay easier. I don't see how you can find the height without knowing the height to begin with in order to calculate final velocity of sqrt(2gh).

B is easy, it's larger due to the basketball exerting a higher force (because of it's larger mass), so it isn't a violation of energy since the basketball has more energy due to it's higher mass. (1/2mv^2) m = bigger for b ball and now tennis ball is moving with that energy only with a smaller mass of itself so it's v will be higher.

I just don't know how to do A. Any help is greatly appreciated. Thanks


Attempt at #2) Same problem as A, I usually do it with numbers, If anyone can help with that, I'd be grateful. I don't get how you can just predict things like that without knowing any numbers :/
 

Answers and Replies

  • #2
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Without numbers you do it the same way you would do with.
You are given the mass of the tennis ball mt, the mass of the basketball mb and the initial height h. So you can use the same equations as before. Just don't insert numbers.
 
  • #3
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Well for the first one I tried that and got too big of an equation, can you help? I did
tb = tennis ball and bb= basketball

v=sqrt (2as)
v=sqrt (19.82h)

u(tb) = - sqrt (19.82h)
u(bb) = sqrt (19.82h)

1/2m(tb)u(tb)^2 + 1/2m(bb)u(bb)^2 = 1/2 m(tb)v(tb)^2 + 1/2m(bb)v(bb)^2

m(tb)u(tb) + m(bb)u(bb) = m(tb)v(tb) + m(bb)v(bb)

v(tb) = (u(tb) (m(tb)-m(bb) + 2(m(bb))(u(bb))/(m(tb)+m(bb))

then KE = 1/2m(tb)v(tb)^2 = PE = m(tb)gh
h=v(tb)^2/19.82

so.. if I were to just put the v(tb) into the KE formula, I'd get h=h, if I were to put what the v(tb) is equal too, it'd just be a really long equation.
How would I simplify more or what would I do??? :S
Thanks
 
  • #4
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As for the second one
I did
delta KE = Wnet
KEf = 0
KEf - KEi = -(nu)mad
-1/2(m+M)v^2 = -(nu)(m+M)d
v=sqrt(-2(nu)ad) - v of block + clay

pi = pf since momentum is conserved

m1v1 +m2v2 = m(1+2) v(1+2)
m1v1 + 0 = (m+M)sqrt(-2(nu)ad)
m1v1 = (m sqrt(-2(nu)ad)) (M sqrt(-2(nu)ad))
v1 = sqrt(-2(nu)ad) (M sqrt(-2(nu)ad))

that seems too long and wrong, any help for this too? thanks :/
 
  • #5
318
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Sometimes equations will be long. Your equation for conservation of kinetic energy and momentum for a) is correct. I assume u denotes the velocities before the bounce and v after the bounce.
If you take a closer look at those two equations you will notice, that you know all quantities, except [tex]v_{tb}[/tex] and [tex]v_{bb}[/tex] as you have two equations for those two unknowns you can solve for them.

And then by equating kinetic and potential energy of the tennis ball as you did, you will find the height.

And please, don't enter a value for g. First, this 9.81 is only an approximation and so it is much more precise to write simply g. And second those numbers do not really look nice.
 
  • #6
318
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Your idea for the second question is correct up to some minor errors.

I'm not sure what you mean by (nu)a . The resulting formula for v is almost correct but you mixed some equations.

From your third last to second last line you did not expand correctly. Look at this again.
 
  • #7
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Ok so for A I did it correctly?

And it is therefore,
1/2((u(tb) (m(tb)-m(bb) + 2(m(bb))(u(bb))/(m(tb)+m(bb)))^2/g = h
?

Also sorry, but (nu) I meant coefficient of friction, since I normally hear it pronounced "nu"
Does that make it correct now?
 
  • #8
318
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I cannot see how you got the result for [tex]v_{tb}[/tex] (7. Equation) but the two equaions above are correct. If you solve them you should get something containing a root for [tex]v_{tb}[/tex].

If nu is the friction coefficient, it's almost ok. You didn't state what a i. And as the normal force comes from gravity a g must have to appear somewhere. If you the solve this equation correctly it will be ok.
 

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