# tennis ball example/coefficient of restitution

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• July 15th 2006, 07:56 AM
dopi
tennis ball example/coefficient of restitution
a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

i was told to find expressions for the height and velocity of the ball for which i did shown below

Vx = Vo*cos(alpha)

Where :
Vo is the initial velocity
alpha is the launch angle
Vx is the x component of velocity

Vy = Voy*sin(alpha) - g*t

From the pythagorean theorem V = sqrt(Vx^2+Vy^2)

X = Vo*cos(alpha)*t
Y = Vo*sin(alpha)*t - 1/2*g*t^2

Assuming that i started at (x,y) = (0,0) i will then end upat the point (D,h)

that was the first part of the quesiton...but the part im stuck on is ...auusuming that the coefficient of restitution between the tennis ball and the wall is e, i need to write down an expression for te the velocity of the ball immediately after impact......is there anyone that can help ???thankz
• July 15th 2006, 08:22 AM
malaygoel
Quote:

Originally Posted by dopi
a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

i was told to find expressions for the height and velocity of the ball for which i did shown below

Vx = Vo*cos(alpha)

Where :
Vo is the initial velocity
alpha is the launch angle
Vx is the x component of velocity

Vy = Voy*sin(alpha) - g*t

From the pythagorean theorem V = sqrt(Vx^2+Vy^2)

X = Vo*cos(alpha)*t
Y = Vo*sin(alpha)*t - 1/2*g*t^2

Assuming that i started at (x,y) = (0,0) i will then end upat the point (D,h)

that was the first part of the quesiton...but the part im stuck on is ...auusuming that the coefficient of restitution between the tennis ball and the wall is e, i need to write down an expression for te the velocity of the ball immediately after impact......is there anyone that can help ???thankz

I think it will end up at (D,-h).
Let $t_1$ be the total time of flight.
You can use your equation:Y = Vo*sin(alpha)*t - 1/2*g*t^2
If t= $t_1$, then Y=-h.
Rest you can do yourself.

Keep Smiling
Malay
• July 15th 2006, 08:31 AM
dopi
expression/coefficient of restittuon
Quote:

Originally Posted by malaygoel
I think it will end up at (D,-h).
Let $t_1$ be the total time of flight.
You can use your equation:Y = Vo*sin(alpha)*t - 1/2*g*t^2
If t= $t_1$, then Y=-h.
Rest you can do yourself.

Keep Smiling
Malay

i dont understand how that will help me to find an expression for the velocity of the ball immediately after the impact, when the coefficient of restitution between the ball and the wass is e.????
thankz
• July 15th 2006, 09:03 AM
malaygoel
Quote:

Originally Posted by dopi
a tennis ball is hit from a height h above the ground with a speed V and at an angle alpha to the horizontal. it hits a wall at a horizontal distance d away . Air resistance is negligible

i was told to find expressions for the height and velocity of the ball for which i did shown below

Vx = Vo*cos(alpha)

Where :
Vo is the initial velocity
alpha is the launch angle
Vx is the x component of velocity

X = Vo*cos(alpha)*t

You have done good work, I will use two of your equations(see quote).
To find the velocity after impact, we need to know two things:
coefficient of restitution(which we know)
velocity before impact(which is our focus now)

Velocity before impact can be divided into two components:
horizontal component(which you know $V_ocos\alpha$)
vertical component--- for it you need to know the total time of flight[t1](since it depends on t)

You could find t1 using the equation:
X = Vo*cos(alpha)*t
you get $t_1=\frac{D}{V_0cos\alpha}$

Keep Smiling
Malay
• July 15th 2006, 01:46 PM
topsquark
Quote:

Originally Posted by dopi
i dont understand how that will help me to find an expression for the velocity of the ball immediately after the impact, when the coefficient of restitution between the ball and the wass is e.????
thankz

The definition for the coefficient of restitution is:
$KE_f = eKE_0$

So if you know the KE before collision (KE0) you can find the KE after collision (KEf). This also works for multibody collisions where the KE's in the formula are the total KE for the system.

By the way, you say you are looking for the velocity after collision. That isn't correct. You can only get the speed after collision using this equation. What's going to happen to the velocity is anyone's guess because we don't know what effect the collision will have on the individual components of velocity. (The only example I can think of where you CAN get the final velocity is if the collision is "head-on." Then the velocity after impact will be directly opposite the velocity before impact.)

-Dan
• July 16th 2006, 03:09 AM
CaptainBlack
Quote:

Originally Posted by topsquark
The definition for the coefficient of restitution is:
$KE_f = eKE_0$

Not according to my sources, which seem to think it is the ratio of the after
to before velocity component normal surface.

RonL
• July 16th 2006, 03:14 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Not according to my sources, which seem to think it is the ratio of the after
to before velocity component normal surface.

RonL

(sigh) Different books, different definitions. Gotta love it. :mad:

-Dan
• July 16th 2006, 04:33 AM
malaygoel
Heelo Topsquark!
You said: $KE_f=eKE_o$
this implies that e is the ratio of squares of final and initial velocities. Do you want to say that?

Keep Smiling
Malay
• July 16th 2006, 08:23 AM
topsquark
Quote:

Originally Posted by malaygoel
Heelo Topsquark!
You said: $KE_f=eKE_o$
this implies that e is the ratio of squares of final and initial velocities. Do you want to say that?

Keep Smiling
Malay

Well, that's how I learned it anyway, I think. (Sigh) I went back to the book I thought I had gotten it from (Serway's intro Physics book) and couldn't even find it! Seeing as I can't find it in my Graduate Mechanics book either, I obviously learned it from some out of book notes. I could simply be remembering it wrong. :(

-Dan
• July 16th 2006, 09:40 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Not according to my sources, which seem to think it is the ratio of the after
to before velocity component normal surface.

RonL

I propose we do an experiment. Let us throw a ball gently on the patio so
it bounces and has forward motion. If Topsquark is right the ball will stop
bouncing with no residual velocity along the patio. If I am right it will continue
rolling along the patio with approximately the same speed as the original
horizontal component of the velocity (some loss to friction and conversion
of translational to rotational energy), yes?

I've done this experiment, and so know the answer :D

RonL

PS my spell checker wants to change TopSquark to pipsqeak! :eek:
• July 17th 2006, 04:03 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I propose we do an experiment. Let us throw a ball gently on the patio so
it bounces and has forward motion. If Topsquark is right the ball will stop
bouncing with no residual velocity along the patio. If I am right it will continue
rolling along the patio with approximately the same speed as the original
horizontal component of the velocity (some loss to friction and conversion
of translational to rotational energy), yes?

I've done this experiment, and so know the answer :D

RonL

PS my spell checker wants to change TopSquark to pipsqeak! :eek:

Good point. ("pipsqueak" Shakes his head in despair!)

-Dan
• July 20th 2006, 05:06 AM
malaygoel
Quote:

Originally Posted by CaptainBlack
I've done this experiment, and so know the answer :D

I haven't performed the experiment, but let me know the answer.

Malay
• July 20th 2006, 05:14 AM
topsquark
Quote:

Originally Posted by malaygoel
I haven't performed the experiment, but let me know the answer.

Malay

The point is, using my definition, the ball would eventually stop even over a frictionless surface. (e is always less than 1 in reality) If the ball had a horizontal component of velocity Netwon's 1st says it should always be in motion. A contradiction, so my definition can't be correct.

-Dan
• July 20th 2006, 05:16 AM
malaygoel
Quote:

Originally Posted by topsquark
The point is, using my definition, the ball would eventually stop even over a frictionless surface. (e is always less than 1 in reality) If the ball had a horizontal component of velocity Netwon's 1st says it should always be in motion. A contradiction, so my definition can't be correct.

-Dan

It means that coefficient of restitution gives the ratios of final and initial normal components of velocities.
Right???

Malay
• July 20th 2006, 05:27 AM
topsquark
Quote:

Originally Posted by malaygoel
It means that coefficient of restitution gives the ratios of final and initial normal components of velocities.
Right???

Malay

Yes. According to CaptainBlack (and I no longer have any reason to doubt his definiton) the coefficient of restitution is a ratio between the final and initial "vertical" components of the velocity. (Vertical referring to a normal to the surface of impact.)

-Dan
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last