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Bouncing balls

  • Thread starter shanie
  • Start date
  • #1
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Homework Statement


A ball is dropped from a height of 1.75m. It bounces on the floor and reaches a maximum height of 1.35m above the floor. The ball's weight is 98.5g.
a) What is the ball's velocity just before and just after the bounce?
b) How large is the impulse that the ball gets in the bounce?
c) Assume that the ball is in contact with the floor for 15 ms. What is the ball's average acceleration in the bounce?


Homework Equations


a) mgh=mv2/2, where the heights are 1.75 and 1.35 respectively.
b) I = Δp =mv2-mv1
c) average acceleration = Δv/Δt

The Attempt at a Solution


Setting downward direction to negative, I've solved a) and b) in the following manner,

a) using the energy principles so that velocity before bounce, v1=(-5.863m/s)
and velocity after bounce, v2=5.149m/s
b) Taking the signs of the velocities into consideration: I=(0.985*5.15)-(0.985*-5.86)=0.985*[5.15-(-5.86)]≈10.8 kgm/s
c) I'm not sure whether Δv is 1)[5.863-5.149]
or 2) [5.149-(-5.863)],
because the first alternative gives an acceleration (setting Δt to 0.015s) of 47.6m/s2
And the second alternative gives a=734 m/s2.

In my opinion the second average acceleration seems excessively large, and so I'm unsure of which I should choose.. And furthermore, are the calculations in a) and b) correct? I could really use some help, thanks!
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi shanie,

Homework Statement


A ball is dropped from a height of 1.75m. It bounces on the floor and reaches a maximum height of 1.35m above the floor. The ball's weight is 98.5g.
a) What is the ball's velocity just before and just after the bounce?
b) How large is the impulse that the ball gets in the bounce?
c) Assume that the ball is in contact with the floor for 15 ms. What is the ball's average acceleration in the bounce?


Homework Equations


a) mgh=mv2/2, where the heights are 1.75 and 1.35 respectively.
b) I = Δp =mv2-mv1
c) average acceleration = Δv/Δt

The Attempt at a Solution


Setting downward direction to negative, I've solved a) and b) in the following manner,

a) using the energy principles so that velocity before bounce, v1=(-5.863m/s)
and velocity after bounce, v2=5.149m/s


These look okay to me, if you have chosen downwards to be negative (which shows up later in the problem).

b) Taking the signs of the velocities into consideration: I=(0.985*5.15)-(0.985*-5.86)=0.985*[5.15-(-5.86)]≈10.8 kgm/s
The procedure here is correct, but the numbers are not right. The mass is not 0.985kg.

c) I'm not sure whether Δv is 1)[5.863-5.149]
or 2) [5.149-(-5.863)],
because the first alternative gives an acceleration (setting Δt to 0.015s) of 47.6m/s2
And the second alternative gives a=734 m/s2.
Although only one is right, I want to point out in your alternative 1 you left off the minus sign from 5.863m/s. So your alternative 1 should be: ( (-5.863) - 5.149), giving a negative number with the same magnitude as alternative 2.

However, [itex]\Delta v[/itex] means [itex]v_{\rm final}-v_{\rm initial}[/itex], so if you're looking for the [itex]\Delta v[/itex] of the bounce, which would be correct?
 
  • #3
23
0
Thank you so much alphysicist!
I realized the mass is 0.0985kg, giving an impulse of 1.08 kgm/s. Also for c) I'd say now it's rather obvious that the second alternative is the correct one, since vfinal = 5.149 m/s and vinitial=5.863m/s, giving \Delta v=11.01m/s, and the acceleration must be positive because the ball is bouncing upwards in the positive direction. Hence, the acceleration is 734 m/s^2?
 

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