Download RRHS P 12

RRHS PHYSICS 12
UNIT 1 VECTORS
© JIM BURKE
RRHS PHYSICS 12
TABLE OF CONTENTS
Module 1.1 Displacement and Velocity Vectors ....................................................................... 3
1.1.1 Introduction to 2D Vectors ...................................................................................... 3
1.1.2 Displacement Vectors and Vector Algebra .......................................................... 12
1.1.3 Velocity Vectors.................................................................................................... 24
1.1.4 Module Summary ................................................................................................. 32
Module 1.2 Force Vectors.......................................................................................................... 33
1.2.1 Dynamics Problems in 2D .................................................................................... 33
1.2.2 Inclined Planes ..................................................................................................... 39
1.2.3 Module Summary ................................................................................................. 46
Module 1.3 Equilibrium .............................................................................................................. 47
1.3.1 Translational Equilibrium ...................................................................................... 47
1.3.2 Torque and Rotational Equilibrium ....................................................................... 55
1.3.3 Module Summary ................................................................................................. 65
Module 1.4 2D Collisions ........................................................................................................... 66
1.4.1 Conservation of Momentum ................................................................................. 66
1.4.2 Elastic and Inelastic Collisions ............................................................................. 70
RRHS Physics
Page 2
Module 1.1 Displacement and Velocity Vectors
1.1.1 Introduction to 2D Vectors
In unit 2, you discussed two kinds of quantities --- vectors and scalars. A scalar is a
quantity that has only magnitude (size); it does not have a direction. For example,
temperature and mass have no direction associated with them. A vector is a quantity
that has both magnitude and direction. For example, displacement, velocity,
acceleration, force, and momentum are all quantities for which it is important to know
the direction. A vector quantity is denoted by placing an arrow over the symbol ( d );
when typing, a vector quantity can be denoted by using boldface (d). In this manual,
we will use the arrow notation.
In units 2 and 3, you talked briefly about vectors in one dimension. In one dimension,
we assigned directions to vectors using positive and negative signs. In this unit, we
will be extending that analysis to two dimensions. A vector in two dimensions is not
just a single number. In two-dimensional space a vector is actually represented with
two numbers, one for each dimension. You have used an x  y coordinate system in
math, and you know that two numbers are needed to specify a position on one of
these graphs. Likewise, two coordinates are needed to specify a vector in twodimensional space. These coordinates are referred to as components. Consider the
diagram below, where east is represented by the positive x direction and north is
represented by the positive y direction.1
1
In physics we usually use compass directions in conjunction with a traditional x-y plane since we are
dealing with the actual motion of objects in real space.
RRHS Physics
Page 3
d
Figure 1: Vector Components
The position vector d in Figure 1 represents a step in space from the origin to some
point whose location is given by the ordered pair (d x , d y ) - d x and d y are known as
the components of d . As with one dimensional vectors, the direction of components
can be specified with a positive or negative sign. The symbol d represents both of
these components together. We will use the symbols d x and d y to refer to the
magnitude of the components; the symbols d x and d y will be used to include the
direction by using a positive or negative sign.
Instead of representing a two-dimensional vector by using both of its components, it
is often convenient to represent a vector by an arrow that indicates the direction of
the vector. The vector d shown in Figure 1 can then be described using a magnitude
d (the ``length'' of the vector) and an angle  (the direction of the vector). Note that
if we know the magnitude d and the angle  , we can use sin  and cos functions
to solve for d x and d y in the above diagram.
Trigonometry Review
Consider the following right angle triangle:
RRHS Physics
Page 4
Figure 2: Right Angle Triangle Trigonometry Review
Where
 The hypotenuse is the side across from the right angle (the longest side).
 The opposite side is the side across from the angle θ.
 The adjacent side is the side joining the angle θ and the right angle.
cos  
adjacent
hypotenuse
sin  
opposite
hypotenuse
tan  
opposite
adjacent
Also, remember that the lengths of the sides can be related to one another using the
Pythagorean Theorem
c 2  a 2  b2
or
c  a 2  b2
where c is the hypotenuse of the triangle.
Example
In Figure 1, assume that the angle in the diagram is 30.0 and the magnitude of the
position vector is 7.2 cm. Calculate the components d x and d y .
Solution
RRHS Physics
Page 5
By definition, components must be perpendicular to one another 2 ; therefore, the
above diagram is a right angle triangle and we can apply the basic trigonometric
functions. Since we know the hypotenuse of the triangle (7.2 cm), we can use the
cosine function to calculate the adjacent side of the triangle ( d x ) and the sine
function to calculate the opposite side of the triangle ( d y ):
adj
hyp
d
cos   x
d
d
cos 30.0  x
7.2
d x  6.2cm
cos  
sin  
sin  
sin 30.0 
opp
hyp
dy
d
dy
7.2
d y  3.6cm
In other words, the location identified by the position vector identified by d can also
be reached by going 6.2 cm to the east and 3.6 cm to the north. When calculating
components, we will often need to identify the direction of the components using
positive or negative signs. East (positive x-direction) and north (positive y-direction)
will be considered positive. West (negative x-direction) and south (negative ydirection) will be considered negative. For the example above, since the directions of
the components are east and north, both should be positive.
In the above example, we drew a sketch of the vector and used trigonometry to
calculate the components. Vectors can also be drawn using scale diagrams, where a
protractor can be used to orient the vector correctly and an appropriate scale can be
used to represent the vector. For example, a scale of 1 cm for every 5 m can be
used; a 30 m displacement vector would then be drawn with an arrow that is 6 cm
long. The components can then be drawn in and measured with a ruler.
When drawing a vector, the arrow indicates the direction of the vector. The arrow is
referred to as the head of the vector and the tail is at the other end.
2
Since they are parallel to the axes in the coordinate system.
RRHS Physics
Page 6
Figure 3: Vector head and tail
RRHS Physics
Page 7
Check Your Learning
Calculate the components for each of the following vectors:
RRHS Physics
Page 8
Expressing Direction
There are different conventions for describing the direction of a vector. For the
examples that follow, assume that   30o in Figure 1.
1. In math, you may have described vector directions as a counterclockwise
rotation from the positive x-coordinate (east using compass directions). In
Figure 1, the direction of the vector would then be 30o . Using this convention,
north would be 90o and south would be 270o .
2. Bearings are another way of expressing directions and are often used in
navigation. In this system, north is 0o and all directions are measured
clockwise from this reference direction. In this system, the direction of the
vector in our diagram would be 60o .
3. The last convention we will discuss is the one that we are going to use. This
convention describes a direction as a rotation from one of the four reference
directions (north, east, south, west). The direction of the vector in our diagram
would now be 30o north of east. (or 30o N of E) This means that a vector that
was pointed east was rotated 30o toward the north. This convention is
convenient because there is no ambiguity about what the reference direction (
0o ) is. The direction in the diagram could also be expressed as 60o east of
north. A slightly different way of expressing 30o north of east would be to say
E 30o N - this can be interpreted as “go east and then rotate 30o toward the
north” for the proper vector direction. The textbook that accompanies this
manual uses this last convention.
RRHS Physics
Page 9
Check Your Learning
State the direction for each of the following vectors:
RRHS Physics
Page 10
1.1.1 In Class or Homework Exercise
1. Break the following vectors into components:
a. 45 km in a direction 25 S of E;
b. 74 km, 35 E of N
2. A bug crawled 34.0 cm E, then 48.5 cm S. What is the bug’s displacement
from his starting point?
3. A ship leaves its home port expecting to travel to a port 500. km due south.
Before it can move, a severe storm comes up and blows the ship 100. km due
east. How far is the ship from its destination? In what direction must the ship
travel to reach its destination?
RRHS Physics
Page 11
1.1.2 Displacement Vectors and Vector Algebra
Since position and displacement vectors can be easily visualized and directly
represented using vector diagrams, it is these types of vectors that will be used in
this section to introduce vector addition and subtraction; the information here can,
however, be applied to any kind of vector.
Since vectors are not single numbers, our usual laws of algebra cannot be applied to
them; in other words, we cannot simply add the magnitude of two vectors together to
obtain a total magnitude. Vector algebra requires the use of a vector diagram.
Vector Addition
What does it mean to add two vectors? Consider displacement vectors a and b
which represent successive displacements of a person walking.
a
3
b
The addition of these two displacements (the total displacement) should tell us where
the person is at the end of his journey relative to where he started. We will use c to
represent the total displacement:
c  a b
To help visualize this, we must draw a vector addition diagram. Since the person
completed displacement a and then completed displacement b , we draw the second
vector where the first one ends.
3
To simplify notation, a and
represent displacement.
RRHS Physics
b are being used here instead of the standard d with subscripts to
Page 12
b
a
Figure 4:
a b
Vectors can always be moved around (by sliding them) when drawing a vector
diagram, as long as their direction remains the same.
The total displacement c (from where he started to where he ended) can then be
drawn.
c
b
a
Figure 5:
a b  c
This diagram represents the vector equation a  b  c . When we add two scalars
together, the answer that we get is referred to as a sum. When we add two vectors
together we get what is referred to as a resultant vector. So when we say that
a  b  c , c is the resultant vector that goes from where we started to where we
ended.
It is important to note that the vector equation a  b  c tells us how to draw the
vector diagram shown in Figure 5, but that is all it should be used for. Numbers
cannot be substituted into a vector equation to be solved algebraically.
When drawing a vector addition diagram, it is important to remember two things:
1. The vectors being added must be drawn head to tail.
2. The resultant vector goes from the tail of the first vector to the head of the
second vector (from where you started to where you ended).
RRHS Physics
Page 13
Check Your Learning
Given the following vectors, draw a vector diagram to represent each of the following
vector equations (where x is an unknown vector in each case):
c
a
b
a. a  b  x
b. b  c  x
c. a  b  c  0
d. a  x  b
e. x  c  a
RRHS Physics
Page 14
It is possible to obtain the magnitude and direction of the resultant from the diagrams
that were completed in the Check Your Learning by drawing scale diagrams with a
ruler and protractor; however, we will be using an algebraic approach to solve this
type of problem by looking at the components of the vectors. If we redraw Figure 4
and add the components (as dotted lines) of each vector, we get
b
a
Figure 6: Vector Addition
Remember that the vector a actually represents the components (ax , a y ) ; the other
vector b represents the components (bx , by ) . If we add these two vectors, we are
actually adding their components. So a  b will give (ax  bx , ay  by ) , and the diagram
can be redrawn to look like this:
b
a
Figure 7: Vector Addition
The only difference between these two diagrams is that the component vectors have
been moved to show the x components together and the y components together –
the original vectors a and b have not been changed. If we add the resultant vector c
to the diagram we get
c
b
a
Figure 8: Vector Addition with Resultant
RRHS Physics
Page 15
Notice that the vector c represents the sum of the components (ax  bx , ay  by ) . We
now have one large right angle, so we can again use the Pythagorean theorem and
our trigonometric functions to find the magnitude and direction of c .
c
Figure 9: Resultant Vector with Components
Since we now have a single right angle triangle, we can use the Pythagorean
theorem
c  (x)2  (y)2
to find the magnitude of c and the angle  can be found using
tan  
y
x
Example
A person walks 5.0 km east and then 8.0 km in a direction 75 N of E. What is his
displacement?
Solution
d1  5.0km E
d 2  8.0km, 75 N of E
dt  ?
We know that the total displacement is the sum of the individual displacements,
dt  d1  d2
This is a vector equation which will be used to draw the vector diagram:
RRHS Physics
Page 16
dt
d 2 y
d 2
d 2 x
d1
The vector d1 is already in one of our 4 component directions (east), so we do not
need to break this vector up into components. We do have to break d 2 into
components, since it has both north and east components:
d 2 x
8.0
d 2 x  2.1km
cos 75 
sin 75 
d 2 y
8.0
d 2 y  7.7 km
dt
d 2
Since we know that the resultant vector has an x-component of 7.1 km and a ycomponent of 7.7 km, we can now find the magnitude and direction:
dt  dtx2  dty2
 (7.1) 2  (7.7) 2
using Pythagorean theorem
 10. km
and
RRHS Physics
Page 17
tan  
dty
dtx
7.7
7.1
 47

The total displacement is
dt  10. km, 47 N of E
RRHS Physics
Page 18
Check Your Learning
A person walks 4.0 km south and then 7.2 km in a direction 21 o W of N. What was
the person’s displacement?
RRHS Physics
Page 19
Vector Subtraction
Just like subtraction of two scalars is really the same as adding a negative scalar (
5  3 is the same as 5  (3) ), the subtraction of two vectors a  b is the same as
a  (b ) ; but (b ) just means (bx , by ) ; in other words, we are just changing the
direction of the vector b and instead of adding the components of the two vectors
we subtract them. Using the same vectors as our previous example,
a
b
a and b
Figure 10: Vectors
a  b  c is the same as a  (b )  c so the two vectors that we must add are a and
b :
b
a
Figure 11: Vectors
a and b
Notice that b is simply pointing in the direction opposite that of b . The vector
diagram for a  (b )  c now looks like
a
c
b
Figure 12: a  (b )  c
RRHS Physics
Page 20
The vector diagram can now be analyzed using components in the same way that it
was for the vector addition problems. The resultant vector c can also still be
represented in component form
ax
ay
a
b y
b
c
bx
Figure 13: a  (b )  c with components
where, in this case, x  ax  bx and y  a y  by .
RRHS Physics
Page 21
Check Your Learning
Given the following vectors, draw a vector diagram to represent each of the following
vector equations (where x is an unknown vector in each case):
c
a
b
a. a  b  x
b. b  c  x
c. a  x  b
d. x  c  a
RRHS Physics
Page 22
1.1.2 In Class or Homework Exercise
1. A delivery truck travels 18 blocks north, 16 blocks east, and 10 blocks south.
What is its final displacement from the origin?
2. A hiker leaves camp and, using a compass, walks 4 km E, 6 km S, 3 km E, 5
km N, 10 km W, 8 km N, and 3 km S. At the end of three days, the hiker is
lost. Compute how far the hiker is from camp and which direction should be
taken to get back to camp.
3. A car is driven 30.0 km west and then 80.0 km southwest. What is the
displacement of the car from the point of origin (magnitude and direction)?
4. An explorer walks
d 2 x
d1
22.0 km north, and then
walks in a direction 60.0
south of east for 47.0 km.
d 2 y
d 2
a. What
d ty
dt
distance has he travelled?
b. What is his
displacement from the
origin?
d tx
c. What
displacement vector must
he follow to return to his original location?
5. A man walks 3.0 km north, 4.5 km in a direction 40. east of north, and 6.0
km in a direction 60. south of east. What is his displacement vector?
6. After the end of a long day of travelling, Slimy the Slug is 255 cm east of his
home. If he started out the day by travelling 90.0 cm in a direction 25o east of
north in the morning, how far did he travel in the afternoon (and in what
direction) to get to his final location?
RRHS Physics
Page 23
1.1.3 Velocity Vectors
You saw in the previous section that an object's position in two dimensions is given
by two coordinates (d x , d y ) . Remember from Unit 2 that velocity is the change in
position, or displacement, over time:
v
d
t
Since velocity is calculated using displacement and has the same direction as the
displacement, velocity is also a vector which has two components (vx , v y ) .
vx 
d x
t
vy 
d y
t
As was discussed in Unit 2, there is no such thing as absolute velocity; the velocity of
an object is always relative to some frame of reference. Consider the example of a
dog on a boat. The boat is moving north at 7 m/s relative to the shore. Now suppose
that the dog is moving north at 2 m/s relative to the boat. In other words, the dog is
moving 2 m/s faster than the boat. How fast is the dog actually moving? It depends
on your point of view. To someone on the boat, the dog is moving at 2 m/s; however,
to somebody on the shore, the dog is moving its 2 m/s plus the boat's 7 m/s (since
they are moving in the same direction), which is 9 m/s.
The situation is similar in two dimensions. Suppose that a boat is crossing a body of
water at 5.0 m/s relative to the water (we will use the symbol vbw to represent the
velocity of the boat with respect to the water).4 If the water is not moving, a person
on the shore sees the boat moving at 5.0 m/s relative to the shore as well. Now
suppose that the body of water is a river flowing perpendicular to the boat at 3.5 m/s
as measured by someone on the shore ( vws ).
The person on the shore now sees the river carrying the boat downstream at 3.5 m/s,
but also sees the boat moving across the river at 5.0 m/s. Just like the dog on the
boat, the person on the shore sees the addition of the two velocities, so the velocity
of the boat with respect to the shore is given by
4
Using this notation, the first subscript identifies the object that is moving, the second subscript
identifies the frame of reference with respect to which it is moving
RRHS Physics
Page 24
vbs  vbw  vws
Remember, however, that these quantities are vectors and must therefore be added
as vectors! (as was described in the last section – in other words, we must draw a
vector diagram).
By using subscripts according to the convention described above we see that the
inner subscripts on the right-hand side of the equation are the same as each other
(in this case, w) and the outer subscripts on the right-hand side of the equation are
the same as the subscripts for the resultant vector on the left vbs . This can be used
as a check if you are not sure if you are adding the proper vectors.
Since they are vectors, remember, these velocities must be added as vectors – they
must be drawn head to tail:
vws
vbw
Figure 14:
vbs
vbs  vbw  vws
The resultant vector (the velocity actually observed by someone on the shore) is the
vector vbs . This resultant velocity has two components (one across the river and one
down the river). Note that the component across the river is the same as the original
velocity of the boat that was directed across the river; therefore, the boat will cross
the river in the same amount of time with the river flowing as without!
Example
A boat that has a speed of 5.0 m/s in still water heads north directly across a river
that is 250 m wide. The velocity of the river is 3.5 m/s east.
a. What is the velocity of the boat with respect to the shore?
b. How long does it take the boat to cross the river?
c. How far downstream does the boat land?
d. What heading (direction) would the boat need in order to land directly across
from its starting point?
Solution
a.
RRHS Physics
Page 25
vbw  5.0m / s N
vws  3.5m / s E
vbs  ?
To find the velocity of the boat with respect to the shore, we must add the
velocity of the boat with respect to the water and the velocity of the water with
respect to the shore:
vbs  vbw  vws
vws
vbw
3.5m / s
5.0m / s
vbs
vbs
or
Using the Pythagorean Theorem,
2
2
vbs  vbw
 vws
 (5.0) 2  (3.5) 2
 6.1m / s
Since we are finding velocity, we must also find the direction
3.5
5.0
 35
tan  
vbs  6.1m / s,35 E of N
b. Since the boat is going at a constant velocity, we can use
d
t
Notice that both the displacement and velocity are vectors, and must have the
same direction; therefore, if we use the displacement of 250 m (which is north)
than we must also use the component of the velocity that is north, which is the
boat’s velocity with respect to the water.
v
RRHS Physics
Page 26
d  250m N
vbw  5.0m / s N
t ?
d
t
250
5.0 
t
t  50. s
v
c. Similarly, if we want to know how far downstream (east) the boat travels we
must use the component of the velocity in this direction. The boat will be
travelling for the same amount of time downriver that it took to cross the river.
vws  3.5m / s E
t  50. s
d  ?
d
t
d
3.5 
50.
d  180m
v
d. If the boat is to land directly across from its starting point, than the velocity of
the boat with respect to the shore must be north.
vbs  vbw  vws
Using this equation, and knowing that the resultant vector
our vector diagram will look like this
vbs
must be north,
vws
vbw
RRHS Physics
vbs
Page 27
sin  
vws
vbw
3.5
5.0
  44

The boat must head 44 W of N . Notice that this is not the same as the angle
that the boat would travel downstream if no corrective action is taken.
RRHS Physics
Page 28
Check Your Learning
In the example just completed, how long will it take the boat to cross the river in part
(d)? In other words, how long will it take to cross the river when corrective action is
taken so that the boat lands directly across from its starting point? How does this
answer compare with the time obtained in part (a) of the example?
RRHS Physics
Page 29
1.1.3 In Class or Homework Exercise
1. An airplane is travelling 1000. km/h in a direction 37 o east of north.
a. Find the components of the velocity vector.
b. How far north and how far east has the plane travelled after 2.0 hours?
2. A motorboat whose speed in still water is 8.25 m/s must aim upstream at an
angle of 25.5 o (with respect to a line perpendicular to the shore) in order to
travel directly across the stream.
a. What is the speed of the current?
b. What is the resultant speed of the boat with respect to the shore?
3. Kyle wishes to fly to a point 450 km due south in 3.0 h. A wind is blowing from
the west at 50.0 km/h. Compute the proper heading and airspeed that Kyle
must choose in order to reach his destination on time.
4. Diane aims a boat that has a speed of 8.0 m/s in still water directly across a
river that flows at 6.0 m/s.
a. What is the resultant velocity of the boat with respect to the shore?
b. If the stream is 240 m wide, how long will it take Diane to row across?
c. How far downstream will Diane be?
5. An airplane whose airspeed is 200. km/h heads due north. But a 100. km/h
wind from the northeast suddenly begins to blow. What is the resulting velocity
of the plane with respect to the ground?
6. An airplane is heading due north at with an airspeed of 300. km/h. If a wind
begins blowing from the southwest at a speed of 50.0 km/h, calculate
a. the velocity of the plane with respect to the ground, and
b. how far off course it will be after 30.0 min if the pilot takes no corrective
action.
c. Assuming that the pilot has the same airspeed of 300. km/h, what
heading should he use to maintain a course due north?
d. What is his new speed with respect to the ground?
7. A swimmer is capable of swimming 1.80 m/s in still water.
a. If she aims her body directly across a 200.0 m wide river whose current
is 0.80 m/s, how far downstream (from a point opposite her starting
point) will she land?
b. What is her velocity with respect to the shore?
c. At what upstream angle must the swimmer aim if she is to arrive at a
point directly across the stream?
8. A pilot wishes to make a flight of 300. km northeast in 45 minutes. If there is to
be an 80.0 km/h wind from the north for the entire trip, what heading and
airspeed must she use for the flight?
RRHS Physics
Page 30
9. A car travelling at 15 m/s N executes a gradual turn, so that it then moves at
18 m/s E. What is the car's change in velocity?
10. A plane's velocity changes from 200. km/h N to 300. km/h 30.0 W of N. Find
the change in velocity.
11. A plane is flying at 100. m/s E. The pilot changes its velocity by 30.0 m/s in a
direction 30.0 N of E. What is the plane's final velocity?
12. A ferryboat, whose speed in still water is 2.85 m/s, must cross a 260 m wide
river and arrive at a point 110 m upstream from where it starts. To do so, the
pilot must head the boat at a 45 o upstream angle. What is the speed of the
river's current?
RRHS Physics
Page 31
1.1.4 Module Summary
In this module you have learned:
o To represent two dimensional vectors using two methods:
 components
 magnitude and direction
o How to add and subtract displacement vectors using vector diagrams and
components:
 When adding vectors, they must be drawn head to tail. The resultant
vector goes from the tail of the first vector to the head of the last vector.
 Subtracting vectors is the same operation as adding a negative vector,
where a negative vector points in the direction opposite the direction of
the original vector.
o How to use vector diagrams and components to calculate relative velocities.
RRHS Physics
Page 32
Module 1.2 Force Vectors
1.2.1 Dynamics Problems in 2D
In Unit 3, you solved many problems involving horizontal and vertical forces by
applying Newton's 2nd Law to different situations using free body diagrams. This will
now be extended to situations where the forces are no longer solely in the horizontal
or vertical directions. Remember that Newton's 2nd Law ( Fnet  ma ) is a vector
equation, since it states the relationship between acceleration and net force, both of
which are vectors. This means that the acceleration and the net force will be in the
same direction; therefore, if we want to use scalar algebra to solve a problem, we
must use this equation in only one dimension at a time ( x or y ).
In the diagram below, a man is pulling a box with a rope that makes an angle  with
the ground.
Note that the expected acceleration (horizontal) for this box and the applied force are
neither parallel nor perpendicular, so Newton's 2nd Law cannot be applied yet. A free
body diagram for this box would look like this:
Notice that although the normal, friction, and gravity forces are all solely in the x
(horizontal) or y (vertical) directions, the force of the man pulling is not. This can be
fixed if we break this force up into its x and y components.
RRHS Physics
Page 33
In other words, pulling with a force Fp at an angle of  is the same as applying two
separate forces of Fx to the right and Fy upward. As can be seen in the diagram
above, all of the forces are now either in the x or y direction if we replace Fp with
its components Fpx and Fpy . We can now analyze the forces in each dimension
using Newton's 2nd Law.
Vertical: We will take up as the positive direction; therefore, FN and Fpy will both be
positive and Fg will be negative. Remember, only vertical forces can be used to
calculate a vertical acceleration.
ma y  Fy
ma y  FN  Fpy  Fg
may  FN  Fpy  Fg
and
0  FN  Fpy  Fg
FN  Fg  Fpy
since the vertical acceleration is zero (the box is not accelerating up or down). Notice
that FN  Fg . Because we often know Fg and Fpy , we can solve for FN and use it in
our calculation of F f (remember that Ff   FN , where  is the coefficient of friction).
Horizontal: We will take the right to be positive; therefore, Fpx will be positive and
F f will be negative. Remember, only horizontal forces can be used to calculate a
horizontal acceleration.
max  Fx
max  Fpx  Ff
max  Fpx  Ff
RRHS Physics
Page 34
This can then be used with the horizontal acceleration. These are not equations to
be memorized and applied to all problems!!! This is a sample analysis of a typical
free body diagram involving forces at an angle. As always when dealing with force
problems, analysis should always start with a free body diagram.
Example
A 52.0 kg sled is being pulled along a frictionless horizontal ice surface by a person
pulling a rope with a force of 235 N. The rope makes an angle of 35.0o with the
horizontal. What is the acceleration of the sled?
Solution
First we need a free body diagram:
m  52.0kg
Fp  235 N
  35.0
ax  ?
We must first calculate the components for the applied force:
Fpx  Fp cos 
 (235) cos 35.0
 193N
Fpy  Fp sin 
 (235) sin 35.0
 135 N
Horizontal:
RRHS Physics
Page 35
max   F x
max  Fpx
(52.0)ax  193
ax  3.71m / s 2
Since there was no friction involved in this problem, we did not need a calculation of
the normal force; since there was no vertical acceleration and the normal force was
not needed, we were not required to do an analysis of the vertical forces.
RRHS Physics
Page 36
Check Your Learning
A 52.0 kg sled is being pulled along a horizontal surface by a person pulling a rope
with a force of 235 N. The rope makes an angle of 35.0 o with the horizontal. If the
coefficient of friction between the sled and the surface is 0.25, what is the
acceleration of the sled?
RRHS Physics
Page 37
1.2.1 In Class or Homework Exercise
1. A 25.0 kg sled is accelerating at 2.3 m/s2. A force of 300.0 N is pulling the sled
along a rope that is being held at an angle of 35o with the horizontal. What is
the coefficient of friction?
2. A 15.0 kg sled is being pulled along a horizontal surface by a rope that is held
at a 20.0o angle with the horizontal. The tension in the rope is 110.0 N. If the
coefficient of friction is 0.30, what is the acceleration of the sled?
3. A man pushes a 15 kg lawnmower at constant speed with a force of 90.0 N
directed downward along the handle, which is at an angle of 30.0o to the
horizontal. What is the coefficient of friction?
4. A sled is being pulled by a rope that makes an angle of 27 o with the
horizontal. The coefficient of friction between the sled and the ground is 0.30.
If the tension in the rope is 245 N and the acceleration of the sled is 1.2 m/s 2,
what is the mass of the sled?
5. A 55.0 kg rock is being pulled along a horizontal surface at a constant speed.
The coefficient of friction between the rock and the surface is 0.76. If the rope
pulling the rock is at a 40.0o angle with the horizontal, with what force is the
rock being pulled?
6. A 40.0 kg iceboat is gliding across a frozen lake with a constant velocity of 14
m/s E, when a gust of wind from the southwest exerts a constant force of 100.
N on its sails for 3.0 s. With what velocity will the boat be moving after the
wind has subsided? Ignore any frictional forces.
7. Two tow trucks attach ropes to a stranded vehicle. The first tow truck pulls
with a force of 25000 N, while the second truck pulls with a force of 15000 N.
The two ropes make an angle of 15.5o with each other. Find the resultant
force on the vehicle.
RRHS Physics
Page 38
1.2.2 Inclined Planes
Up until now, we have dealt only with dynamics situations where the motion has
been either horizontal or vertical, and where we have calculated all of the
components of our vectors in horizontal and vertical planes. We are now going to
apply force vectors and Newton's second law to an inclined plane (a ramp). On an
inclined plane, the acceleration is neither horizontal nor vertical. It is parallel to the
plane.
If we place a box on a ramp (ignoring friction for now), as in Figure 15,
Figure 15: Ramp
it can be observed that there are only two forces acting on the box - the normal force
FN (which is perpendicular to the surface of the plane) and the force of gravity Fg .
Drawing a free body diagram, we get
Figure 16: Ramp Free Body Diagram
Notice that these vectors exist in two dimensions and are not in component form
(they are not either parallel or perpendicular to one another). In order to apply
Newton's second law, we want to analyze the forces one dimension at a time.
Instead of using our usual coordinate system containing horizontal and vertical axes,
RRHS Physics
Page 39
it makes more sense in this situation to rotate our axes so that they are
perpendicular and parallel to the surface of the inclined plane (the same direction as
the acceleration). In other words, our x direction will be parallel to the plane and the
y direction will by perpendicular to the plane.
Since the normal force is already perpendicular to the plane (and any applied force
along with the force of friction will usually be parallel to the plane), only the force of
gravity must be broken up into components.
Figure 17: Ramp Free Body Diagram with Components
This can be done as shown in Figure 18 (where the Fg vector from the previous
diagram has been enlarged).
Figure 18: Gravity Components
The angle  in the top of the triangle is the same angle as the slope of the inclined
plane (try showing this using geometry). Using trigonometry, it can be found that the
two components are
Fgx  mg sin 
(1.1)
Fgy  mg cos 
(1.2)
and
RRHS Physics
Page 40
The two components of gravity can be thought of as having different roles in this
situation: Fgx is down (and parallel) to the ramp and is trying to accelerate the box
down the ramp; Fgy is pushing the box onto the ramp and is responsible for keeping it
in contact with the ramp.
If we apply Newton’s Second Law to the perpendicular forces we get
ma y  Fy
ma y  FN  Fgy
ma y  FN  Fgy
Since there is no acceleration perpendicular to the plane we now have
m(0)  FN  Fgy
FN  Fgy
where Fgy can be found using equation (1.2). If friction is present, the normal force
can then be used in this calculation. Again notice that FN  Fg .
Similarly, the parallel forces can be used to obtain an expression for the parallel
acceleration on the inclined plane
max  Fx
max  Fgx
where Fgx
can be found using equation (1.1). Notice that this is just a simple
analysis where friction and other external forces have not been included; if present,
these would have to be considered in the force analysis. Again, it is extremely
important to draw a free body diagram at the start of the problem!
Example
A 1200 kg car is on an icy (frictionless) hill that is inclined at an angle of 12 o with the
horizontal. What is the acceleration of the car down the hill?
Solution
Since there is no friction, the only forces involved are gravity and the normal force:
RRHS Physics
Page 41
Since the acceleration that we are looking for is parallel to the inclined plane, we will
rotate our axes so that the x-axis is parallel to the inclined plane.
m  1200kg
  12
ay  0
ax  ?
Parallel Forces
If we use down the ramp as positive,
max   F x
max  Fgx
max  mg sin 
ax  (9.80)(sin12)
 2.0m / s 2
Although we were given the mass of the car, notice that it cancelled out in this
problem and was not needed.
RRHS Physics
Page 42
Check Your Learning
A 1200 kg car is on an icy hill that is inclined at an angle of 12 o with the horizontal.
As the car starts sliding, the driver locks the wheels. The coefficient of friction
between the locked wheels and the icy surface is 0.14. What is the acceleration of
the car down the hill?
RRHS Physics
Page 43
1.2.2 In Class or Homework Exercise
1. An 18.0 kg box is released on a 33.0o incline and accelerates at 0.300 m/s2.
What is the coefficient of friction?
2. A box (mass is 455 g) is lying on a hill which has an inclination of 15.0o with
the horizontal.
a. Ignoring friction, what is the acceleration of the box down the hill?
b. If there is a coefficient of friction of 0.20, will the box slide down the hill?
If so, at what acceleration?
3.
4.
5.
6.
c. How much force is required to push the box up the ramp at a constant
speed?
A 165 kg piano is on a 25.0 ramp. The coefficient of friction is 0.30. Jack is
responsible for seeing that nobody is killed by a runaway piano.
a. How much force (and in what direction) must Jack exert so that the
piano descends at a constant speed?
b. How much force (and in what direction) must Jack exert so that the
piano ascends at a constant speed?
A car can decelerate at -5.5 m/s2 when coming to rest on a level road. What
would the deceleration be if the road inclines 15o uphill?
A sled is sliding down a hill and is 225 m from the bottom. It takes 13.5 s for
the sled to reach the bottom. If the sled’s speed at this location is 6.0 m/s and
the slope of the hill is 30.0 , what is the coefficient of friction between the hill
and the sled?
A 5.0 kg mass is on a ramp that is inclined at 30o with the horizontal. A rope
attached to the 5.0 kg block goes up the ramp and over a pulley, where it is
attached to a 4.2 kg block that is hanging in mid air. The coefficient of friction
between the 5.0 kg block and the ramp is 0.10. What is the acceleration of this
system?
Figure 19: Diagram for Question 6
RRHS Physics
Page 44
7. A force of 500.0 N applied to a rope held at 30.0o above the surface of a ramp
is required to pull a box weighing 1000.0 N at a constant velocity up the plane.
The ramp has a base of 14.0 m and a length of 15.0 m. What is the coefficient
of friction?
Figure 20: Diagram for Question 7
8. If a bicyclist (75 kg) can coast down a 5.6o hill at a steady speed of 7.0 km/h,
how much force must be applied to climb the hill at the same speed?
RRHS Physics
Page 45
1.2.3 Module Summary
In this module you learned that
o Force vectors can be broken into components so that dynamics situations can
be analyzed using free body diagrams and Newton’s Second Law.
o Situations involving inclined planes (ramps) can be analyzed by rotating the
coordinate system so that the x-axis is parallel to the ramp and the y-axis is
perpendicular to the ramp. The components for the force of gravity can then
be given by
Fgx  mg sin 
Fgy  mg cos 
RRHS Physics
Page 46
Module 1.3 Equilibrium
1.3.1 Translational Equilibrium
You saw in Unit 3 that if two equal but opposite forces are applied to an object, the
net force is zero and the object is said to be in equilibrium. A body in equilibrium at
rest in a particular reference frame is said to be in static equilibrium; a body moving
uniformly at constant velocity is in dynamic equilibrium. We will be dealing with
mainly static equilibrium in this module, although the net force is zero in both cases.
We will now extend our discussion of equilibrium to two dimensions.
The type of equilibrium discussed in Unit 3 is known as translational equilibrium.
Consider a mass being supported in midair by two ropes.
Figure 21: Hanging Mass
The mass is stationary; therefore, it is obviously not accelerating either sideways or
up and down. The net force must therefore be zero and the object is said to be in
translational equilibrium.
RRHS Physics
Page 47
Figure 22: Hanging Mass Free Body Diagram
As can be seen by the free-body diagram, there are three forces acting on the mass.
As we said, the net force acting on the mass must be zero; therefore,
FT 1  FT 2  Fg  0 . Remember, these are vectors so they must add as vectors5 to be
zero, as shown in the following vector diagram:
Figure 23: Hanging Mass Vector Diagram
Note that our vector diagram starts and ends at the same point; therefore, the
resultant vector (the net force) is zero.
Since force is a vector, the components of the net force on a body in equilibrium
must each be zero, so
Fx  0
(1.3)
and
5
When drawn head to tail, they must return to the starting point.
RRHS Physics
Page 48
Fy  0
(1.4)
Looking at the components in the x and y direction separately, this tells us that in
the x direction
 Fx  0
F1x  F2 x  0
F1x  F2 x  0
F1x  F2 x
and in the y direction
F
y
0
F1 y  F2 y  Fg  0
F1 y  F2 y  Fg  0
F1 y  F2 y  Fg
The requirement that the net force be zero (which provides translational equilibrium)
is only the first condition for equilibrium. The second condition will be discussed in
section 5.3.2.
Equilibrant Force
If the vector sum of all of the forces acting on an object is not zero, there will be a
net force in some direction. There is a single additional force that can be applied to
balance this net force. This additional force is called the equilibrant force. The
equilibrant force is equal in magnitude to the sum of all of the forces acting on the
object, but opposite in direction.
Example
A 20.0 kg sack of potatoes is suspended by a rope. A man pushes sideways with a
force of 50.0 N and maintains this force so that the sack is in equilibrium. What is the
tension in the rope and what angle does the rope make with the vertical?
Solution
While the man is pushing sideways, the rope will be at an angle as shown in the
picture below:
RRHS Physics
Page 49
m  20.0kg
Fp  50.0 N
FT  ?
Fg  mg
 (20.0)(9.80)
 196 N
Our free body diagram will look like this:
Since the bag is in equilibrium, the net force must be zero; therefore, FT  Fg  Fp  0 .
RRHS Physics
Page 50
Since the force of gravity and the force of the person pushing are perpendicular, the
Pythagorean Theorem can be used to find the magnitude of the tension and using
trigonometry will allow us to calculate the angle.
FT  Fg2  Fp2
 (196) 2  (50.0) 2
 202 N
RRHS Physics
tan  
Fp
Fg
50.0
196
  14.3

Page 51
Check Your Learning
Joe wishes to hang a sign weighing 750.0 N so that cable A attached to the store
makes a 30.0 angle as shown in the picture below. Cable B is attached to an
adjoining building and is horizontal. Calculate the necessary tension in cable B.
RRHS Physics
Page 52
1.3.1 In Class or Homework Exercise
1. Find the unknown mass in the diagram below:
2. A sign with a mass of 165 kg is supported by a boom and a cable, as shown
in the diagram below. The cable makes an angle of 36o with the boom. Find
the magnitude of the force exerted by the boom and the cable on the sign.
Ignore the mass of the boom.
3. Find the tensions FT 1 and FT 2 in the two strings indicated:
4. In the diagram shown below, block A has a mass of 10.5 kg and block B has a
mass of 52.6 kg. The string runs from Block B to the wall. The segment of
string from block B to point P on the string is horizontal. The friction between
block B and the table is unknown. Find the minimum coefficient of friction
between block B and the table that would prevent block B from moving.
RRHS Physics
Page 53
5. Three students are pulling ropes that are attached to a car. Barney is pulling
north with a force of 235 N; Wilma is pulling with a force of 175 N in a direction
23o E of N; Betty is pulling with 205 N east. What equilibrant force must a
fourth student, Fred, apply to prevent acceleration?
6. Your mother asks you to hang a heavy painting. The frame has a wire across
the back, and you plan to hook this wire over a nail in the wall. The wire will
break if the force pulling on it is too great, and you don't want it to break. If the
wire must be fastened at the edges of the painting, should you use a short
wire or a long wire? Explain.
7. When lifting a barbell, which grip will exert less force on the lifter's arms: one
in which the arms are extended straight upward from the body so that are at
right angles to the bars, or on in which the arms are spread apart so that the
bar is gripped closer to the weights? Explain.
RRHS Physics
Page 54
1.3.2 Torque and Rotational Equilibrium
Until now, we have treated all objects as point sources when drawing our free body
diagrams; in other words, we were not concerned with the where on the object the
force was being applied. In many situations, however, it makes a difference on which
part of the object we apply a force.
Even if all of the forces acting on an object balance, it is possible for the object not to
be in total equilibrium. Consider a board where equal forces are applied at opposite
ends of the board, but one up and one down.
Obviously, even though the forces are equal and opposite and the board as a whole
should not move up or down, the board will begin to spin. It is not in rotational
equilibrium. Rotational equilibrium refers to the situation where there is no rotary
motion. To examine this more, we must introduce the notion of a torque.
Torque
A torque has the same relationship to rotation as force does to linear movement. It
can be thought of as a twisting force. To measure the rotating effect of a torque, it is
necessary to choose a stationary reference point for the measurements (the pivot
point). This pivot point can be chosen arbitrarily, since the point of rotation is often
not known until the rotation begins.6
The size of a torque depends on two things:
1. The size of the force being applied (a larger force will have a greater effect)
2. The distance away from the pivot point (the further away from this pivot, the
greater the effect).
A line drawn from the pivot to the force that is providing the torque is known as the
torque arm. It is along this torque arm that the distance should be measured.
A torque  is the product of a force multiplied by a distance from the pivot.
6
If there is a natural pivot point (for example, on a see-saw) then it usually makes sense to choose
this as the pivot point. However, it is not necessary to do so.
RRHS Physics
Page 55
  F r
(1.5)
where it is only the component of the force that is perpendicular to the torque arm
that contributes to the torque (try opening a door by pushing parallel to the door – it
does not work!).
As can be seen from equation (1.5), the units for torque are N·m , if the force F is in
newtons and the distance r from the pivot is in metres. You may recall from Unit 4
that a N·m was defined as a joule; however, this unit is not called a joule when
calculating torques as the force and the distance are perpendicular, not parallel (as
they are when calculating work).
Example 1
Suppose that you are trying to open a door that is 70.0 cm wide. You are applying a
force of 68 N 10.0 cm from the outer edge of the door, but you are pushing at an
angle of 75o from the surface of the door. What torque are you applying on the door?
Solution
Figure 24: Overhead view of door in Example 1
As can be seen in the overhead view of the door, the force is being applied 60.0 cm
from the natural pivot point, which is the hinge. Remember also, that it is only the
component of the force that is perpendicular to the torque arm (the door) that
contributes to the torque.
r  0.600m
F  68 N
  75
 ?
RRHS Physics
Page 56
  F r
 Fy r
 Fr sin 
 (68)(0.600) sin 75
 39 N  m
RRHS Physics
Page 57
Rotational Equilibrium
While forces were described using up, down, left, right, etc., torques are described
using the terms clockwise and counterclockwise. A clockwise torque added to an
equal (in magnitude) counterclockwise torque will be zero. Rotational equilibrium is
attained if the sum of all of the torques is zero.
  0
(1.6)
This is the second condition for static equilibrium. Just as the components of a force
are positive or negative depending on their direction, so is a torque. If a clockwise
torques is considered to be positive, then a counter-clockwise torque must be
considered to be negative; therefore, equation (1.6) can also be expressed by saying
that the sum of all of the clockwise torques must equal the sum of all of the counterclockwise torques.
 cw   ccw  0
 cw   ccw
(1.7)
As we have seen, there are two conditions for static equilibrium:
1. The sum of the forces is zero (providing translational equilibrium), and
2. The sum of the torques is zero (providing rotational equilibrium).
It is important to remember that when calculating the torques, all distances must be
measured from the pivot point. Any convenient location can be chosen for the pivot
point.
An equilibrant force must provide both translational and rotational equilibrium. When
finding an equilibrant force to satisfy both of these conditions, it is necessary to find
both the force itself (magnitude and direction) and the location of application.
Centre of Gravity One of the forces often involved in calculating the torques on an
object is the force of gravity. Before dealing with torques, we were not usually
concerned with the location of the force on a body, but for calculating torques, this is
important. Where does gravity act on a body? Of course, it acts on every particle in
the body, but there is a point called the centre of gravity (cg) where the entire force of
gravity can be considered to be acting. The center of gravity is the point at which we
could apply a single upward force to balance the object. For a mass with a uniform
distribution of mass (such as a ruler), the centre of gravity would be at the geometric
center (the middle of the ruler).
Example 2
RRHS Physics
Page 58
A 2.0 kg board serves as a see-saw for two children. One child has a mass of 30.0
kg and sits 2.5 m from the pivot point. At what distance from the pivot must a 25.0 kg
child sit on the other side to balance the see-saw? Assume that the board is uniform
and centred over the pivot.
Solution
mb  2.0kg
m1  30.0kg
Fg1  m1 g
m2  25.0kg
 (30.0)(9.80)
r1  2.5m
 294 N
Fg 2  m2 g
 (25.0)(9.80)
 245 N
r2  ?
Drawing a free body diagram, we have
If we choose our pivot point here to be the centre of the board, then there is no
torque due to either the force exerted by the pivot or the weight of the board, since
both of these forces act on the pivot point and the torque arm is zero. The only
torques present will be due to the weight of the two children. Since we already know
both of these forces, we do not have to consider translational equilibrium in this
problem.
 cw   ccw
Fg 2 r2  Fg1r1
245r2  (294)(2.5)
r2  3.0m
The second child must be 3.0 m away from the pivot on the other side of the board.
Example 3
RRHS Physics
Page 59
A 4.0 m platform with a uniform distribution of mass has a 3.2 kg box 0.80 m from the
left end. The mass of the platform is 2.0 kg. Calculate the size and location of the
required equilibrant force.
Solution
The equilibrant force is the single force that will provide equilibrium (both
translational and rotational). In this case, the direction of the equilibrant force is
obviously upward since all of the forces on the platform are downward.
Since the platform has a uniform distribution of mass, the center of gravity is at the
center of the platform. We can guess at the approximate location of the upward
equilibrant force.
In this situation, there is no natural (or fixed) pivot point – the platform could rotate
about a number of points. In this situation, it is completely arbitrary where we pick the
pivot point. For this problem, we will choose the left end to be the pivot point (as
indicated by the X in the diagram).
mb  3.2kg
m p  2.0kg
rb  0.80m
rp  2.0m
Fgb  mb g
 (3.2)(9.80)
 31.4 N
Fgp  m p g
 (2.0)(9.80)
 19.6 N
req  ?
Translational Equilibrium
To satisfy translational equilibrium, the forces must balance – in this case, the total
force up must equal the total force down.
RRHS Physics
Page 60
Fy  0
0  Feq  Fgb  Fgp
Feq  Fgb  Fgp
 31.4  19.6
 51N
Rotational Equilibrium
To satisfy rotational equilibrium, the torques must balance – the clockwise torques
must equal the counter-clockwise torques. In this example, the forces of gravity on
the platform and on the box provide clockwise torques; the upward equilibrant force
will provide a counter-clockwise torque.
  0
 cw   ccw  0
 cw   ccw
Fgb rb  Fgp rp  Feq req
(31.4)(0.80)  (19.6)(2.0)  (51)req
req  1.3m
Feq  51N up, 1.3 m from the left end
RRHS Physics
Page 61
Check Your Learning
A uniform 1500 kg bridge, 20.0 m long, supports a 2200 kg truck whose centre of
mass is 5.0 m from the right support column as shown in the diagram below.
Calculate the force on each of the vertical support columns.
RRHS Physics
Page 62
1.3.2 In Class or Homework Exercise
1. A person is trying to open a door that is 90.0 cm wide. If there is a spring on a
door 5.0 cm from the hinges which exerts a force of 60.0 N to keep the door
closed,
a. How much force must be used to open the door if the force is applied at
the outer edge of the door?
b. How much force must be used if the force from the person is applied
15.0 cm from the hinges?
2. A 60.0 kg person is sitting 1.2 m from the pivot on a see-saw. A 50.0 kg
person is sitting 0.90 m away from the pivot on the other side. Where must a
22.0 kg child sit to balance the see-saw?
3. A long board is holding a person in the air. The person has a mass of 75.0 kg
and is located 2.0 m from one end. The 10.0 m board has a mass of 10.0 kg,
and its center of gravity is located 4.0 m from the same end as the person.
The board is being held up by two students, one at either end. What force is
required by each student to hold the board up?
4. Find the equilibrant force (magnitude, direction, and location) in the following
diagram.
5. Find the equilibrant force (magnitude, direction, and location) which will
stabilize the following beam:
6. In the following diagram, determine the magnitude, direction, and point of
application of the necessary equilibrant force.
RRHS Physics
Page 63
7. A 5.0 m long ladder leans against a wall at a point 4.0 m above the ground.
The ladder is uniform and has a mass of 12.0 kg. Assuming that the wall is
frictionless (but the ground is not) determine the force exerted on the ladder
by the wall.
8. A sign with a mass of 165 kg is supported by a 35 kg boom (with a uniform
distribution of mass and a length of 1.6 m) and a cable, as shown in the
diagram below. The cable makes an angle of 36o with the boom. Find the
magnitude of the force exerted by the cable.
9. Calculate the forces F1 and F2 that the supports exert on the diving board
shown below when a 50.0 kg person stands at its tip.
a. ignoring the mass of the board
b. If the board has a mass of 40.0 kg (uniformly distributed)
RRHS Physics
Page 64
1.3.3 Module Summary
In this module you learned that
o An object is said to be in static equilibrium if it is in both translational
equilibrium and rotational equilibrium.
o Translational Equilibrium is achieved when the net force is zero:
Fx  0
Fy  0
o Torque can be calculated using the equation
  F r
o Rotational Equilibrium is achieved when the net torque is zero:
  0
RRHS Physics
Page 65
Module 1.4 2D Collisions
1.4.1 Conservation of Momentum
You learned in grade 11 that the total momentum of an isolated system remains
constant. Also, if you remember from grade 11, momentum is a product of mass and
velocity ( p  mv ). Since velocity is a vector, so is momentum. This vector nature of
momentum becomes extremely important in two dimensional collisions.
When you analyzed one dimensional collisions, you could show that in an isolated
system the momentum of each object before the collision added up to equal the total
momentum after the collision. This still applies in two dimensional collisions, but
remember that momentum is a vector so it must be added as a vector!! For a
collision involving two objects in one dimension, you would write
pa  pb  pa  pb
(1.8)
ma va  mb vb  ma va  mbvb
(1.9)
or, since p  mv ,
where primed quantities mean after the collision and unprimed mean before the
collision. The vector nature of the momentum was addressed in one dimensional
situations using positive or negative values for the velocities.
In two dimensions, the vector nature of momentum does not allow simple algebraic
operations using equation (1.9). Although you can still express the conservation of
momentum using equations (1.8) and (1.9), special attention must be paid to the
vector nature of momentum. To add momentum vectors in two dimensions, a
vector diagram must be drawn. Equation (1.9) can only be used algebraically if
you first break the vectors into components and then apply the equation in each
dimension.
Consider the example of a ball moving to the right that collides with another ball at
rest.
If the collision is not head on, the two balls will go in different directions after the
collision.
RRHS Physics
Page 66
Just as with one dimensional collisions, the sum of all of the momentum vectors after
the collision ( pa and pb ) is equal to the total of the momentum vectors before the
collision ( pa ).
pa  pa  pb
(1.10)
Since momentum is a product of mass (a scalar) and velocity (a vector), the
momentum vector for an object will be in the same direction as the velocity vector of
the object; however, remember that it is momentum that is conserved, not velocity –
Do not draw a velocity vector diagram when solving these problems! The
momentum vector diagram for equation (1.10) would look like this:
where p t is really just p a , since there is only one momentum vector before the
collision.
The individual momentum vectors can be found using the formula p  mv . We can
now use our usual methods of component analysis for solving vector problems.
If we draw our components into the momentum vector diagram, we see that the
momentum is conserved in each dimension.
RRHS Physics
Page 67
In other words, the sum of the x components of momentum before the collision are
equal to the sum of the x components of momentum after the collision.
pa  pax  pbx
 pax  pbx
where the momentum components can be found using the appropriate velocity
components ( pax  ma vax and pbx  mb vbx ).
Similarly the sum of the y components of momentum before the collision are equal to
the sum of the y components of the momentum after the collision. Since the original
y momentum is zero in this example, the y momentum after the collision is still zero.
0  pay  pby
0  pay  pby
RRHS Physics
Page 68
1.4.1 In Class or Homework Exercise
1. A collision between two vehicles occurs at a right angled intersection. Vehicle
A is a car of mass 1800 kg travelling at 60. km/h north. Vehicle B is a delivery
truck of mass 3500 kg initially travelling east at 45 km/h. If the two vehicles
remain stuck together after the impact, what will be their velocity after the
impact? How much kinetic energy was lost in the collision?
2. A radioactive nucleus at rest decays into a second nucleus, an electron, and a
neutrino. The electron and neutrino are emitted at right angles and have
momenta of 8.6 1023 kg m/s and 6.2 1023 kg m/s. What is the magnitude
and direction of the momentum of the recoiling nucleus?
3. A collision investigator is called to an accident scene where two vehicles
collided at a right-angled intersection. From skid marks, the investigator
determined that car A, mass 1400 kg was travelling 50. km/h west before
impact. The two vehicles remained stuck together after impact and the
velocity of the cars after impact was 10. km/h in a direction 30.0 W of N.
a. What was the mass of vehicle B?
b. How fast was car B travelling before the accident?
4. Two streets intersect at a 40. angle. Car A has a mass of 1500 kg and is
travelling at 50. km/h. Car B has a mass of 1250 kg and is travelling 60. km/h.
If they collide and remain stuck together, what will be the velocity of the
combined mass immediately after impact?
5. A billiard ball of mass 0.400 kg moving with a speed of 2.00 m/s strikes a
second ball, initially at rest, of mass 0.400 kg. The first ball is deflected off at
an angle of 30.0o with a speed of 1.20 m/s. Find the speed and direction of the
second ball after the collision.
6. Billiard ball A is moving at a speed of 3.0 m/s east when it strikes an equal
mass ball B at rest. The two balls are observed to move off at 45o angles to
A’s original direction – ball A goes northeast and ball B goes southeast. What
are the speeds of the two balls after the collision?
7. A grenade of mass 10.0 kg explodes into 3 pieces in the same plane, two of
which A (5.0 kg) and B (2.0 kg), move off as shown. Calculate the velocity of
the third piece C.
RRHS Physics
Page 69
1.4.2 Elastic and Inelastic Collisions
Elastic Collisions
As you learned in grade 11, an elastic collision is one in which no kinetic energy is
lost; the total kinetic energy of the particles before the collision is the same as the
total kinetic energy of the particles after the collision. For a two body collision, this
would be expressed as
1
1
1
1
ma va2  mb vb2  ma va2  mbvb2
2
2
2
2
(1.11)
Remember that energy is not a vector; therefore, it is only the magnitude of the
velocity that is used in (1.11).
Consider the special case where particle b is initially at rest. We now have
1
1
1
ma va2  ma va2  mb vb2
2
2
2
If the mass of each particle is the same, then after cancelling the mass and the factor
of one half, our conservation of energy equation (1.11) reduces to
va2  va2  vb2
(1.12)
which is really an expression of the pythagorean theorem. Since the masses are
equal, the velocity vectors are proportional to the momentum vectors. A velocity
vector diagram in this situation7 would therefore show that the vectors va and vb
would add to give the vector va . Since the magnitudes of these vectors are related
by the pythagorean theorem, the vector diagram must be a right angle triangle.
In other words, va and vb (and pa and pb ) are perpendicular to one another; after
this collision, the two particles move off at right angles to one another. Remember,
though, that this is only true for the special case where the two objects have the
same mass, the collision is elastic, and one of the particles is initially at rest.
Inelastic Collisions
An inelastic collision is one in which the kinetic energy is not conserved; some of the
energy is transformed into other types of energy, such as thermal energy. A
completely inelastic collision is one in which the objects stick together; some energy
is lost, but a completely inelastic collision does not mean that all of the energy is lost.
In this type of collision, it may be possible to calculate the amount of energy lost by
comparing the total initial kinetic energy with the total final kinetic energy.
7
A velocity vector diagram can be applied here only because the masses are all the same; therefore,
every velocity vector is multiplied by the same factor to obtain the corresponding momentum vector
RRHS Physics
Page 70
1.4.2 In Class or Homework Exercise
1. A proton travelling with speed 8.2 105 m/s collides elastically with a stationary
proton. One of the protons is observed to be scattered at a 60.o angle. At what
angle will the second proton be observed, and what will be the velocities of
the two protons after the collision?
2. Two cars collide at an intersection. The first car has a mass of 925 kg and
was travelling north. The second car has a mass of 1075 kg and was
travelling west. Immediately after impact, the first car had a velocity of 52.0
km/h, 40.0o north of west, and the second car had a velocity of 40.0 km/h,
50.0o north of west. What was the speed of each car prior to the collision?
3. Sphere A of mass 2.0 kg is travelling at 10.0 m/s west and approaches sphere
B of mass 5.0 kg travelling at 5.0 m/s N. After the collision, sphere A moves
away at 4.0 m/s in a direction 35o N of E.
a. What is the final velocity of sphere B?
b. Was the collision elastic?
4. A particle of mass m travelling with a speed v collides elastically with a target
particle of mass 2m (initially at rest) and is scattered at 90.0 .
a. At what angle does the target particle move after the collision?
b. What are the particles' final speeds?
c. What fraction of the initial kinetic energy is transferred to the target
particle?
5. A billiard ball is moving North at 3.00 m/s, and another is moving East with a
speed of 4.80 m/s. After the collision (assumed elastic), the second ball is
moving North. What is the final direction of the first ball, and what are their
final speeds?
6. A billiard ball of mass ma  0.40 kg strikes a second ball, initially at rest, of
mass mb  0.60 kg. As a result of this elastic collision, ball A is deflected at an
angle of 30o and ball B at 53o . What is the ratio of their speeds after the
collision?
RRHS Physics
Page 71