Energy Speed of Ball Question

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A BALL OF MASS 9M IS DROPPED FROM REST FROM A HEIGHT H = 5.0 METERS ABOVE THE GROUND. IT UNDERGOES A PERFECTLY ELASTIC COLLISION WITH THE GROUND AND REBOUNDS. AT THE INSTANT THAT THE BALL REBOUNDS, A SMALL BLOB OF CLAY OF MASS M IS RELEASED FROM REST FROM THE ORIGINAL HEIGHT H, DIRECTLY ABOVE THE BALL. THE CLAY BLOB WHICH IS DESCENDING, EVENTUALLY COLLIDES WITH THE BALL, WHICH IS ASCENDING. ASSUME THAT G = 10 M/S2, THAT AIR RESISTANCE IS NEGLIGIBLE, AND THAT THE COLLISION PROCESS TAKES NEGLIGIBLE TIME.

A) DETERMINE THE SPEED OF THE BALL IMMEDIATELY BEFORE IT HITS THE GROUND.

B) DETERMINE THE TIME AFTER THE RELEASE OF THE CLAY BLOB AT WHICH THE COLLISION TAKES PLACE.

C) DETERMINE THE HEIGHT ABOVE THE GROUND AT WHICH THE COLLISION TAKES PLACE.

D) DETERMINE THE SPEEDS OF THE BALL AND THE CLAY IMMEDIATELY BEFORE THE COLLISION.

E) IF THE BALL AND CLAY BLOB STICK TOGETHER ON IMPACT, WHAT IS THE MAGNITUDE AND DIRECTION OF THEIR VELOCITY IMMEDIATELY AFTER THE COLLISION?


Any help would be great, thanks!
 

Answers and Replies

  • #2
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Please... anyone?
 
  • #3
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A)Integrate G once with respect to time to find veloicity as a function of time, and twice to find position. Use the position function you've found (y = 5t^2) to find the time it collides with the ground, then use that to find velocity.
B)The clay falls at the same rate the ball did, but now the ball is rising, and has acceleration - g. Find their velocity's and position seperately, as you did before. Set the postions equal to each other to find the time they collide.
C)Use that time in the formula for position of either of them to find the position.
D)Use that time in the velocity formula you found for both of them.
E)Use energy conservation, Ebefore = Eafter to find the velocity's after, and the formula for kinetic energy, 1/2 * m*v^2.

and by the way, caps lock is really annoying.
 
  • #4
Hi,
I have been given the same exact question for a homework assignment of mine and I'm not understanding how to find the time exactly. Could someone please be a little more specific than pi-r8, even though he did a great job, on part b. Is there a way to do it without calculus, possiby conservation of momentum? I'll be able to do the rest if I get b.
BTW, what would the right answer be?

Thanks. :)
 
  • #5
Doc Al
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part b

No need for calculus if you can assume knowledge of basic kinematic relationships. What kinematic formula tells you how far an object falls as a function of time?
 
  • #6
Well I understand that you can use d = vit + 1/2at^2, but when I set that equal to the negative distance of the other item, I get 1s. That is wrong though, right?
 
  • #7
Doc Al
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Well I understand that you can use d = vit + 1/2at^2, ..
That's the correct relationship (don't forget the initial height). But in part b they are talking about the collision of the clay blob with the rebounding ball (not with the ground). Write expressions for the positions of the blob and the ball as functions of time. When they collide, their positions are the same--solve for the time.
 
Last edited:
  • #8
That's the correct relationship. But in part b they are talking about the collision of the clay blob with the rebounding ball (not with the ground). Write expressions for the positions of the blob and the ball as functions of time. When they collide, their positions are the same--solve for the time.

Right, so what I ended up doing was:
position of the blob = 1/2at^2
position of the ball = vit + 1/2at^2
Now since the ball and blob are going in the opposite direction, I make one of them negative.
-1/2at^2 = vit + 1/2at^2 (I solved for vi in part a)
5t^2 = 10t - 5t^2
10t^2 = 10t
t = 1

One second can't be correct though so I don't see where I'm going wrong.
 
  • #9
That's the correct relationship (don't forget the initial height). But in part b they are talking about the collision of the clay blob with the rebounding ball (not with the ground). Write expressions for the positions of the blob and the ball as functions of time. When they collide, their positions are the same--solve for the time.



Oooh, let me try with the inital height.
 
  • #10
Doc Al
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One second is the time it would take to fall the entire 5m, so you know that's not right. You can't just "make one negative"--the only difference between the two equations for position should be the initial height and the initial speed. So rewrite those two position functions and set them equal.

Hint: Do it systematically. Pick a sign convention and a coordinate system: I suggest calling the ground y = 0 and "up = positive, down = negative".
 
  • #11
One second is the time it would take to fall the entire 5m, so you know that's not right. You can't just "make one negative"--the only difference between the two equations for position should be the initial height and the initial speed. So rewrite those two position functions and set them equal.

Hint: Do it systematically. Pick a sign convention and a coordinate system: I suggest calling the ground y = 0 and "up = positive, down = negative".

So is it that t = .5 seconds?

Because:
position of blob = 1/2at^2 + 5
position of ball = vit + 1/2at^2
set them equal and discover 5 = vit
5 = 10t
t = 1/2

??? Is that at all right?

I know I'm a nussance.
 
  • #12
Doc Al
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Looks good to me.
 
  • #13
how would you do numbers 3, 4, and 5 after finding the time for number 2 in .5 seconds????
 

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