a tennis ball rebounds straight up from the ground with speed 4.8 m/sec. how high will it climb? it...
TRANSCRIPT
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