Download 05 Momentum Chapters 5_-_momentum_combined

TableofContents
Momentum ............................................................................................................................................... 132 Impulse .................................................................................................................................................. 133 Conservation of Momentum .................................................................................................................... 137 Elastic vs. Inelastic Collisions ................................................................................................................. 137 Bouncing Collisions ............................................................................................................................... 139 Explosions.............................................................................................................................................. 141 Glancing Collisions ................................................................................................................................ 143 Momentum Wrap‐up ............................................................................................................................... 146 AP Physics – Momentum
Momentum is a common word – you hear it every so often, but not like every day, normally.
Sportscasters will say that a team has a “lot of momentum”. Or a news anchor person might
say “the need for federal regulation of the qualification of physics teachers is gaining
momentum”. A kind of related thing would be the idea that something is momentous (which
would mean that it was important).
Momentum sounds important too. If something has momentum, well, that’s got to be a big deal.
It has MOMENTUM!
Well, forget all that! In physics momentum is simply the velocity of an object multiplied by its
mass.
When something is at rest it has a certain quality which is very different from the one it has when
it is moving. You would feel safe stepping in front of a locomotive and pushing on its nose – if
it were at rest. But you would not want to do this if it was moving. Especially if it was moving
fast. This is because of its momentum.
Momentum can be thought of as sort of like inertia in motion. Recall that inertia is the property
of matter that is responsible for Newton’s first law. It resists changes in a system’s motion.
Anyway, the thing with momentum is that when things are moving, the inertia seems to get
amplified or something. See, when something gets underway, the faster it goes the more it
seems to want to keep moving.
Here’s an important thing about momentum:
Momentum is a vector quantity.
Momentum in mathematical terms is:
p  mv
Momentum Equation
Where p is momentum, m is mass, and v is velocity.
Momentum has units of:
kg  m
s
or
N s
To have momentum, an object must have a velocity - be moving. A strand of spider web silk
drifting on the air currents has more momentum than a mile long coal train that is waiting on a
siding.

What is the momentum possessed by a car with a mass of 2 350 kg if it is traveling at 125
km/h?
The velocity must be expressed in units of m/s, so we do the conversion:
131
125
km  1 h  1 000 m 


 = 34.72 m/s
h  3 600 s  1 km 
m

p  mv  2 350 kg  34.72  
s

8160
kg  m
s
Because momentum is a vector, we can always break it down into its horizontal and vertical
components:
p x  mvx
p y  mv y
Impulse: Today, we use F = ma as a quick definition of the second Law of motion. But
Newton did not express it in that form. Instead he said that a force was the rate of change of
momentum. Instead of F = ma he said:
F
change in momentum
time interval
or
F
p
t
It turns out that this is the same as F = ma – if you do some algebra type fiddling around.
F
p

t
mv  mvo
t
Recall that
v  vo  at
So
F
m  v  vo 
t

m  v  vo 
t
We can plug this into the equation we developed:

m   vo  at   vo 
t
F
m  vo  at  vo 
t
 ma
F  ma
Thus:
Now let’s go back to Newton’s original equation:
F
p
t
Which is:
Multiply both sides by t:
F t  p
F t  m  v  vo 
The quantity on the left, Ft, is called impulse.
F t  Impulse
132
The AP Test has this equation which is used for impulse:
J  F t  p
Here J is the impulse, FT, and p is the change in momentum.
The impulse is the applied force multiplied by the time over which it acts. This quantity is equal
to the change in momentum of a system.
We can say that impulse changes momentum.
We can look at two events, both involving a car, a 1 200 kg car. We’ve got this car that is
traveling along at 75 mph on the interstate. That’s about 125 km/h or roughly 35 m/s. We want
to stop the car. This means that the car will undergo a change in momentum of:
kg  m
 m
p  mv  1 200 kg  35   42 000
s
s

This will be the change in momentum no matter what. Now we’ll bring in the impulse, J recall
that it is really (Ft). If we use the brakes to come to a nice, controlled stop, it will take a fairly
long time to stop the car – maybe 12 seconds. But if we run into a humungous, massive boulder
that has rolled down the mountain onto the roadway, the car will come to a stop in an extremely
short amount of time. Less than a second.
In both cases the car will have the same change in momentum. It will also undergo the same
impulse. The difference will be in the time and force that it takes to stop the car.
Using the brakes, the force acting on the car will be small. Using the boulder, the force that acts
on the car will be enormous. Can you see why this is so?
Let’s use some numbers with our car example.
Case 1: Using the brakes:
F  42 000
J  F t  p
F
kg  m  1 

 
s  12.0 s 
p
t

m  v  vo 
t
3 500 N
Case 2: Hitting the boulder:
F  42 000
kg  m  1 

 
s  0.010 s 
4 200 000 N
Stopping the car with brakes requires 3 500 N, stopping the car with the boulder has a force of
over four million Newtons! Thee force is about 12 000 time bigger! This is why stopping a car
133
with brakes is a safe, rational thing to do, but running into a boulder to effect the same velocity
change is crazy.
So, if we apply a large force for a short time we can generate a momentum change of the same
magnitude as having a small force for a long time.
F

t

F
t
What impulse is required to stop a 0.25 kg baseball traveling at 52 m/s? (b) If the ball is in
the fielder’s mitt for 0.10 seconds as it is being stopped, what average force was exerted on
the ball?
(a)
J  F t  p

m 

J  m  v  vo   0.25 kg  0   52   
s 


(b) Now we can find the force:
F  13 kg
m 1 

 
s  0.10 s 
F t  p
F
13 kg
m
s
p
t
130 N
Legal Trivia:




Jaguar images and costumes were outlawed by the Catholic Church in the seventeenth
century because of their association with Indian religion, militia, and politics.
The minimum age for marriage of Italian girls was raised by law to 12 years in 1892.
Japanese bowing carries different meanings at different angles.
- A bow at an angle of five degrees means "Good day" (simple greeting).
- A bow at an angle of fifteen degrees is also a common salutation, a bit more formal it
means "Good morning."
- A bow at an angle of thirty degrees is a respectful bow to indicate appreciation for a kind
gesture.
- A bow at a forty-five-degree angle is used to convey deep respect or an apology.
The State of Nevada first legalized gambling in 1931. At that same time, the Hoover Dam
was being built and the federal government did not want its workers (who earned 50 cents
an hour) to be involved with such diversions, so they built the town of Boulder City to
house the dam workers. To this day, Boulder City is the only city in Nevada where
gambling is illegal. Hoover Dam is 726 feet tall and 660 feet thick at its base. Enough
rock was excavated in its construction to build the Great Wall of China. Contrary to old
wives' tales, no workers were buried in the dam's concrete.
134















Japanese rules for the proper use of chopsticks are many. Improper use includes
wandering the chopsticks over several foods without decision, and is called mayoibashi.
The unforgivable act of licking the ends of chopsticks is called neburibashi. Lack of
chopstick etiquette is strictly taboo.
King James VI and the Privy Council issued an edict in 1603 banning the use of the
surname MacGregor.
The penalty for conviction of smuggling in Bangladesh is death.
In Russia, buying carnations or roses is a prerequisite for a first date. They must be given
in odd numbers, because flowers given in even numbers are reserved for funerals.
A 1989 law in Florida forbids the release of more than ten lighter-than-air balloons at a
time. This is to protect marine creatures that often mistake balloons for food and can
suffer intestinal injuries if they eat the balloons.
A bride stands to the groom’s left at a wedding so that his sword hand would be free.
Apparently Anglo-Saxon brides were often kidnapped before a wedding and brawls were
common. That’s also why the best man stands with the groom; the tribe’s best warrior was
there to help the groom defend the bride.
A charming wedding custom in early Yorkshire, England, involved a plate holding
wedding cake. It was thrown out of the window as the bride returned to her parental home
after the wedding. If the plate broke, she would enjoy a happy future with her husband. If
the plate remained intact, her future was bleak.
A couple living together for two years in Russia is considered married. This is called a
citizen marriage.
A law in Illinois prohibits barbers from using their fingers to apply shaving cream to a
patron's face.
The U.S. interstate highway system requires that 1 mile in every 5 must be straight. These
sections can be used as airstrips in a time of war or other emergencies.
The United States Supreme Court once ruled Federal income tax unconstitutional. Income
tax was first imposed during the Civil War as a temporary revenue-raising measure.
The wedding custom of “something old, something new, something borrowed, something
blue” dates to the Middle Ages. The “old” thing was a personal gift from the bride’s mom
to make a bond to the bride’s old life. The “new” item symbolized hope for the future and
the newly formed family. The “borrowed” item was a gift from a happily married woman
that would carry some of the woman’s happiness into the new marriage. The something
“blue” came from two sources that had similar meanings. To ancient Romans, maidens
wore blue to show fidelity and modesty, and to Christians, blue was linked to the purity of
the Virgin Mary.
Tourists need to be aware that, when traveling in Germany, a screwing gesture at one's
head is a strong symbol, meaning "You're crazy." Often used by drivers on the autobahn
to comment on the driving skills of other travelers, this gesture can get you arrested. The
same gesture is used in Argentina.
Trap regulations in California for lobster fisherman require an escape port for undersized
lobsters.
A local ordinance in Atwoodville, Connecticut, prohibits people from playing Scrabble
while waiting for a politician to speak.
135
AP Physics – Conservation of Momentum
We’ve studied the law of conservation of energy and found it to be a very powerful thing in the
world of physics. Another very powerful conservation law has to do with momentum. This is
the law of conservation of momentum.
Momentum is conserved in an isolated system.
“Isolated system” means that there are no external forces acting on the thing.
The type of interaction that involves changes to momentum that we will deal with are called
collisions. (Although not all of them are what we think of as a collision, as we shall see.)
The law of conservation of momentum means that for a collision in an isolated system,
momentum must be conserved. This means that the total momentum of the system before the
collision must be equal to the total momentum after the collision.
p  p'
Here p is the momentum before the event and p ' is the
momentum after the event.
Collisions can be very complicated. They can get so complicated that they can’t really be solved
as a matter of fact. To simplify things, we will deal with collisions involving two bodies. We
will also deal with one-dimensional collisions (like between balls rolling along a track), or else
two dimensional collisions that occur at right angles.
This helps out a great deal. Imagine trying to deal with a violent collision between two cars.
Before the collision you have two bodies in motion that have momentum. After the collision
you’ve got thousands of them – all the bits and pieces of the cars flying off in every imaginable
direction.
Collisions can be classified according to the energy interaction that takes place:
Elastic collision  kinetic energy is conserved
Inelastic collision  kinetic energy is not conserved
Perfectly inelastic collision  objects stick together and have the same velocity.
Application of Conservation of Momentum:
On the AP Test, you will only be
given the equation for momentum, p = mv. From this you will have to derive the formula for a
two-body collision. Here’s how to do it.
We begin with a two-body collision where each object has some velocity prior to the collision.
Both objects have momentum equal to the mass times the velocity.
136
The total momentum before the collision is:
p  m1v1  m2v2
After the collision the velocities will have changed. The total momentum will be:
p '  m1v1 ' m2v2 '
We know that the momentum before must equal the momentum after:
p  p'
so
m1v1  m2v2  m1v1 ' m2v2 '
So here is an equation which will work for any two-body collision:
m1v1  m2v2  m1v1 ' m2v2 '
Perfectly Inelastic Collisions:
These are collisions where the two objects stick
together after they collide.
The key thing to remember here is that after the collision both objects stick together and have the
same velocity.

We see a 15 000 kg railroad car moving along the track at 2.5 m/s. It collides and couples
with a stationary 12 500 kg car. What is the new velocity of the two cars?
p  m1v1
p '  m1v1 ' m2v2 '
The velocity of each is the same (they’ve joined together into a “single body”) so we can
simplify the equation for the final momentum:
p '   m1  m2  v '
Now we set the two equations equal to each other:
m1v1   m1  m2  v '
And we get our equation. Now we solve for the final velocity:
v' 
m1v1
 m1  m2 
m
1

 15 000 kg  2.5 
s  15 000 kg  12 500 kg 


1.4
m
s
137

2 mud balls collide in a perfectly elastic collision. One has a mass of 4.0 kg, the second has a
mass of 3.5 kg. The first one has an initial velocity of 3.4 m/s, the second has an initial
velocity of – 4.8 m/s. What is their velocity after the collision?
p  m1v1  m2v2
p '   m1  m2  v '
m1v1  m2v2   m1  m2  v '
m
m


4.0 kg  3.4   3.5 kg  4.8 
s
s


vf 
4.0 kg  3.5 kg
v' 
m1v1  m2v2
 m1  m2 

 0.43
m
s
The negative sign means that they are going in the opposite direction from what our first mud
ball was doing. We chose its direction as the positive direction.
Bouncing Collisions: When two objects collide and bounce off each other, there are two
possible energy interactions. Kinetic energy can be conserved or not conserved. If it is
conserved, then we have a perfectly elastic collision. The other possibility is that some of the
kinetic energy is transformed into other forms of energy. This is mostly what happens in the real
world.
The type of problem you will be tasked with solving will have two objects bouncing off each
other. You will know three of the four velocities and solve for the fourth. You may be asked
whether kinetic energy is conserved and, if it isn’t, how much kinetic energy is lost.

Two balls hit head on as shown, what is the final velocity of the second ball if the first one’s
final velocity is –1.50 m/s?
v2 = 1.30 m's
v1 = 2.30 m's
m2 = 1.85 kg
m1 = 1.50 kg
The general conservation of momentum equation is:
m1v1  m2v2  m1v1 ' m2v2 '
All we have to do is solve for v2’.
v2 ' 
m1v1  m2v2  m1v1 '
m2
138
v2 ' 

1.50 kg   2.30 ms   1.85 kg   1.30 ms   1.50 kg   1.50 ms 



1.85 kg




1.78
m
s
Two balls roll towards each other and collide as shown. The second’s ball velocity after the
collision is 3.15 m/s. (a) What is the velocity of the first ball after the collision? (b) How
much kinetic energy is lost during the collision?
v2 = 8.60 m's
v1 =4.75 m's
m2 = 2.85 kg
m1 = 2.50 kg
(a)
v1 ' 
m1v1  m2v2  m1v1 ' m2v2 '
v1 ' 
m1v1  m2v2  m2v2 '
m1
 2.50 kg   4.75 ms    2.85 kg   8.60 ms    2.85kg   3.15 ms 



2.50 kg




 8.64
(b) Difference in kinetic energy before and after collision:
1
1
2
2
K  m1  v1   m2  v2 
2
2
Kinetic energy before the collision:
2
1
m 1
m


K   2.50 kg   4.75    2.85 kg   8.60 
s 2
s
2


2
1
1
2
2
K '  m1  v1 '  m2  v2 '
2
2
Kinetic energy after the collision
2
1
m 1
m


K '   2.50 kg   8.64    2.85 kg   3.15 
s 2
s
2


K  K ' K
 133.6 J  107.5 J
 133.6 J

2
 107.5 J
26.1 J
What happened to this energy? Where did it go? Well, it was converted into some other form of
energy, most likely heat.
139
m
s
Explosions:
The final type of collision is the explosion. Not like you get with a bomb. That
would be one very hard problem. No, we want us a nice, simple two body explosion. An
example you ask? Okay, you stand on a skateboard and throw a bowling ball away from you.
This would be a simple two-body explosion.
The idea of an explosion is that you have two bodies that are at rest – this means that they have
no momentum, don’t it? Anyway, the explosion takes place and the bodies end up moving away
from each other. Because momentum is conserved, their final momentum, when added together,
must still equals zero; the momentum they started out with.

A 23 kg girl is standing on a low friction 2.0 kg cart. She is at rest. She then throws a heavy
8.9 kg rock, giving it a speed of 7.5 m/s. What is the final speed of the girl/cart system?
m1v1  m2v2  m1v1 ' m2v2 '
v1 ' 
Solve for the speed of the girl:
v1 ' 
m2v2 '
m1
0  m1v1 ' m2v2 '
becomes:
m2v2 '
m1
1
m 

 8.9 kg  7.5  
s   23 kg  2.0 Kg


 

 2.7
m
s
The negative sign means that the girl is going in the opposite direction from the rock.
AP Test Question:
Two identical objects A and B of mass M move on a one-dimensional,
horizontal air track. Object B initially moves to the right with speed vo. Object A initially
moves to the right with speed 3vo, so that it collides with object B. Friction is negligible.
Express your answers to the following in terms of M and vo.
a. Determine the total momentum of the system of the two objects.
p  m1v1  m2v2
b.
 M  3v0   Mv0 
4 Mv0
A student predicts that the collision will be totally inelastic (the objects stick together on
collision). Assuming this is true, determine the following for the two objects immediately
after the collision.
i. The speed
ii. The direction of motion (left or right)
m1v1  m2v2   m1  m2  v '
v' 
M v0  M  3v0 

M M
4 Mv0
2M
v' 

m1v1  m2v2
m1  m2
2 v0
140
Moves right (positive direction).
When the experiment is performed, the student is surprised to observe that the objects separate
after the collision and that object B subsequently moves to the right with a speed 2.5 vo .
c.
Determine the following for object A immediately after the collision.
i. The speed
ii. The direction of motion (left or right)
m1v1  m2v2  m1v1 ' m2v2 '
v1 ' 
v1 ' 
Mv0  M  3 v0   M  2.5 v0 
M
m1v1  m2v2  m2v2 '
m1
 4 v0  2.5 v0

1.5 v0
Object A move to the right
d.
Determine the kinetic energy dissipated in the actual experiment.
1
1
K  m1v12  m2v2 2
2
2
1
2 1
K  M  3v0   Mv0 2
2
2

1
2 1
2
K '  m1  v1 '  m2  v2 '
2
2

1
 M 9 v0 2  v0 2
2
1
2
2
K '  M 1.5v0    2.5v0 
2


 5 Mv0 2

1
 M 2.25 v0 2  6.25 v0 2
2
K  K ' K  5 Mv0 2  4.25 Mv0 2 

 4.25M v0 2
0.75 Mv0 2
141
Glancing collisions: This type of collision takes place in two dimensions.
Collisions on a
pool table are good examples of this type of collision. Momentum still has to be conserved – it’s
the law for crying out loud! But the problems can become fairly complex. One technique to
keep track of what’s going on is to break things into x and y components:
 px   px'
 P   P'
y
y
v1f
v1i
O
O
Before
v2f
After
We will deal with greatly simplified collisions. Basically with right angles and one of the bodies
at rest at the beginning.

An 8.00 kg mass moving east at 15.0 m/s strikes a 10.0 kg mass that is at rest. The 8.00 kg
mass ends up going south at 4.00 m/s. (a) What is the velocity of the second ball?
v2’

v1
Before
v1’
After
(a) We analyze the momentum in the x and y directions.
The x direction:
m1v1x  m2v2 x  m1v1x ' m2v2 x '
becomes
m1v1x  m1v2 x '
142
This is because the second body has no initial velocity so it has no initial momentum in either the
x or y direction. After the collision, the first body has momentum only in the y direction, so the x
direction momentum after the collision involves only the second body.
v2 x ' 
m1v1x
m2
m  1

 8.00 kg 15.0  
s   10.0 kg


m
  12
s

Now let’s look at the y direction:
m1v1 y  m2v2 y  m1v1 y ' m2v2 y '  0  m1v1 y ' m2v2 y '
The first body’s initial motion is only in the x direction, therefore it has no initial momentum in
the y direction. The second body is at rest at the beginning so the total initial y direction
momentum is zero.
m2v2 y '  m1v1 y '
v2 y ' 
m

  8.00 kg   4.00 
s


10.0 kg
m1v1 y '
m2
 3.20
m
s
Now we can solve for the velocity of the second body using the Pythagorean theorem.
2
v2  v2 f x  v2 f y
2
2
m 
m

 12.0    3.20 
s 
s

Now we find the angle :
Trigonometry is just the thing to find the old angle:
tan  
v2 y '
v2 x '
m
s

m
12.0
s
3.20
2

12.4
v2y’
m
s
v2’

  14.9o
v2x’
143

A 2.0 kg block is sliding on a smooth table top. It has a totally inelastic collision with a 3.0
kg block that is at rest. Both blocks, now moving, hit the spring and compress it. Once the
blocks come to rest, the spring restores itself and launches the blocks. They slide off the
right side of the table. The spring constant is 775 N/m. Find: (a) the velocity of the two
blocks after the collision. (b) The distance the spring is compressed. (c) The velocity of the
blocks after they leave the spring. (d) The distance the blocks travel before they hit the deck
after the leave the tabletop. (e) The kinetic energy of the blocks just before they hit the
v = 10.0 m/s
1.2 m
deck.
(a)
m1v1  m1v1 ' m2v2 '
v' 
(b)
m1v1
 m1  m2 
1 2 1 2
mv  kx
2
2
m
1

 2.0 kg 10.0 
s   2.0 kg  3.0 kg 

xv
m
k
 4.0
m
s

 2.0 kg  3.0 kg 
775
kg  m
4.0

m
s
0.32 m
s2 m
(c) If energy is conserved in the spring, they should have the same speed when they leave as
they had going into the spring. So they should be moving at 4.0 m/s to the right.
(d) Find the time to fall:
1
2y
y  at 2 t 
2
a
x  vt
 4.0
2 1.2 m 
m
9.8 2
s
m
 0.495 s  
s
 0.495 s
2.0 m
(e) The kinetic energy of the blocks at the bottom (just before they hit) must equal the potential
energy at the top of table plus the kinetic energy the blocks had before they began to fall.
1
K '  mgy  mv 2
2
m
1
m


 5.0 kg  9.8 2  1.2 m    5.0 kg   4.0 
s
2
s 


2

99 J
144
Examples of Conservation of Momentum:
When you fire a rifle, you experience the consequences of the conservation of momentum. The
bullet is fired out of the thing at a very high velocity. Because momentum must be conserved,
the rifle must end up going in the opposite direction. We call this backwards motion the
“recoil”.
Rockets, which we discussed using Newton’s third law, can also be explained using the
conservation of momentum. The rocket is an explosion type event. It contains a large amount of
fuel. The fuel is ignited and blasts out of the rocket nozzle at a high velocity. The rocket must
accelerate in the opposite direction so that momentum can be conserved.
AP Physics – Momentum AP Wrapup
There are two, and only two, equations that you get to play with:
p  mv
This is the equation for momentum.
J  F t  p
This is the equation for impulse. The equation sheet uses, for some reason, the
symbol J for impulse (the Physics Kahuna has never seen this anywhere else. Oh
well.)
Here is what you are supposed to be able to do.
A. Systems of Particles, Linear Momentum
1.
Impulse and Momentum: You should understand impulse and linear momentum so you
can:
a. Relate mass, velocity, and linear momentum for a moving body, and calculate the
total linear momentum of a system of bodies.
Just use the good old momentum equation.
b. Relate impulse to the change in linear momentum and the average force acting on a
body.
Just use the impulse equation. We did several of these problems. If you have a
collision, time, and a force in a problem, think’ impulse’.
2. Conservation of Linear Momentum, Collisions
145
a. You should understand linear momentum conservation so you can:
(1) Identify situations in which linear momentum, or a component of the linear
momentum, is conserved.
The momentum of an isolated system (no outside forces) is conserved in any
interaction. Period.
(2) Apply linear momentum conservation to determine the final velocity when two
bodies that are moving along the same line, or at right angles, collide and stick
together, and calculate how much kinetic energy is lost in such a situation.
This is the good old inelastic collision problem. We’ve done a bunch. If the
collision happens at right angles, then you are going to have to look at components
of momentum (which you can do since momentum is a vector, right?) Anyway, the
idea is that after the collision both bodies have the same velocity. Therefore:
m1v1  m2v2   m1  m2  v f
If the two bodies are moving at right angles, then it’s a bit more complicated. The
vertical momentum is conserved and the horizontal momentum is conserved, so you
can write equations for the conservation of momentum in the x and y directions.
Then use these to solve for whatever unknown you’ve been presented with.
(3) Analyze collisions of particles in one or two dimensions to determine unknown
masses or velocities, and calculate how much kinetic energy is lost in a collision.
Pretty much all the problems that you will see will involve finding one unknown
velocity. Body A moving at velocity v has an eleastic collision with body A which is
at rest. Body B ends up with a velocity of vf, etc. That sort of thing.
Let’s look at a few test problems. The first one we’ll look at is off 2001 test.

An incident ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of
negligible friction shown above. It makes a head-on collision with a target ball B of mass
0.50 kg at rest at the edge of the table. As a result of the collision, the incident ball rebounds,
sliding backwards at 0.70 m/s immediately after the collision.
146
(a) Calculate the speed of the 0.50 kg target ball immediately after the collision.
We use conservation of momentum to solve this one.
m1v1  m1v1 ' m2v2 '
m1v1  m1v1 '  m2v2 '
 m 
m
1.0 kg 1.4   0.70  
s 
s 

v2 ' 
0.50 kg

0.42
v2 ' 
m1  v1  v1 '
m2
m
s
The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally
and initially strikes the floor at a horizontal displacement d from the point of collision.
(b) Calculate the horizontal displacement d.
This is a projectile motion problem. We know the ball’s horizontal velocity and the height of
the table, so we can easily find the horizontal distance it travels as it falls.
Time to fall: y 
1 2
at
2
t
2y
a

2 1.2 m 
m
9.8 2
s
 0.495 s
The ball has a horizontal velocity of 0.42 m/s (which we just figured out), so the distance d is
simply:
x  vt
 0.42
m
 0.495 s  
s
0.208 m
In another experiment on the same table, the target ball B is replaced by target ball C of mass
0.10 kg. The incident ball A again slides at 1.4 m/s, as shown above left, but this time makes a
glancing collision with the target ball C that is at rest at the edge of the table. The target ball C
strikes the floor at point P, which is at a horizontal displacement of 0.15 m from the point of the
collision, and at a horizontal angle of 30 from the + x-axis, as shown above right.
(c) Calculate the speed v of the target ball C immediately after the collision.
147
This sounds very hard, angles and all that stuff, right? Except we know the distance it traveled
(1.5 m) and we know how long it is in the air before it hits – same as the previous problem time.
So this is a ridiculously simple problem. Using the horizontal distance and the time to fall we
can find the horizontal velocity, which is the velocity it began with, which is the velocity right
after the collision.
Find v:
v
x
t

0.15 m
0.495 s

0.30
m
s
(d) Calculate the y-component of incident ball A's momentum immediately after the collision.
We know that momentum is conserved in the x and y directions. So we can sum momentum in
the y direction. We know that this momentum has to add up to be zero as the ball A had no
initial momentum in the y direction.
0  ma vay  mc vcy
 ma vay  mc vc  sin  

m

  0.10 kg    0.30  sin 30o
s

vay 
0.10 kg


vay 
0.15
mc vc  sin  
ma
m
s
148
From 1996:

Two identical objects A and B of mass M move on a one-dimensional, horizontal air track.
Object B initially moves to the right with speed vo. Object A initially moves to the right with
speed 3vo, so that it collides with object B. Friction is negligible. Express your answers to
the following in terms of M and vo.
a.
Determine the total momentum of the system of the two objects.
We simply add up the momentum of each object to get the total momentum.
ptot  mAv A  mB vB  M  3v0   Mv0 
b.
4 Mv0
A student predicts that the collision will be totally inelastic (the objects stick together
on collision). Assuming this is true, determine the following for the two objects
immediately after the collision.
i.
The speed.
This is an inelastic collision. It is a simple thing to solve:
mAv A  mB vB   mA  mB  v f
vf 
ii.
M  3 v0   Mv0
M  M 

4 Mv0

2M
vf 
mAv A  mB vB
 mA  mB 
2v0
The direction of motion (left or right).
The combined object moves right. All values are positive.
When the experiment is performed, the student is surprised to observe that the objects
separate after the collision and that object B subsequently moves to the right with a speed
2.5 v0 .
c.
Determine the following for object A immediately after the collision.
i.
The speed of object A after the collision.
mAv A  mB vB  mAv A ' mB vB '
m Av A  mB vB  mB vB '  mAv A '
149
vA ' 
m Av A  mB vB  mB vB '
mA
v A '   3 v0   v0   2.5 v0  
ii.

M  3 v0   Mv0  M  2.5 v0 
M
1.5 v0
The direction of motion (left or right).
Object A move right. All values are positive.
d.
Determine the kinetic energy dissipated in the actual experiment.
The kinetic energy dissipated is the change in kinetic energy. So we solve this question
by finding the kinetic energy before the collision and the kinetic energy after the
collision. The difference in the two is the energy dissipated.
1
1
2
2
K i  m  v A   m  vB 
2
2
1
1
2
2
K i  M  3vo   M  vo 
2
2
K i  5Mvo 2
K dissipated  Ki  K f
1
1
2
2
K f  m  v A '   m  vB ' 
2
2
1
1
2
2
K f  M  2.5vo   M 1.5vo 
2
2
K f  4.25Mvo 2
 5Mvo 2  4.25Mvo 2 
0.75Mvo 2
150
From 1992:

A 30-kilogram child moving at 4.0 meters per second jumps onto a 50-kilogram sled that is
initially at rest on a long, frictionless, horizontal sheet of ice.
a.
Determine the speed of the child-sled system after the child jumps onto the sled.
You immediately recognize that this is an inelastic collision.
m1v1   m1  m2  v f
b.
vf
 m
30 kg   4 

m1v1
 s


 m1  m2   30 kg  50 kg 

1.5
m
s
Determine the kinetic energy of the child-sled system after the child jumps onto the
sled.
1
K  mv 2
2
2
1
m

  80 kg  1.5  
s
2

1
  m1  m2  v 2
2
90 J
After coasting at constant speed for a short time, the child jumps off the sled in such a
way that she is at rest with respect to the ice.
c.
Determine the speed of the sled after the child jumps off it.
 m1  m2  v  m2v2
v2 
d.
m2 
 m1  m2  v
v2
 30 kg  50 kg  1.5 ms 

50 kg


2.4
m
s
Determine the kinetic energy of the child-sled system when the child is at rest on the
ice.
2
K total
e.
1
1
m

 m2v2 2   50 kg   2.4  
2
2
s

144 J
Compare the kinetic energies that were determined in parts (b) and (d). If the energy
is greater in (d) than it is in (b), where did the increase come from? If the energy is
less in (d) than it is in (b), where did the energy go?
The child must decelerate in order to stop. So a force negative to the motion of the sled is
applied. This force is applied through a distance resulting in negative work. The sled
receives an equal and opposite (positive) force through a distance resulting in positive work.
151
Work is added to the sled, and work is a change in energy (Work energy theorem). So the
energy of the sled is increased.
From 1995:

As shown, a 0.20-kilogram mass is sliding on a horizontal, frictionless air track with a speed
of 3.0 meters per second when it instantaneously hits and sticks to a 1.3-kilogram mass
initially at rest on the track. The 1.3-kilogram mass is connected to one end of a massless
spring, which has a spring constant of 100 Newtons per meter. The other end of the spring is
fixed.
a.
Determine the following for the 0.20-kilogram mass immediately before the impact.
i.
Its linear momentum.
p   0.20 kg  3.0 m s  
p  mv
ii.
kg  m
s
Its kinetic energy.
1
K  mv 2
2
b.
0.60
K
1
 0.20 kg  3.0 m s 2 
2
0.90 J
Determine the following for the combined masses immediately after the impact.
i.
The linear momentum
pf 
p f  pi
ii.
0.60
kg  m
s
The kinetic energy
p f   m1  m2  v f
vf 
pf
 m1  m2 
Kf 

0.60 kg  m s
 0.40 m s
 0.20 kg  1.3 kg 
1
 m1  m2  v f 2
2

1
 0.20 kg  1.3 kg  0.4 m s 2 
2
0.12 J
152
After the collision, the two masses undergo simple harmonic motion about their position at
impact.
c.
Determine the amplitude of the harmonic motion.
2K f
1
U s  K f  kx 2 x 
2
k

2  0.12 N
N

100 m

 m




0.050 m
d. Determine the period of the harmonic motion.
T  2
m
k
 2
 0.20 kg  1.3 kg 
kg  m 

100 2

s m 


0.77 s
From 1994:

A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough
horizontal section YZ. Block A of mass M is released from rest at point X, slides down the
curved section of the track, and collides instantaneously and inelastically with identical block
B at point Y. The two blocks move together to the right, sliding past point P, which is a
distance l from point Y. The coefficient of kinetic friction between the blocks and the
horizontal part of the track is . Express your answers in terms of M, l, , R, and g.
a.
Determine the speed of block A just before it hits block B.
Ui  K f
b.
1
mv 2
2
v
2 gh

2 gR
Determine the speed of the combined blocks immediately after the collision.
pi  p f
c.
mgh 
m1v1   m1  m2  v f
vf 
M 2 gR
M  M 

1
2 gR
2
Determine the amount of kinetic energy lost due to the collision.
153
K  K Before  K After
1
 M
2

2 gR

2

1
  2M  

2

1
MgR
 2 gR 
K  M  2 gR   M 
  MgR 
2
2
 4 
d.
2
1
MgR
2
The specific heat of the material used to make the blocks is c. Determine the
temperature rise that results from the collision in terms of c and the other given
quantities. (Assume that no energy is transferred to the track or to the air surrounding
the blocks.)
Q  mcT
e.

2 gR 

2 
T 
Q
mc
1
 1
  MgR 
2
 2Mc
gR
4c

Determine the additional thermal energy that is generated as the blocks move from Y
to P
Work done by friction = heat generated
W  fs
f  N
W   Ns    2 Mg  l

2 Mgl
154