A tennis ball rebounds straight up from the ground with speed 4.8 m/sec.

How high will it climb?

atvv +=0

It will climb until its speed drops to zero!

final speed = 0g = 9.8 m/sec2

sec

smsmt

490.0

)/8.9()/8.4( 2

=−−=

and in that much timeit will rise a distance

2

21

0attvd +=

22

21 )490.0)(/8.9(

)490.0)(/8.4(

ssm

ssmd

−

=

m d 176.1=

orsince time up = time down

the distance it falls from rest in 0.49 sec:

22

21 )490.0)(/8.9( ssmd =

2

21 atd =

m d 176.1=

A tennis ball rebounds straight up from the ground with speed 4.8 m/sec.

How high will it climb?

2

21 mv

It will climb until all its kinetic energy has transformed into gravitational potential energy!

mgh=

h g

v =2

2

)/8.9(2

)/8.4(2

2

sm

sm h =

= 1.176 meter

A skier starts from rest down a slope with a 33.33% grade (it drops

a foot for every 3horizontally).h

3hThe icy surface is nearly frictionless. How fast is he traveling by the bottom of the 15-meter horizontal drop?

W

The long way: The accelerating force down along the slope is in the same proportion to the

weight W as the horizontal drop h is to the hypotenuse he rides:

F

10

1

109 22==

+=

hh

hh

hWF

WF10

1= ga

10

1=

The slope is meters hh 434.479 22 =+ long

So it takes him2

21 atd = adt /22 =

sec ghght 533.5/20)10//()10(2 ===

During which he builds to a final speed:

secm atvvg

hg /146.17))((0 20100

=+=+=

A skier starts from rest down a slope with a 33.33% grade (it drops

a foot for every 3horizontally).h

3hThe icy surface is nearly frictionless. How fast is he traveling by the bottom of the 15-meter horizontal drop?

W

The short way: The gravitational potential energy he has at the top of the slope will convert

entirely to kinetic energy by the time he reaches the bottom.

F

2

21 mv mgh =

gh v 22 =

msm v 15)/8.9(2 2=

= 17.146 m/sec

h

3h 3h

2h

½h

1.5hh

The bottom of the run faces a slope with twice the grade (steepness).

Neglecting friction, the skier has just enough energy to coast how high?

A

B

C

D

We know rubber tires are easily deformedby the enormous weight of the car theysupport…but not permanently. They

regain their round shape when removed.

This “spring-like” resiliency explainsthe rebound of all sorts of balls.

Racquetballrebounding

from concrete.

Tennis ballrebounding

from concrete.

Airtrack bumper carts:

Notice that if the spring bumbersreverberate (ring) this would haveto represent some energy that did

not get returned to forward motion!

This represents a fractional loss in kinetic energy!

A purely ELASTIC COLLISION is defined as one which conserve kinetic energy.

Unlike the stored potential of the compressed bumpers, this is not a “potential” energy that can ever be recovered as kinetic energy.

Look how a racquetballstill undulates after leaving the floor!

These vibrationsare a wasted form ofenergy!

Notice this tennis ball’s rebound:

less thanvelocity the ballhit the floor with!

Velocityon

rebound

If the collision with the floor is not perfectly elastic, theball will not bounce as high as the point of its release!

Notice this is not the same as a ball falling back from the height it was thrown to!

Dropped with

potential energymg(h1)

…equal tothe kinetic

energy itstrikes

ground with

Golf Ball

h1

rebounding with a slightly smaller kinetic energy

equal to the potential energymg(h2) it will climb up to.

h2

outgoing kinetic energyincoming kinetic energy

1

2

1

2

h

h

mgh

mgh==

Where did the “missing” energy go?

2

121

2

221

1

2

mv

mv

h

h=

Golf balls (and many industrial materials)are rated on their “coefficient of restitution”

outgoing speedincoming speed

Since our ratio: 2

1

2

2

v

v=

2

1

2

⎟⎟⎠

⎞⎜⎜⎝

⎛=

v

v

outgoing speedincoming speed

1

2

h

h=

golf ball 0.858billiard ball 0.804hand ball 0.752tennis ball 0.712hollow plastic ball 0.688glass marble 0.658wooden ball 0.603steel ball bearing 0.597

Results in fact are different for rebounding off a steel or glass plate.

Examples of some coefficients of restitutionfrom a concrete surface

The coefficient of restitution of thesurface hit (its elasticity) can make

a difference as well.

The elastic netting of a tennis racquetis more “lively” than the clay surface

of a court.

There are interactions that do meet the ideal of totally elastic collisions!

Negatively charged electron scattered by a fast-movingnegatively charged muon

Here a positively charged alpha particle colliding with a proton (hydrogen nucleus)

Nuclear and high energy particle reactions

Electron beam scattering off other electrons.

The head-on collision between a car traveling 65 mph that strays

across the median and strikes an identical model (and total mass) car

traveling 65 mph in the opposite directionis equivalent to a collision between

a parked car of this same model,and a second hitting it head-on at

A) 32.5 mphB) 65 mphC) 97.5 mphD) 130 mph

v v

A rear-end collision between a car traveling 65 mph and a slower

identical model (and total mass) car traveling 35 mph in front of it

does damage roughly equivalent to a collision between a parked car

of this model, and a second hitting from behind at

A) 30 mphB) 35 mphC) 65 mphD) 100 mph

So when rebounding off surfaces that are not stationary

it’s the relative speed between them that matters.

coefficient of restitution = speed of separationspeed of approach

0.55 baseball on wooden bat0.44 softball on wooden bat

A 100 mph fastball is struck by a bat with swing speed of 40 mph.

How fast is the batted ball?

phm

separation of speed

14055.0 =

mph

mph separation of speed

77

14055.0

=×=

But that’s how fast the ball is racing from the bat,

so that 77 mph is “on top of” the bat’s 40 mph swing.

The ball is moving at 77mph+40mph = 117 mph

You’re twirling a valuable pendantat the end of a delicate gold chainwhen a fragile link snaps off.

Viewed from above, it snaps at the position illustrated.Which of the paths shown best represents the likely trajectory of the pendant?

A

B

C

D

E

The only force that acting in this horizontal plane (gravity pulls down, of course, but that’s vertically) was the tension from the chain. With that gone NO FORCE acts on the pendant, so it should, of course, move off in a straight line (continuing FORWARD).

A force acting FORWARD on a moving objectwill increase its speed.

A force acting BACKWARD on a moving objectwill decrease its speed.

What will a force acting NEITHER FORWARD NOR BACKWARD but perpendicular to the direction of motion

do to a moving object?

Steer it in a new direction(without changing its speed).

Crash test dummyseated in

stationary vehicle

Car accelerates:suddenly lurching

forward

Seat accelerates forward, compressing

back cushionagainst driver

Contents of carsettles intothe moving

frame of the car

At ignition (to), explosive fuel combustion carries the rocket from its stationary position near the space outpost by producing a ~steady thrust F over the t seconds of the brief burn.

to to+t to to+t

Select the best graphical representation of the ship’s speed:

to to+t to to+t

How do you tell when you are completely stopped?

Sitting in class, at “rest” in your seatyou are in fact moving, along with your seat,

the lecture room to which its attached, the building, and the ground its anchored in…

The swaying of follicles within the fluid of the chochlea give us our sensation of motion.

But like the hanging fuzzy dice, these only respond to accelerations, not constant velocity.

The sensations of just how our internal organs (heart, stomach) normally hang within the connective tissue that holds it, also change

under accelerations. You DO feel acceleration in the pit of your stomach!

v0= 8 m/sec

a=g

After the 1st 10th of a second

ball has moved out

m s smtvxx

8.0)1.0)(/8( ===Δ

and down 22

212

21 )1.0)(/8.9( ssmgty ==Δ

m 049.0=

but it has also built vertical speed:

secm s sm vy

/98.0)1.0)(/8.9( 2 ==v0= 8 m/sec

vy = 0.98 m/sec

It has turned!

What is YOUR experience as you sit in a turning car?

Its your own inertia (trying to simply move along a straight line) that feels like it pushes you out. It’s the walls pushing inward that hold you in a circle.

Like the car’s fuzzy dicehanging outward on theseSwings is evidence that you’re accelerating inward!

F = 2

Centripetal force:

F

Turning right on level ground relies entirely on friction

A banked curve let’s the car’s own weight help negotiate the turn

B

h

SOME ANSWERS

Question 1