Download The Law of Conservation of Momentum

PHYSICS SEMESTER ONE
UNIT 6
UNIT 6: MOMENTUM
Relationship to Labs: This unit is supported by Lab 5: Conservation of Momentum (RWSL)
The energy section covered several cases; some where energy is conserved and some where the useful
energy is not conserved as some is lost to friction (heat). In particular, we saw that the conservation of
energy is a very useful tool in problem solving. In this section, we will look at momentum which is
another physical property that is conserved.
The linear momentum, p , of a object of mass m moving with a velocity v is defined as the product of
the mass and velocity:
p  mv
or, in component form,
p  px ˆi  p y ˆj  pz kˆ
 mvx ˆi  mv y ˆj  mvz kˆ
Momentum is a vector with units kg·m/s.
An 8 000 kg elephant running at 7 m/s has more momentum than a 50 kg cheetah running at 30 m/s. As
noted above, momentum is one of the physical properties that are conserved. If the elephant and
cheetah were traveling in opposite directions and collided, the momentum of the combined
cheetahphant would be in the direction of the elephant’s original momentum. More on this later.
Let’s look at the interaction of two objects. According to Newton’s 3rd Law, the force on object 2 from
object 1 is equal and opposite the force on object 1 from object 2 during the interaction.
F21  F12
If these are the only forces involved, they are the net forces on the two objects. Newton’s 2nd Law says
that we can write these as
m1a1  m2a2
We can rewrite this as
m1a1  m2a2  0
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PHYSICS SEMESTER ONE
UNIT 6
We can then replace the accelerations with the derivatives or change of the objects’ velocities with
respect to time
m1
dv1
dv
 m2 2  0
dt
dt
or
m1
v1
v
 m2 2  0
t
t
If the masses are constant, we can write this as
d  m1v1 
dt

d  m2 v 2 
dt
0
or
  m1v1 
t

  m2 v 2 
t
0
then, group everything into a single derivative or change
d  m1v1  m2 v 2 
dt
0
or
  m1v1  m2 v 2 
t
0
Now, notice that the terms inside the brackets are the momenta (momentums) of the two objects.
d  p1  p 2 
dt
0
or
  p1  p 2 
t
0
This is true at any time (at all times). The sum of the momenta of the two objects must be constant
(derivative of a constant is zero).
p1  p2  constant
In an isolated system, the momentum after an interaction must be equal to the momentum before the
interaction. This is known as the Law of Conservation of Momentum. This can be written as CoM, C of
M, or C of Mom if you need a shortened form when stating your reasoning for problem solving.
p1i  p2i  p1 f  p2 f
The Law of Conservation of Momentum
The total momentum of an isolated system (system of objects) equals the system’s initial momentum at
all times. Alternately, whenever two or more particles interact, the total momentum of the two or more
particle system is constant.
The isolated system is one where there is no net external force acting on the objects. If a net external
force were present it would change the velocities and momentum of the objects. We will usually
assume that collisions are instantaneous so there is no time for any external force to change the total
momentum of the system. This gets noted a few times because it is important.
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PHYSICS SEMESTER ONE
UNIT 6
If momentum is always conserved, how can a car change its momentum (accelerate) without firing a
mass out the back?
The car pushes on Earth. The sum of the car’s and Earth’s momentum is what is constant. Earth is so
massive and has many other forces acting on it so we don’t perceive any change in its momentum.
Example 1
In a burst of “creativity”, Bob invented the game of ice bowling. The area where the bowlers launch
their balls is covered with frictionless ice. Consider the following case
An ice-bowler with ball stands motionless on the ice, without contact with any other objects.
After the ceremonial wind-up, the bowler launches the 8.0 kg ball with a velocity of 15 m/s i
relative to the ground. Find the velocity of the 85 kg bowler after launching the ball.
v1i = v2i = 0
v2f
v1f
terms:
Before
m1  8.0 kg,
m2  85 kg,
v1i  0,
After
v1 f  15 m/s ˆi, v 2i  0,
v2 f  ?
The initial momentum of the system is
pi  p1i  p 2i
 m1v1i  m2 v 2i
  8.0 kg  0    85 kg  0 
0
The final momentum of the system is
p f  p1 f  p 2 f
 m1v1 f  m2 v 2 f


  8.0 kg  15 m/s ˆi   85 kg  v 2 f
 120 kg  m/s ˆi   85 kg  v 2 f
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PHYSICS SEMESTER ONE
UNIT 6
Using the law of conservation of momentum and rearrange to find the final velocity of the bowler,
pi  p f
0  120 kg  m/s ˆi   85 kg  v 2 f
v2 f 
120 kg  m/s ˆi
 1.41 m/s ˆi
85 kg
The final velocity of the ice bowler is 1.4 m/s in the direction opposite the direction the ball.
Problem 1
A 2.0 kg fish moving with a velocity of 6.0 m/s to the East is swallowed by a 6.0 kg fish moving with a
velocity of 1.0 m/s to the West.
a)
Which of the fish has the larger magnitude (size) of momentum?
b)
Use conservation of momentum to find the final velocity of the fish. Assume that the final
velocity of both fish is the same.
c)
Compare the initial and final mechanical (kinetic) energy of the two fish system.
Problem 2
Anne, Bob and Carla are sliding towards each other on a frictionless ice surface. If Anne has a mass of 55
kg and initial velocity of 4.5 m/s North, Bob has a mass of 85 kg and initial velocity of 3.5 m/s East, and
Carla has a mass of 62 kg and initial velocity of 3.0 m/s West, what is the speed of Anne and Bob if they
cling to each other while Carla travels off at 2.0 m/s North after their collision? State your final answer
in magnitude-direction form.
The type of example in Problem 2 can be done with any combination of vectors. The trick is to
remember that the total momentum immediately before the collision is equal to the total momentum
immediately after the collision. We use the momenta immediately before and after because it doesn’t
give any time for some external force to change the momentum of the system.
Problem 3
Bob sits in the middle of a round, frictionless ice rink. He needs to get to one side as fast as possible.
Bob’s mass is 85 kg. Bob has two rocks, each having a mass of 12 kg. Bob can throw the rocks, either
individually or together at a velocity of 15 m/s with respect to Bob (the difference between Bob’s
velocity and the rock(s)’s velocity is 15 m/s). Which option will give Bob the highest velocity?
a) launch both rocks at once
b) launch the rocks one-at-a-time
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PHYSICS SEMESTER ONE
UNIT 6
The case in problem 3 is very similar to what happens with rocket propulsion. The rocket fires hot
combustion products out of its engines. As the rocket goes higher, it gets lighter. The effect firing a
mass of propellant out the engine gets larger. This method has a few benefits.
a)
The acceleration occurs over several minutes instead of one very short but powerful blast.
Humans can only handle high G-force levels (multiplications of the acceleration due to gravity)
for very short times.
b)
The efficiency of the propulsion is higher. If the rocket and propellant have the same mass
before launch, this method is up to 40% more efficient than the single more powerful blast.
This method of launching many tiny objects is taken to an extreme with ion-propulsion systems for
spacecraft. Ion-propulsion systems accelerate individual ions electrically to very high speeds (~30 km/s).
This is about 10 faster than a typical chemical rocket. Ion-propulsion can use solar energy to accelerate
the ions, so heavy and dangerous high-energy chemicals are not required. Unfortunately, the ionpropulsion systems can only expel a relatively tiny amount of mass at these high speeds. Where a
chemical rocket can produce rocket accelerations on the order of 10 g’s (around 100 m/s2), the ionpropulsion systems can only produce rocket accelerations of around 0.001 g’s (0.01 m/s2). Ionpropulsion thrusters where used on Soviet spacecraft in the 1970’s and 1980’s.
Check out the information on the Dawn spacecraft (http://dawn.jpl.nasa.gov/) which will use ion-thrust
engines to travel the dwarf planet Ceres between Mars and Jupiter. Ion thrust engines have been
proposed for use with very long flights (to other stars) where the spacecraft have plenty of time, and
less room for heavy fuel.
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PHYSICS SEMESTER ONE
UNIT 6
IMPULSE
When we looked at the link between the conservation of momentum and Newton’s 2nd Law, we had
the change in the momentum of a system with respect to time
  p1  p 2 
t
0
This was equivalent to the sum of the forces on the two items
F21  F12
Using similar arguments, we can show that the average rate of change in momentum in a time interval
Δt is equal to the average force on the item in that time interval.
p
F
t
Multiplying both sides by Δt give us
p  Ft
The term Ft is known as the impulse I on the object. Specifically,
If a constant force F acts on an object, the impulse I delivered to the object over a time interval Δt is
I  Ft
The force does not have to be constant between the initial and final times, ti and tf. A more general
expression for the impulse is given by the integral (integral formula, not required for exams).
tf
I   Fdt
ti
The impulse-momentum theory states that the impulse is the change in momentum.
I  p  p f  pi
We can use impulse when a force is involved for a specific time interval.
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PHYSICS SEMESTER ONE
UNIT 6
Example 2
A 0.145 kg baseball traveling with a velocity of 41.4 m/s West collides
with a bat. The bat provides an impulse of 10.9 kg·m/s [21.3° South of
East]. What is the velocity of the baseball just after the hit?
terms:
m  0.145 kg, vi  42.5 m/s W,
I  10.9 kg·m/s  21.3 S of E
The momentum before the hit is
pi  mv
  0.145 kg  41.4 m/s W   12 kg  1.65 m/s 
 6.003 kg  m/s W
 6.003 kg  m/s E
In order to find the final momentum, and then the final velocity, we will need to break the impulse into
component form, I E and I N
I  10.9 kg·m/s  21.3 S of E
 10.9 kg·m/s  21.3 N of E
so
I E  10.9 kg·m/s  cos  21.3 
 10.155 kg·m/s
I N  10.9 kg·m/s  sin  21.3 
 3.959 kg·m/s
I  10.155 kg  m/s E  3.959 kg  m/s N
The impulse is the change in momentum
I  p f  pi
p f  pi  I
   6.003 kg  m/s E   10.155 kg  m/s E  3.959 kg  m/s N 
 4.152 kg  m/s E  3.959 kg  m/s N
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PHYSICS SEMESTER ONE
UNIT 6
The final velocity is
vf 

pf
m
4.152 kg  m/s E  3.959 kg  m/s N
0.145 kg
 28.64 m/s E  27.31 m/s N
The vectors in the question are in magnitude-direction form so we will convert back to that form.
v  vE2  vN2

 28.64 m/s 2   27.31 m/s 2
 39.57 m/s
  tan 1
 tan 1
vN
vE
27.31
 43.64
28.64
The velocity of the ball immediately after the hit is 39.6 m/s -43.6° North of East (or 39.6 m/s 43.6°
South of East).
Problem 4
A 45 kg crate slides along a horizontal, frictionless surface. What is the constant force needed to change
the velocity of the crate from 24 m/s East to 18 m/s North over a span of 5.0 s? Express the solution in
component form.
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PHYSICS SEMESTER ONE
UNIT 6
INELASTIC COLLISIONS (1D OR 2D)
We have already looked at several collisions where the two objects in the collision are lumped into one
item after the collision. After the collision, both objects have the same final velocity. This type of
collision is known as a perfect inelastic collision. Only momentum is conserved in perfect inelastic
collisions.
If we have two objects, object 1 with mass m1 and initial velocity v1i , object 2 with mass m2 and initial
velocity v2i, and a final velocity vf. Conservation of momentum gives us
pi  p f
p1i  p 2i  p1 f  p 2 f
m1v1i  m2 v 2i   m1  m2  v f
Isolating the final velocity
vf 
m1v1i  m2 v 2i
m1  m2
Example 3
Two train cars, mA  7500 kg and mB  8500 kg , traveling along a straight track with initial speeds
vA  1.7 m/s and vB  1.5 m/s . The cars collide and latch together. What is the final velocity of the
joined cars after the collision?
The cars latch together so they have the same final velocity, and we have a perfectly inelastic collision.
We can use the final velocity equation noted above to find the final velocity.
terms:
mA  75000 kg,
m2  8500 kg,
v Ai  1.7 m/s,
vBi  1.5 m/s,
v Af  vBf  v  ?
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PHYSICS SEMESTER ONE
UNIT 6
Using the noted equation (in scalar form)
vf 

mAv Ai  mB vBi
mA  mB
 7500 kg 1.7 m/s   8500 kg  1.5 m/s 
 7500 kg   8500 kg 
0
The final speed of the coupled train cars is zero.
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PHYSICS SEMESTER ONE
UNIT 6
1D ELASTIC COLLISION
Elastic collisions conserve both momentum and kinetic energy. This is a slight change on the
conservation of energy we had in the last section where energy could have different initial and final
forms. Even inelastic collisions conserve energy; the energy gets converted into forms other than kinetic
energy. The important difference for elastic collisions is that the kinetic energy is conserved.
Pool ball collisions are essentially elastic collisions. A golf club strikes a golf ball with an essentially
elastic collision. The word “essentially” is used here because there is some energy lost in the form of
sound and heat. The lost energy is small compared to the kinetic energy.
Just as our equation for the final velocity in a perfectly inelastic collision, we can find similar
relationships for the final velocities in an elastic collision. If we have two objects, object 1 with mass m1
and initial velocity v1i , and object 2 with mass m2 and initial velocity v2i. The scalar velocities are used
here because, in 1D, the direction can be indicated by the sign of the value.
Conservation of momentum gives is expressed as
pi  p f
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
and conservation of energy gives
Ei  E f
E1i  E2i  E1 f  E2 f
1
2
m1v12i  12 m2v22i  12 m1v12f  12 m2v22 f
or, factoring out the ½’s
m1v12i  m2v22i  m1v12f  m2v22 f
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 2
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PHYSICS SEMESTER ONE
UNIT 6
We are looking for two unknown variables, v1 f and v2 f and we have two equations to use to find the
variables. Equation 1 can be rearrange to get
m1v1i  m2 v2i  m1v1 f  m2 v2 f
1
m1v1i  m1v1 f  m2 v2 f  m2 v2i



m1 v1i  v1 f  m2 v2 f  v2i

1'
Equation 2 can be rearranged in a similar manner.
m1v12i  m2v22i  m1v12f  m2v22 f
 2
m1v12i  m1v12f  m2v22 f  m2 v22i




 v1i  v1 f   m2  v2 f  v2i  v2 f  v2i 
m1 v12i  v12f  m2 v22 f  v22i
m1 v1i  v1 f

 2 '
We can use equation 1’ to substitute into equation 2’ (alternatively we can state that the coloured terms
are equal and divide out).

 v1i  v1 f   m2  v2 f  v2i  v2 f  v2i 
m2  v2 f  v2i  v1i  v1 f   m2  v2 f  v2i  v2 f  v2i 
 v1i  v1 f    v2 f  v2i 
 2 '
v1i  v1 f  v2 f  v2i
or
m1 v1i  v1 f

v1i  v2i   v1 f  v2 f

from 1'
 3
This leaves us with two fairly simple equations
m1v1i  m2v2i  m1v1 f  m2v2 f
1
and

v1i  v2i   v1 f  v2 f

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 3
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PHYSICS SEMESTER ONE
UNIT 6
We can then use these to solve for v1f and v2f.
v1 f 
m1  m2
2m2
v1i 
v2i
m1  m2
m1  m2
v2 f 
2m1
m  m1
v1i  2
v2i
m1  m2
m1  m2
and
In the special case when v2i = 0 (teed golf ball), these simplify to
v1 f 
m1  m2
v1i
m1  m2
and
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v2 f 
2m1
v1i
m1  m2
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PHYSICS SEMESTER ONE
UNIT 6
Other special cases:
When m1 >> m2 (read “m1 is much greater than m2”), such as with a golf driver & ball, we can use these
formulas to show there is little change in the heavier object’s velocity.
v1 f  v1i
and
v2 f  2v1i
We can also show when m1 = m2 (Newton’s pendulum) the objects trade their velocities (check out
http://www.ap.smu.ca/demonstrations/index.php?option=com_content&view=article&id=85&Itemid=8
5)
v1 f  v2i
and
v2 f  v1i
These are just simplifications based on certain conditions; all you really need to know is that both
momentum and energy are conserved in an elastic collision.
Classic Demonstration
An example of this is provided on one of the reference websites at the end of
the section.
Consider dropping a stack of three balls with: a basket ball (or bowling ball)
on the bottom; a softball (or pool ball) in the middle; and a ping pong (or golf
ball, superball) ball on top. Labelling the balls from the bottom up, we have
the following relationship between the masses
m1  m2  m3
Let’s say we drop the balls from a height of y = 1.0 m.
The initial speeds are zero so, by the time the basketball touches the floor, the speed of all balls is
v  2ay

from v 2f  vi2  2ad

 2 9.8 m/s 2 1.0 
 4.4 m/s down
or
v  4.4 m/s up
This is the initial speed for all balls before their collisions with the objects below them.
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PHYSICS SEMESTER ONE
UNIT 6
Here is where the fun starts. Assuming we have a perfect collision with Earth ( mE  mbb ), the
basketball will rebound with a speed of
vbb  4.4 m/s up
We can regard the basketball (mass = 600 g) as rebounding from the floor and colliding with the softball
(mass = 190 g). It is a bit more complicated but the end result is very similar. The speed of the softball
after such a collision is
vsbf 

2mbb
m  mbb
vbb  sb
vsbi
mbb  msb
mbb  msb
2  600 g 
 600 g   190 g 
 4.4 m/s  
190 g    600 g  4.4 m/s


 600 g   190 g 
 9.0 m/s
Repeating the arguments for the softball–ping pong ball collision, the speed of the ping pong ball (mass
= 2.7 g) after the collision is
v ppf 

m pp  msb
2msb
vsbf 
v ppi
msb  m pp
msb  m pp
2 190 g 
190 g    2.7 g 
 9.0 m/s  
 2.7 g   190 g  4.4 m/s


190 g    2.7 g 
 22 m/s
If everything goes well, the final speed of the ping pong ball is around 22 m/s or almost 80 km/h, so be
careful when you are demonstrating this! Wear safety goggles.
This is especially true with a super ball. The super ball is heavier and smaller than the ping pong ball.
Combining these properties with the high speed makes it a serious hazard for eye damage.
Wear safety goggles.
This demonstration can be difficult to pull off perfectly. Any slight misalignment of the balls can send
the ping pong ball flying at unexpected angles. Also, the collisions are not perfectly elastic so the final
speed can be reduced somewhat.
There are no specific collision applets at the PHeT website, yet. You can check out the following site for
a set of momentum and collision demos.
www.physics.ucla.edu/demoweb/demomanual/mechanics/momentum_and_collisions/a_set_of_mome
ntum_and_collision_demos.html
This site has the two ball collision along with several other collision demos.
www.wakeforest.edu/physics/demolabs/demos/avimov/bychptr/chptr3_energy.htm
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PHYSICS SEMESTER ONE
UNIT 6
Problem Solving Strategy for 1D Collisions
The following procedure is recommended for solving one-dimensional problems involving between two
objects:
1.
Sketch the problem, labeling velocities and masses. This provides a picture of what is
happening and defines the variables used in the analysis. It also helps ensure that the
velocity components have the appropriate signs.
2.
Apply the rule of conservation of momentum. The total momentum along the direction of
motion (probably x-direction) must be the same before and after the collision. Fill in the
known values. Remember to use appropriate signs on the terms.
3.
If the collision is inelastic, kinetic energy is not conserved so you can’t use conservation of
energy. If the collision is perfectly inelastic, the objects will stick together and have the
same final velocity.
If the collision is elastic, kinetic energy is conserved, and you can equate the initial and
final total kinetic energy.
4.
Solve the system of equations for the unknown variables.
You can find up to two unknown variables with the two equations (C of M, and one of C of E
for an elastic collision or equal final velocities for a totally inelastic collision).
This is great, but if you are just looking for the speed of the objects after the collision then you can use
the equations we derived earlier. Don’t get too attached to those equations unless you plan to
memorize/know them, they are usually not on the exam formula sheet. Check with your
instructor/professor’s notes.
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PHYSICS SEMESTER ONE
UNIT 6
Problem 5
Two identical pool balls move towards each other with velocities of v1i = 15 m/s and v2i = –25 m/s. Find
the velocities of the balls after the head-on elastic collision using the problem solving strategy outlined
above.
Problem 6
A block of mass m1 = 2.40 kg, initially moving to the right at v1i = 4.00 m/s on a frictionless horizontal
track, collides with a spring attached to a second block of mass m2 = 3.20 kg and moving to the left with
a velocity of v2i = –3.00 m/s. The spring has a spring constant of k = 8.00 × 102 N/m.
a)
Determine the velocity of block 2 at the instant when block 1 is moving to the right with a
velocity of v1f = 3.00 m/s..
b)
Determine the spring compression at the time of part a).
GLANCING COLLISIONS
We can’t always line objects up along a track so we have perfect 1D collisions. Automobile collisions,
pool ball collisions and collision in hockey are all primarily constrained to a single 2D plane. Here, we
expand our understanding of collisions to cover the 2D versions.
When we looked at forces and free body diagrams, we solved equations by splitting each vector
equation into vector component equations. Newton’s Laws of Motion are valid for the vector equation
and for each of the individual vector component equations. The same is true with conservation of
momentum. Splitting the vectors into components we have conservation of momentum in the x and y
directions.
p1i  p 2i  p1 f  p 2 f
m1v1i  m2 v 2i  m1v1 f  m2 v 2 f
becomes
p1xi  p2 xi  p1xf  p2 xf
m1v1xi  m2v2 xi  m1v1xf  m2v2 xf
and
p1 yi  p2 yi  p1 yf  p2 yf
m1v1 yi  m2v2 yi  m1v1 yf  m2v2 yf
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PHYSICS SEMESTER ONE
UNIT 6
Let’s look at the simple case where v1i  v1xi ˆi and v 2i  0 .
y
m1, v1i
x
before
m2, v2i = 0
m1, v1f
θ
after
Φ
m2, v2f
Note that we have four unknowns, the magnitudes of the two final velocities and their directions (or the
two component velocities for each of the final velocities). Assuming kinetic energy is conserved as well
as momentum, we only have three equations.
m1v1xi  m2v2 xi  m1v1xf  m2v2 xf
m1v1yi  m2v2 yi  m1v1yf  m2v2 yf
and
1 m v2
2 1 1i
 12 m2v22i  12 m1v12f  12 m2v22 f
This is true for any 2D elastic collision where we are given only the initial conditions. In 2D collision
questions, you must have at least one of the final velocity components, or a magnitude or direction, in
order to solve for the remaining three variables.
When police are looking at collision scenes, they can estimate the final speeds and directions from the
skid marks, amount of damage and proximity to an intersection. In some cases, the initial directions are
obvious from the skid marks or the geometry of the crash (one care was in North-bound traffic and
other was in East-bound traffic). In other cases, they have to rely on witnesses for at least one piece of
the initial conditions in order to make a good estimate for the initial velocities.
In 2D perfectly inelastic collisions, kinetic energy is not conserved but the two objects are stuck together
so you only need the two velocity components for a single velocity. You can find the components with
the two conservation of momentum equations.
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PHYSICS SEMESTER ONE
UNIT 6
Problem Solving Strategy for 2D Collisions
The following procedure is recommended for solving one-dimensional problems involving between two
objects:
1.
Choose a coordinate axis that is in the direction of motion of at least one of the objects either
before or after the collision. This will reduce the number of variables to consider.
2.
Sketch the problem, and label velocities and masses. This provides a picture of what is
happening and defines the variables used in the analysis. It also helps ensure that the
velocity components have the appropriate signs.
3.
Apply the rule of conservation of momentum in the x and y-directions. The total momentum
in the x-direction must be the same before and after the collision. The total momentum in the
y-direction must be the same before and after the collision. Fill in the known values.
Remember to use appropriate signs on the terms.
4.
If the collision is inelastic, kinetic energy is not conserved so you can’t use conservation of
energy.
If the collision is perfectly inelastic, the objects will stick together and have the same final
velocity.
If the collision is elastic, kinetic energy is conserved, and you can equate the initial and final
total kinetic energy. Energy is a scalar so there are no component directions for the
conservation of energy, there is only one equation.
5.
Solve the system of equations for the unknown variables.
You can find up to three unknown variables with the three equations (C of Mx, C of My, and
one of C of E for an elastic collision or equal final velocities for a totally inelastic collision).
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PHYSICS SEMESTER ONE
UNIT 6
Problem 7
In a game of pool, a player wishes to sink the “11” ball in the corner pocket. In order to “find” the
corner, the “11” ball must travel at an angle of 55° relative to the direction of the cue ball. If the initial
speed of the cue ball is 5.0 m/s, and the “11” ball is stationary. Find the final speed of the two balls and
angle of the cue balls. Assume the collision is elastic, both balls have a mass of 0.17 kg, and the “11”
finds the corner pocket.
We could have solved the equations using component form for the velocities. Both ways require some
serious algebra. The end result would be the same. With the magnitude-direction form, it helps to
know that sin 2   cos 2   1 . At least one quadratic equation will show up where one solution is the
initial value and the other is the final value.
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PHYSICS SEMESTER ONE
UNIT 6
CENTRE OF MASS
A very common assumption in physics is that we can treat a complex system of masses as a single
concentrated mass at a single point. The free body diagrams have the all forces acting on a single point.
The gravitational force is between the centres of planets. In this section, we look at the physics behind
this assumption in more detail.
The location where the mass can be treated as concentrated is known as the centre of mass. In one
dimension, the location of the centre of mass is
xCM 
m1 x1  m2 x2
m1  m2
for two items
and
xCM 
 mi xi
i
for several items
 mi
i
where mi and xi are the mass and position for the ith item.
The total mass of the system is
M   mi
i
Using this, the centre if mass becomes
xCM 
 mi xi
i
M
Or, if we let the masses mi go to zero
xCM  lim
mi  0
 mi xi
i
M

1
M
 xdm
This gives us the ability to analyze objects with complex shapes by just looking at the movement of the
centre of mass.
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PHYSICS SEMESTER ONE
UNIT 6
Example
Consider an object consisting of two masses connected by a light (0 mass) rigid rod. Mass 1 is more than
mass 2, specifically m1 = 4.0 m2. If x1 = 1.0 m and x2 = 2.0 m, the centre of mass is
xCM 


m1 x1  m2 x2
m1  m2
 4.0m2 1.0 m   m2  2.0 m 
4.0m2  m2
 6.0 m2  m
5.0 m2
 1.2 m
The centre of mass is closest to the heavier mass.
If you threw this object, the centre of mass would follow the parabolic arc of projectile motion even if
the object was rotating.
In 2 or 3 dimensions, the centres of mass can be defined in a similar way with the components
yCM 
zCM 
 mi yi
i
M
or
yCM 
1
M
 ydm
or
zCM 
1
M
 zdm
 mi zi
i
M
Defining the position vector r  xˆi  yˆj  zkˆ , the centre of mass can be written
rCM  xCM ˆi  yCM ˆj  zCM kˆ
with the components defined as above, or
rCM 
 mi ri
i
M
or
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rCM 
1
rdm
M
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PHYSICS SEMESTER ONE
UNIT 6
Example
An object consists of two masses, m1  1.0 kg is located at r1  1.0 m ˆi , and m2  2.0 kg is located
at r2  2.0 m ˆj . Find the centre of mass of the object.
For the x-direction
xCM 

m1 x1  m2 x2
m1  m2
m2
1.0 kg 1.0 m    2.0 kg  0 
1.0 kg    2.0 kg 
 0.33 m
r2
r1
m1
In the y-direction
yCM 

m1 y1  m2 y2
m1  m2
1.0 kg  0    2.0 kg  2.0 m 
1.0 kg    2.0 kg 
 1.33 m
The centre of mass is at rCM  xCM ˆi  yCM ˆj  0.33 m ˆi  1.33 m ˆj = 0.33 m i + 1.33 m j.
m2
centre of mass
r2
rCM
r1
m1
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PHYSICS SEMESTER ONE
UNIT 6
Example (Don’t worry if you haven’t covered this material in Calculus yet)
Find the centre of mass of a 1.0 m rod (between x1 = 0 and x2 = 1.0 m), where the mass per unit of length
is
dm
 (1.0  2.0 x) kg/m
dx
This is a 1D case so we want
1
xdm
M
xCM 
The total mass is
M   dm
Using the derivative, we can rewrite dm with
dm  (1.0  2.0 x)dx
so
M 
1.0
0
1.0  2.0 x  dx
 
 1.0  1.0    0  0 
1.0
 x  x2
0
2
2
 2.0 m
and
xCM 


1 1.0
x 1.0  2.0 x  dx
2.0 0
1
2.0

1 
2.0 
1
2

x 2  32 x3
1 1.02
2

1.0
0
 
 32 1.03 
1 02
2

 13 03 

7
 6
2.0
7
12
 0.58 m
The centre of mass of the rod is at xCM = 0.58 m.
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PHYSICS SEMESTER ONE
UNIT 6
Motion of a System of Particles
I mentioned earlier that if you threw an object, the centre of mass would follow the parabolic arc of
projectile motion even if the object was rotating. The object will appear to rotate about one point as it
flies. This point is the centre of mass. The centre of mass does not have to be on the object.
The centre of mass of a system of particles of combined mass M moves like an equivalent particle of
mass M would move under the influence of the net external force on the system.
Relationships
Taking the time derivative of the centre of mass gives the velocity of the centre of mass.
v CM 
drCM
dt
 1
d
M
 



 mi ri 

i
in the numerator
dt
dri
dt
1
M
 mi
1
M
 mi vi
i
ri is the only variable
i
The velocity of the centre of mass has the same form with respect to the individual masses as the
position of the centre of mass.
Multiplying this by the total mass of the system, we see the total momentum of a system is the
momentum of the centre of mass.
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PHYSICS SEMESTER ONE
UNIT 6
pCM  MvCM
M
1
M
 mi vi
i
  mi v i
i
  pi
i
Taking the second derivative of the position (first derivative of the velocity) of the centre of mass gives
the acceleration of the centre of mass. The acceleration of the centre of mass is the net force on the
system divided by the total mass (just like Newton’s second law for a single object).
aCM 
dv CM
dt
 1
d
M
 
 mi vi 
1
M
 mi
dv i
dt
1
M
 miai
1
M
 Fi





i
in the numerator
dt
i
v i is the only variable
i
i
Multiplying this by the total mass of the system, we see that sum of the forces on the system is the total
external force on centre of mass. This sum equals the external forces on the centre of mass because
Newton’s 3rd Law dictates that all forces between masses within the group cancel out (equal and
opposite).
FCM  MaCM
  mi ai
i
  Fi
i
This sum equals the external forces on the centre of mass because Newton’s 3rd Law dictates that all
forces between masses within the group cancel out (equal and opposite).
FCM   Fi , external
i
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PHYSICS SEMESTER ONE
UNIT 6
All of our ideas for the dynamics of a single object apply to the centre of mass.
Example
Earlier, we looked at the 2D collision between two pool balls.
Before Collision
m1  m2  0.17 kg, v1i  5.0 m/s ˆi and v2i  0
The initial momentum of the centre of mass of the system is
pCMi   mi vi
i


  0.17 kg  5.0 m/s ˆi   0.17 kg  0 
 0.85 kg  m/s ˆi
After Collision
v1 f  4.1cos  35  m/s ˆi  4.1sin  35  m/s ˆj
 3.4 m/s ˆi  2.3 m/s ˆj
v 2 f  2.3cos  55  m/s ˆi  2.3sin  55  m/s ˆj
 1.6 m/s ˆi  2.3 m/s ˆj
The final momentum of the centre of mass of the system is
pCMf   mi vi
i



  0.17 kg  3.4 m/s ˆi  2.3 m/s ˆj   0.17 kg  1.6 m/s ˆi  2.3 m/s ˆj

 0.85 kg  m/s ˆi
 pCMi
The momentum of the centre of mass is unchanged because there is no external force on the system.
Similarly, in the example with Bob throwing one or two rocks in order to get off the pond, the centre of
mass is at the initial location of the person and rocks. Throwing the rocks causes Bob and rocks to move
but the centre of mass is unchanged because there is no net external force on the system.
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PHYSICS SEMESTER ONE
UNIT 6
Consider fireworks launched from some point on the ground.
explosion consists of forces from within
fireworks, no external force besides gravity
boom!
all particles initially together
particles take individual paths,
path of centre of mass unchanged
path of the centre of mass is that of a
single particle under projectile motion
Once the fireworks canister is in flight, there are no external forces acting on the system. The explosion
is an internal force. The force on each particle from another particle is balanced by the equal and
opposite forces. The individual particles fly out in all directions, each following its own path. The net
force on the system is zero so the centre of mass continues to follow the initial parabolic path. This
simplification ignores air resistance. We could use denser objects which are less affected by air
resistance, or launch the fireworks into space (no air) to make it more realistic.
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PHYSICS SEMESTER ONE
UNIT 6
SUMMARY
Momentum
p  mv
p  px ˆi  p y ˆj  pz kˆ  mvx ˆi  mv y ˆj  mvz kˆ
p1  p2  constant
Law of Conservation of Momentum, C of Mom
p1i  p2i  p1 f  p2 f
Impulse
I  Ft , I  p  p f  pi
Collisions
m1v1i  m2 v 2i   m1  m2  v f



m1v1i, x  m2v2i, x   m1  m2  v f , x 

m1v1i, y  m2v2i, y   m1  m2  v f , y 
perfectly inelastic collision (items stick together)
m1v1i  m2 v 2i  m1v1 f  m2 v 2 f


m1v1i , x  m2v2i , x  m1v1 f , x  m2v2 f , x 


m1v1i , y  m2v2i, y  m1v1 f , y  m2v2 f , y 

m1v12i  m2v22i  m1v12f  m2v22 f

perfectly elastic collision, energy conserved
remember v 2  vx2  v y2 for 2D vectors
Centre of Mass
M   mi
i
rCM 
vCM 
 miri
i
M
 xCM ˆi  yCM ˆj  zCM kˆ , xCM 
drCM
1

dt
M
 mi xi
i
M
, yCM 
 mi yi
i
M
, zCM 
 mi zi
i
M
 mi vi
i
pCM  MvCM   mi vi   pi
i
aCM 
dvCM
1

dt
M
i
1
1
 miai  M  Fi  M  Fexternal 
i
i
i
Fnet
M
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PHYSICS SEMESTER ONE
UNIT 6
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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