Chapter 7
Momentum
7.1
What is Momentum?
Momentum is a physical quantity which is closely related to forces. We will learn about this
connection a little later. Remarkably momentum is a conserved quantity. This makes momentum
extremely useful in solving a
6.4
Equations of Motion
This section is about solving problems relating to uniformly accelerated motion. Well rst
introduce the variables and the equations, then well show you how to derive them, and after
that well do a couple of examples.
u = starting v
time axis
Area
=
=
=
1
bh
2
1
1s 2m/s
2
1m
Step 3 : Determine the total displacement of the car
Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from t = 14s to t = 15s) the velocity of the
velocity (m/s)
4
15
5
time (s)
12
2
Answer:
Step 1 : Decide how to tackle the problem
We are asked to calculate the displacement of the car. All we need to remember
here is that the area between the velocitytime graph and the time axis gives us the
disp
acceleration (m/s2 )
constant positive acceleration.
Step 8 : Acclerationtime graph  46 seconds
For the nal 2 seconds the object is traveling with a constant velocity. During this
time the gradient of the velocitytime graph is once again zero, and thu
3. Describe the behaviour of the object
velocity (m/s)
Step 3 : Velocitytime graph  02 seconds
For the rst 2 seconds we can see that the displacement remains constant  so the
object is not moving, thus it has zero velocity during this time. We can rea
b) shows the graph for an object moving at a constant acceleration. In this case the acceleration is positive  remember that it can also be negative.
acceleration (m/s2 )
We can obtain the velocity of a particle at some given time from an acceleration ti
6.3.2
VelocityTime Graphs
Look at the velocitytime graph below:
velocity
B
v
t
A
time
This is the velocitytime graph of a cyclist traveling from A to B at a constant acceleration,
i.e. with steadily increasing velocity. The gradient of this graph is ju
Using the rule of Pythagoras:
s
=
=
2
(30m) + (40m)
2
50m in the direction f rom A to C
Step 4 : Now we can determine the average velocity from the displacement and the
time
v
=
=
50m
10s
m
5 in the direction f rom A to C
s
For this cyclist, his velocity
Displacement
Velocity
Acceleration
Figure 6.3: A Relationship Between Displacement, Velocity and Acceleration
6.3.4
Worked Examples
Worked Example 25
Relating displacement, velocity, and accelerationtime graphs
displacement (m)
Question: Given the disp
Step 2 : The velocity during the last 3 seconds
For the last 3 seconds we can see that the displacement stays constant, and that the
gradient is zero. Thus = 0m/s
v
Worked Example 28
From an accelerationtime graph to a velocitytime graph
Question: Given
include in the nal answer the direction of the objects motion
Worked Example 32
Calculating Momentum 1
Question: A ball of mass 3kg moves at 2m.s1 to the right. Calculate the balls
momentum.
Answer:
Step 1 : Decide what information has been supplied
The
6.5
Important Equations and Quantities
Quantity
Displacement
Velocity
Distance
Speed
Acceleration
Units
Symbol Unit
s
,
u v
s
v
a

Base S.I. Units
m + direction
m.s1 + direction
m
m.s1
m.s1 + direction
Table 6.1: Units used in Rectilinear Motion
125
v1
s1
u2
a1 , a 2
balcony
u1
s2
v2
ground
where the subscript 1 refers to the upward part of the balls motion and the subscript
2 refers to the downward part of the balls motion.
Step 2 : Decide how to tackle the problem
First the ball goes upwards with g
We can nd s1 by using equation 6.4
2
v1
=
s1
u2 + 2a1 s1
1
2
v1 u 2
1
2a
(0m/s)2 (10m/s)2
2(10m/s2 )
5m
=
=
=
Step 6 : For Stage 2, decide what information is supplied
For Stage 2 we have the following quantities:
u2
=
0m/s
v2
a2
=
=
t2
s2
=
=
?
10m/s2
?
Step 6 : Do the calculation
v
=
=
m
m
+ (8 2 )(4s)
s
s
m
32
s
0
Step 7 : Time at half the distance: Find an equation to relate the unknown and known
quantities
Here we have the quantities s, u and a so we do this in 2 parts, rst using equation
6.4 to calc
Step 4 : Do the calculation
Substituting in the values of the known quantities this becomes
a
=
=
=
2(725m 100 m 10s)
s
102 s2
2(275m)
100s2
m
5.5 2
s
Step 5 : Quote the nal answer
m
m
The racing car is accelerating at 5.5 s2 , or we could say it is dece
then equation 6.2 becomes
s
=
=
=
u + u + at
t
2
2ut + at2
2
1
ut + at2
2
Equation 6.4
This equation is just derived by eliminating the time variable in the above equation. From
Equation 6.1 we know
vu
t=
a
Substituting this into Equation 6.3 gives
s
2as
where v is the change in velocity, i.e. v = v u. Thus we have
a
=
v
=
vu
t
u + at
velocity (m/s)
Equation 6.2
In the previous section we saw that displacement can be calculated from the area between a
velocitytime graph and the timeaxis. For uniformly a
So, here weve shown that an object traveling at 10m/s for 5s has undergone a displacement
of 50m.
Important: The area between a velocitytime graph and the time
axis gives the displacement of the object.
a)
velocity
velocity
Here are a couple more velocit
displacement (m)
50
A
C
s
t
time (s)10
This graphs shows us how, in 10 seconds time, the cyclist has moved from A to C. We know
y
the gradient (slope) of a graph is dened as the change in y divided by the change in x, i.e x .
In this graph the gradient of
Friction
We all know that Newtons First Law states that an object moving without a force acting on it
will keep moving. Then why does a box sliding on a table stop? The answer is friction. Friction
arises from the interaction between the molecules on the
!
A: Both the car and
the person travelling at
the same velocity
5.5.2
B: The cars turns but
not the person
C: Both the car and
the person are travelling at the same velocity again
Second Law
Denition: The resultant force acting on a body results in
an ac
5.5.1
First Law
Denition: Every object will remain at rest or in
uniform motion in a straight line unless it is made
to change its state by the action of an external force.
For example, a cement block isnt going to move unless you push it. A rocket in spa
Step 1 : Firstly, draw a force diagram for the block
Fnormal = 100N
Fapplied = 75N
Fweight = 100N
Be careful not to forget the two forces perpendicular to the surface. Every object
with mass is attracted to the centre of the earth with a force (the object
Step 2 : Determine the tension in the chain
ii) Since the engine in sketch A is stationary, the resultant force on the engine is zero.
Thus the tension in the chain exactly balances the weight of the engine,
Tchain
=
=
W
2000N
Step 3 : Force diagram for s
Equilibrium
Question: A car engine of weight 2000N is lifted by means of a chain and pulley
system. In sketch A below, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a rope. The
en
50o
40o
T2
T1
W
Each rope exerts a force on the object in the direction of the rope away from the object.
These tension forces are represented by T1 and T2 . Since the object has mass, it is attracted
towards the centre of the earth. This weight is repres
F1
F3
Resultant of
F1 and F2
Equilibrant of
F1 and F2
F2
In the gure the resultant of F1 and F2 is shown in red. The equilibrant of F1 and F2 is then
the vector opposite in direction to this resultant with the same magnitude (i.e. F3 ).
F1 , F2 and
The easiest way to determine this resultant force is to construct what we call a force diagram.
In a force diagram we represent the object by a point and draw all the force vectors connected
to that point as arrows. Remember from Chapter ? that we use the