# Ball hits wall and bounces off

1. Nov 10, 2007

### ~christina~

1. The problem statement, all variables and given/known data

A machine designed for quality testing in a handball factory propels the 0.15kg balls toward a wall with a velocity of 9.5m/s at an angle of 60 deg frome the normal to the wall. A typical ball rebounds at 70 deg from the normal at a speed of 9.0m/s.
The impact point of the balls is 0.5m above the floor

a) what impulse does a typical ball deliver to the wall

b) what average force acts on the wall if the ball is in contact with the wall 4.9ms?

c) what is the maximum height that the ball reaches after leaving the wall?

d) How long does it take to reach that height?

e) Find the horizontal distance from the wall to the point where the rebounding
ball lands on the floor

f) How long does it take the ball to travel from the wall to the floor?

g) Find the velocity of the ball (magnitude and direction just before it hits the floor

h) Draw a free body diagram for the ball at the highest point of it’s trajectory

2. Relevant equations
$$I= \Delta P = p_f- p_i= mv_f- mv_i$$

$$\Delta P/ \Delta t= Faverage$$

3. The attempt at a solution

a) what impulse does a typical ball deliver to the wall

$$\Delta P= mv_f- mv_i$$
$$p_i= m(vf_x- vi_x)= \Delta P_x$$

$$m(vf_y- vi_y)= \Delta P_x$$

m= 0.15kg
v1= 9.5m/s
v2= 9.0m/s
theta 1= 60 deg
theta 2= 70 deg

$$\Delta P_x=$$0.15kg(9.0cos70 - 9.5cos120)= 1.17$$\hat{}i$$

$$\Delta P_y=$$0.15kg(9.0sin70 - 9.5sin120)= 0.034$$\hat{}j$$

$$I= \Delta P= 1.17\hat{}i, 0.034\hat{}j$$
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b) average force acts on wall if ball is in contact with wall for 4.9 ms?

$$Faverage= \Delta P / \Delta t$$

~I'm not sure if this is correct but since I have the x and y component of the impulse to get the change of momentum value to plug into the F average equation I think I need to get the magnitude.

$$\Delta P= \sqrt{} P^2 _x + P^2 _y$$

$$\Delta P= (1.17)^2 + (0.034)^2 = 1.17kg*m/s$$

t= 4.9ms= 4.9 x 10^-3s

$$Faverage= \Delta P / \Delta t= (1.17kg*m/s) / 4.9 x10^-3s = 238.8N$$

c.) Max height the ball reaches after leaving the wall

t= ?
dy= _____?
Voy= 9.0m/s
theta= 70 deg

Vy= 0 ==> at max height

Vy= Voy + at

t= - Voy/ a

$$Sy= Soy + Voy(t) + .5 at^2$$

$$Sy= Voy(-Voy/ a) + .5 a(-Voy/ a)^2$$
$$Sy= -Voy^2 / a + .5 a (-Voy/ a)^2$$

$$Sy= - (9.0 sin 70) ^2 / (-9.81m/s^2) + .5 (-9.81m/s^2) (- 9.0sin70 /(-9.81m/s^2) )^2$$

Sy= 7.291 - 3.645 = 3.645s

Sy= 3.645s

d) time it takes to reach that height

Vy= Voy + att= -Voy/ a

t= -9.0 sin 70 / (-9.81m/s^2)

$$t max= 0.8621s$$

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e) horizontal distance from point where rebounding ball lands on floor

~I'm not quite sure about this since I would think that the max height time would be doubled to see when it reaches the same point again from the rebound on the wall however that wouldn't be when it reaches the floor.
I was thinking that I should see what time is to reach the horizontal distance of 0 since the initial doy= 0.5m so

Soy= 0.5m
Sy= 0 m
t= ?
tmax= 0.8621s

dx= ?

$$Sy= Soy + Voy (t) + 0.5 a*t^2$$

0= 0.5 +(9.0sin 70) t + (-4.9) t^2

[-8.457 +/- $$\sqrt{} (8.457)^2 - 4(0.5*-4.9)$$] / 2*-4.9

(-8.457 +/- 9.011)/ -9.8

t = 1.78 s ===> to reach the ground

Sx= Sox + Vx t

Sx= 0 + 9.0 cos 70 (1.78s)

Sx= 5.479m

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f) how long does it take for the ball to travel from the wall to the floor?

found this in last part

t= 1.78s

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g) find velocity of the ball (magnitude and direction just before it hits the floor)

~ how do I determine the magnitude and direction just before it hits the floor?

What time do I use?

I do know that it takes 1.78 sec for the ball to go from the wall to the floor but what is the time right before that?

Is it 1.77s ?

____________________________________________

h) free body diagram of ball at highest point
I think that the ball would only have the Vx as the force on the ball only.

_______________________________________________________________

Could someone please tell me if what I did in the other parts of this long problem are correct?

And also I need help on the last part g) where I have to find the magnitude and velocity of the ball right before it hits the ground.

THANK YOU

2. Nov 10, 2007

### catkin

What a beautifully presented post! So disciplined and well laid out it's hard to believe any of it is wrong (and it would take so long to check I'm daunted -- a good reason for not making long posts).

As for part g, "just before" means an infinitesimal, vanishingly small time before so you can use the same time as when it does touch the floor, 1.78. Immediately after that the force from the floor starts affecting it and changing its velocity.

3. Nov 10, 2007

### ~christina~

okay...I guess I'd better not go and do that ..sad..they say they prefer if I show work though..show too much and...get the same response as not showing any work at all.

Thanks for your help on the last part though

PS If anyone is nice enough to check my work please do (I'd greatly appreciate it)

Last edited: Nov 10, 2007
4. Nov 11, 2007

### catkin

You're right and it's not fair but, hey, who said life was fair?!

a) OK

b) OK

c) Not the easiest method. For these velocity, acceleration, and time problems there are 5 variables. Normally 3 are given and one is asked for so the most direct solution is to use an equation that does not use the 5th. I know these as the SUVAT equations (but many people here use a different terminology). See http://en.wikipedia.org/wiki/SUVAT_equations

In this question you are not given time and are not asked for it (until part d). Here's the SUVAT equation without time.
$$v^2 = u^2 + 2as$$
where
v = final velocity
u = initial velocity
a = acceleration
s = displacement

In your terminology that would be
$$V_{fy}^2 = V_{iy}^2 + 2a(S_y - S_{y0})$$

It's easier to leave out the $S_{y0}$ here and add it at the end. A lot of the skilll in working solutions is making them as eas as possible; that usually means making them very simple and elegant.

d) OK

e) You need to use a vertical calculation to determine when the ball reaches the floor, starting with zero velocity from the height you calculated in c ...

f) and g) will change with a re-worked e)

h) Right idea but isn't Vx a velocity and the question asks for a force?

I'm glad I found time to go through this because you are obviously keen to learn, judging from the work you put in, and there are a couple of mistakes for you to learn from.

5. Nov 11, 2007

### ~christina~

Hm..for part

e) I get the same answer 1.783 sec for time to reach the floor from the wall...

I went and did this:
t max= 0.8630s

Voy= 0
Vy= 0
Soy= 4.15m ====> used the height which included the added part for the max height
Sy= 0
a= -9.8m/s

t= ?

Sy= Soy + Voy (t) + 0.5 a t^2

$$\sqrt{} -4.15/ -4.9 = t$$

t= .920s to reach the floor from max height

t from wall to floor = 0.8630s + 0.920s= 1.783s

~did I do something wrong I get the same value 1.783 just off by my old answer by 0.003s??~

Plugging in the values for the distance

Sx= Sox + Vxt
Sox= 0
Sx= ?
Vx= 9.0 cos 70= 3.08m/s
t= 1.783s

Sx= 3.09m/s(1.783)

Sx= 5.49m

Part f)
t= 1.783s

Part g)

~actually they said that I have to find the magnitude and direction of the velocity when the ball hits the ground...not the force

Vx= 3.078 since it doesn't change and

Vy= Voy+ at

Vy= 9.0sin 70 + -9.8(1.783)

Vy= -9.02m/s

V= $$\sqrt{} (3.078m/s)^2 + (-9.02m/s)^2= 9.53m/s$$

direction = > $$theta= tan^{-1}$$ (-9.02m/s / 3.078m/s ) = -71.16 deg

so velocity is

-9.02m/s -71.16 deg

Thanks for all your help catkin I really, really appreciate you going through my work

6. Nov 12, 2007

### catkin

Going step by step ...

e)

OK to 4.15 m above floor (the correct answer to part c)

OK to t = .920s to reach the floor from max height

OK to adding part d answer (I haven't checked part d itself)

No to ~did I do something wrong I get the same value 1.783 just off by my old answer by 0.003s??~. Now I've looked more closely at what you did
$$Sy= Soy + Voy (t) + 0.5 a*t^2$$
was perfectly correct. Sorry for not looking closely enough the first time. Not only is your method correct, it is better than my method because it does not rely on the earlier answers so will not be wrong if there was an earlier error.

Mmm to "Sx= 3.09m/s(1.783)" because one quantity has units and the other doesn't. The numbers themselves are correct. Personally I find it confusing to put units in numerical calculations but the choice is a matter of personal taste.

f)

The question is a bit strange; I don't see how part e can be solved without getting this answer first.

g)

Looks fine (I haven't checked the numbers)

h)

7. Nov 12, 2007

### ~christina~

h) isn't h to draw a free body diagram of the ball at the max height??

at the max height the velocity component only includes the Vx and no Vy and no gravity acceleration at that point if I'm not incorrect.

8. Nov 12, 2007

### catkin

As I understand free body diagrams they only have forces, not velocities ... maybe I'm wrong; maybe you have been taught differently.

9. Nov 12, 2007

### ~christina~

Oh..well I'm not sure..then..what force would be on the ball...
I know there isn't any gravity..

there can't be no force on it...

10. Nov 12, 2007

### catkin

Why isn't there any gravity?

11. Nov 12, 2007

### ~christina~

Um..oh..yep there is gravity...and I think that would be it

I guess it's confusing me since I get mixed up and start thinking velocity is a force..