A tennis ball rebounds straight up from
the ground with speed 4.8 m/sec.
How high will it climb?
It will climb until its
speed drops to zero!
v  v0  at
g = -9.8 m/sec2
final speed = 0
t  - (4.8m / s) ( -9.8m / s )
 0.490 sec
2
and in that much time
it will rise a distance
2
d  v0t  12 at
d  (4.8m / s )(0.490s )
-
2
2
1
(9.8m / s )(0.490s )
2
d  1.176 m
or
since time up = time down
the distance it falls
from rest in 0.49 sec:
d  at
1
2
2
d  12 (9.8m / s 2 )(0.490s )2
d  1.176 m
A tennis ball rebounds straight up from
the ground with speed 4.8 m/sec.
How high will it climb?
It will climb until all its kinetic energy has
transformed into gravitational potential energy!
1
2
mv 2
= mgh
2
v
 h
2g
2
(4.8m / s )
h 
2
2(9.8m / s )
= 1.176 meter
h
A skier starts from rest down a slope with
a 33.33% grade (it drops
a foot for every 3
F
horizontally).
W
3h
The icy surface is nearly frictionless. How fast is he
traveling by the bottom of the 15-meter horizontal drop?
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:
The long way:
F
h
h



2
2
W
h 10
9h  h
1
F
W
a
10
1
10
1
g
10
h  9h  47.434 meters long
2
2
So it takes him d  1 at
t
 2d / a
2
The slope is
2
2
t  2( 10h) /( g / 10 )  20h / g  5.533 sec
During which he builds to a final speed:
v  v0  at  0  (
g
10
)(
20 h
g
)  17.146 m / sec
h
A skier starts from rest down a slope with
a 33.33% grade (it drops
a foot for every 3
F
horizontally).
W
3h
The icy surface is nearly frictionless. How fast is he
traveling by the bottom of the 15-meter horizontal drop?
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.
The short way:
mgh 
v
v 
2
1
2
mv
2
 2 gh
2
2(9.8m / s )15m
= 17.146 m/sec
D
C 2h
B
A
h
1.5h
h
½h
3h
3h
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?
We know rubber tires are easily deformed
by the enormous weight of the car they
support…but not permanently. They
regain their round shape when removed.
This “spring-like”
resiliency explains
the rebound of all
sorts of balls.
Racquetball
rebounding
from
concrete.
Tennis ball
rebounding
from
concrete.
Airtrack bumper carts:
Notice that if the spring bumbers
reverberate (ring) this would have
to represent some energy that did
not get returned to forward motion!
Unlike the stored potential of the compressed
bumpers, this is not a “potential” energy
that can ever be recovered as kinetic energy.
This represents a fractional
loss in kinetic energy!
A purely ELASTIC COLLISION is defined
as one which conserve kinetic energy.
Look how a
racquetball
still undulates
after leaving
the floor!
These
vibrations
are a
wasted
form of
energy!
Notice this tennis ball’s rebound:
Velocity
on
rebound
less than
velocity
the ball
hit the
floor with!
If the collision with the floor is not perfectly elastic, the
ball 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!
Golf Ball
Dropped
with
potential
energy
mg(h1)
h1
…equal to
the kinetic
energy it
strikes
ground
with
h2
rebounding with a slightly
smaller kinetic energy
equal to the potential energy
mg(h2) it will climb up to.
mgh2 h2
outgoing kinetic energy


incoming kinetic energy
mgh1 h1
Where did the “missing” energy go?
Golf balls (and many industrial materials)
are rated on their “coefficient of restitution”
outgoing speed
incoming speed
Since our ratio:
h2
h1

2
1
mv2
2
2
1
mv1
2
outgoing speed 
incoming speed

v2
2
v1
h2
h1
2
 v2 
 
v 
 1
2
Examples of some
coefficients of restitution
from a concrete surface
golf ball
billiard ball
hand ball
tennis ball
hollow plastic ball
glass marble
wooden ball
steel ball bearing
0.858
0.804
0.752
0.712
0.688
0.658
0.603
0.597
Results in fact are different for
rebounding off a steel or glass plate.
The coefficient of restitution of the
surface hit (its elasticity) can make
a difference as well.
The elastic netting of a tennis racquet
is more “lively” than the clay surface
of a court.
There are interactions that do meet
the ideal of totally elastic collisions!
Nuclear and high energy particle reactions
Electron beam scattering off other electrons.
Negatively charged electron
scattered by a fast-moving
negatively charged muon
Here a positively charged
alpha particle colliding with
a proton (hydrogen nucleus)
v
v
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 direction
is equivalent to a collision between
a parked car of this same model,
and a second hitting it head-on at
A)
B)
C)
D)
32.5 mph
65 mph
97.5 mph
130 mph
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)
B)
C)
D)
30 mph
35 mph
65 mph
100 mph
So when rebounding off surfaces that are not stationary
it’s the relative speed between them that matters.
speed of separation
coefficient of restitution = speed of approach
0.55 baseball on wooden bat
0.44 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?
speed of separation
0.55 
140 mph
speed of separation  0.55  140 mph
 77 mph
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 pendant
at the end of a delicate gold chain
when a fragile link snaps off.
Viewed from above,
it snaps at the
position illustrated.
Which of the paths
shown best represents
B
the likely trajectory
of the pendant?
A
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).
C
D
E
A force acting FORWARD on a moving object
will increase its speed.
A force acting BACKWARD on a moving object
will 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 dummy
seated in
stationary vehicle
Car accelerates:
suddenly lurching
forward
Seat accelerates
forward, compressing
back cushion
against driver
Contents of car
settles into
the 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.
Select the best graphical representation
of the ship’s speed:
to
to+t
to
to+t
to
to+t
to
to+t
How do you tell when you are completely stopped?
Sitting in class, at “rest” in your seat
you 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
x  v x t  (8m / s )(0.1 s )  0.8 m
and down
y  gt  (9.8m / s )(0.1s)
1
2
2
2
1
2
 0.049 m
but it has also built vertical speed:
2
v y  (9.8 m / s )(0.1 s )  0.98 m / sec
v0= 8 m/sec
vy = 0.98 m/sec
It has turned!
2
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 dice
hanging outward on these
Swings is evidence that you’re accelerating inward!
Centripetal force:
2
F=
F
Turning right on level ground relies entirely on friction
A banked curve let’s the car’s own weight help
negotiate the turn
SOME ANSWERS
Question 1
B
h