# Momentum Ball Collision Find Normal Force

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1. Feb 4, 2016

### aimen khattakk

[Note: Thread moved by moderator]

Hi! so I have a question im unable to do and ive been trying it for 3 hours. The question had 2 parts, i figured out part a but cannot figure out part b.

A) A 0.006kg ball is dropped from a height of 4.0m above the floor. Neglecting air resistance, determine the momentum acting on the ball

So i did the following and the answer was correct:

Vf=Vi + 2ad
Vf = 0m/s + 2(-9.81m/s)(4.0m)
i solved for Vf and got 8.858893836m/s down

then solved for momentum
p = (0.06kh)(8.858893836m/s down )
p = 5.31x10^-1 kgm/s down

Now i cant do part B
B) The ball in part A bounces to a height of 3.0 m. if the collision with the floor took 0.024s, determine the normal force acting on the ball.

So i attempted it and this is what i did

F = MΔV/ΔT
F = (0.06kh)(8.858893836m/s down - 0m/s) / 0.024s
F = 22.15N this is the net force right?
So i now tried solving for the normal force knowing that gravity and normal force were acting on the ball so i did
Fnet = Fn - Fg
22.15N = Fn - mg
22.15N = Fn - (-0.5886N)
22.15N-0.5886N = Fn
so Fn = 21.56N

But the answer sheet i have says the answer is 42N so im so lost

2. Feb 4, 2016

### Nathanael

ΔT is the time interval on which the normal force acted.
ΔV is the change in velocity over this same time interval.
Are you then saying that right when the ball loses contact with the ground it has a speed of 0?

Also, if you want to be exact, you should know that F in your equation is actually the net force (not just the normal force).

3. Feb 4, 2016

### aimen khattakk

Im aware that F is the net force. thats why i did Fnet = Fn - Fg and solved for Fg later. and the initial speed of 0 was the speed before the ball was let go (in the previous question) do i need to solve for initial speed before the ball hit the ground?

4. Feb 4, 2016

### Nathanael

Oh, good. Sorry for not noticing that.
You already did, that is the 8.86m/s value, isn't it? What you haven't considered is the speed of the object just as it leaves the ground.

5. Feb 4, 2016

### aimen khattakk

the 8.86m/s is the final velocity that was calculated in the previous question in order to find the momentum of the ball just before it hits the ground. Now i need to find the normal force acting on the ball after it bounces to a height of 3.0m. so once the ball hits the ground isnt my initial velocity 0? so do i need to calculate the Vf given the new height?

6. Feb 5, 2016

### Nathanael

I think I see your confusion:
The ball will not stop in the exact moment that it touches the ground (wouldn't that be strange to go from 8.86m/s to 0 in a single instant?). In reality the ball will deform (even if just slightly) so it will still be moving down even though it is touching the ground. (In reality, though, the normal force is not constant, so things are more subtle. This problem wants you to treat the normal force as being constant.)

The initial and final velocities you are concerned with in your net force equation are the velocities at the moment it first touches the ground and the moment it loses contact with the ground. (This is because these are the moments when the normal force begins, and then finishes, acting on the ball.)

Yes, you will need to know the speed just after the ball loses contact with the ground.

7. Feb 5, 2016

### aimen khattakk

Okay so for the Vi i'll need to use the height that the ball was initially dropped from to find the Vi the moment it hits the ground? so would i do the following:

d=Vit+2at2 and sub in all the values and then do the same for Vf but use a d of 3.0 instead? because thats how high it bounced after it hit the ground?

8. Feb 5, 2016

### Nathanael

Yes, this is the right idea.
This is not the correct equation, though.
(Maybe this was just a typing mistake, though, because you correctly found the initial speed in your original post.)

9. Feb 5, 2016

### aimen khattakk

so the initial speed is the one i found originally? 8.86m/s? and now using that i need to find my final speed once the ball leaves the floor? so would i do

Vf2 = Vi2 + 2ad
Vf2 = 8.86m/s2 + 2(-.981m/s2)(3.0m)
and then solve it for Vf?
i did this and got 4.42m/s for my Vf
then when i put that into my Fnet equation i get an Fnet of 11.07 which makes no sense because
Fnet = Fn-Fg
and Fg=0.06kg x -9.81m/s2
Fnet - Fg = Fn but the answer sheet says 42N and i didnt get anywhere close to 42N

10. Feb 5, 2016

### Nathanael

I think you are getting confused by the words "initial" and "final." It is important to realize the problem takes place in three parts:
the part where the ball is dropped from 4m and falls to the ground,
the part where the ball is in contact with the ground,
the part where the ball rises from the ground to a height of 3m.

The "final speed" in one part is the "initial speed" for the next part.

Your equation [Vf2 = Vi2 + 2ad] will be useful on the first and third parts (where the ball is in free fall).
Your equation [Fnet.average = MΔV/Δt] will be useful on the second part (where the ball is in contact with the ground).

Be careful that you do not mix up the initial/final speed of one part with the initial/final speed of another part.
(This has been the basis of all your mistakes.)

11. Feb 5, 2016

### aimen khattakk

Okay so let me see if i understand. the final velocity that i calculated earlier (8.86m/s) is now the initial velocity for my second part? using THAT initial velocity, i need to find my final velocity for the second? and id find that using [Vf2 = Vi2 + 2ad]? and then use [Fnet.average = MΔV/Δt] to find the fnet?

12. Feb 5, 2016

### Nathanael

The final velocity for the second part is the initial velocity for the third part. You know the final velocity for the third part is zero, so you can use [Vf2 = Vi2 + 2ad] to find the initial velocity for the third part, which is then the final velocity for the second part.

13. Feb 5, 2016

### aimen khattakk

okay so here is all the data i have that im going to lay out

Ball dropped from 4m and falls to ground
vi = 0 and Vf = 8.86 (calculated using [Vf2 = Vi2 + 2ad])

Ball in contact with ground

Vi = 8.86m/s and Vf = Vi of third part which works out to be 7.67m/s

Ball bounces to height of 3m

Vf =0 and Vi = 7.67m/s (calculated using [Vf2 = Vi2 + 2ad] by subbing in 3.0 m as the height, -9.81m/s2 as the acceleration and 0 as the Vf)

So now what im looking for is the velocities at the moment it first touches the ground to the moment it first LEAVES the ground so that would be 8.86m/s and 0? or 8.86 and 7.67?

14. Feb 5, 2016

### Nathanael

Yes, it would be 8.86 m/s and 7.67 m/s. You're very close now. Just remember, the direction of these two speeds is also important.

15. Feb 5, 2016

### aimen khattakk

the direction of the Vi would be downward right? and the direction of the Vf would be upwards? so that would make it 7.67m/s-(-8.86m/s) which is 16.53m/s and then i plug that into the Fnet equation and solve for Fn?

16. Feb 5, 2016

### aimen khattakk

IT WORKED! thank you so so much for the help! i finally got it and i understand it too! thank you!

17. Feb 5, 2016

### aimen khattakk

I just want to ask one more question so i know i understand. The 7.67m/s that we used, is the final velocity when the ball comes in contact with the ground right? not the initial velocity for when it leaves the ground? thank you for all the help!

18. Feb 5, 2016

### Nathanael

It is both. The final velocity after the ball comes in contact with the ground is the same as the velocity of the ball as it leaves the ground.
In other words, the "moment where it leaves the ground" is the same moment as the "final moment of contact." It is the same time.

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