Momentum
CPD Resource for First Teaching
Momentum
In this pack we explore the concept of momentum, which is defined as the product of mass and
velocity
𝑝 = 𝑚𝑣
Momentum is a vector and has the same direction as the velocity of the object. Objects with a larger
mass will have greater momentum than an object with a smaller mass at the same velocity. It can be
used to determine the effect a force has on an object, and is especially useful in the context of
collisions between objects of differing masses.
Momentum is a result of Newton’s laws, and this can be seen by starting with Newton’s second law
𝐹 = 𝑚𝑎 = 𝑚
∆𝑣
∆𝑡
A change in momentum can be written as
∆𝑝 = 𝑚∆𝑣
or when a rate of change is considered
∆𝑝
∆𝑣
=𝑚
∆𝑡
∆𝑡
comparing this expression with the equation for Newton’s second law above, it can be seen that the
rate of change of momentum is a force
∆𝑝
𝐹=
∆𝑡
The change in momentum can be found by determining the area under an F-t diagram.
∆𝑝 = 𝐹∆𝑡
The change in momentum ∆𝑝 is also called impulse.
Conservation of Momentum
Momentum is conserved, momentum cannot be created nor destroyed. In every process the sum of
all changes in momentum must be zero. Conservation of momentum follows from Newton’s third
law: every force has an opposite force of equal magnitude and of the same type.
When two objects interact the forces they exert on each other are the same in magnitude but
opposite in direction. The time the forces act must also be the same.
The change in momentum for each object is therefore also equal in magnitude, but opposite in
direction. The sum of changes in momentum must then be zero. A worked example is given below
for the collision between two vehicles.
Momentum
CPD Resource for First Teaching
Momentum conservation can be derived as follows:
Newton’s third law states that
𝐹2 = −𝐹1
The time that the force acts is the same for both objects, thus the impulse for each object is
∆𝑝1 = ∆𝐹1 ∆𝑡 and ∆𝑝2 = ∆𝐹2 ∆𝑡
The sum of the changes in momentum is
∆𝑝1 + ∆𝑝2 = ∆𝐹1 ∆𝑡 + ∆𝐹2 ∆𝑡
Applying Newton’s third law this becomes
∆𝑝1 + ∆𝑝2 = ∆𝐹1 ∆𝑡 − ∆𝐹1 ∆𝑡 = 0
The sum of changes is zero and momentum is conserved.
Momentum and Kinetic Energy
Kinetic energy E is defined as
1
𝐸 = 𝑚𝑣 2
2
and is a scalar.
Momentum and Kinetic Energy may appear similar on first sight, however both have distinct
applications. Momentum is a vector and gives information on the direction of movement. It is also a
conserved quantity which can be used to calculate the interaction of objects.
In a collision between objects momentum is conserved, but kinetic energy generally is not. While
energy is a conserved quantity, the kinetic energy can be converted to alternative forms (such as
heat or deformation), thus reducing the kinetic energy after a collision.
A good example of the different applications of kinetic energy and momentum is by calculating the
stopping distance and time of an object. An object with mass m is travelling at initial velocity u and
is being brought to rest with a constant force F. The stopping time ∆𝑡 and distance s are to be
determined.
First, to find the stopping time use the change in momentum:
∆𝑝 = 𝐹∆𝑡
𝑚𝑢 = 𝐹∆𝑡
𝑚𝑢
∆𝑡 =
𝐹
Momentum
CPD Resource for First Teaching
To find the stopping distance, use the change in kinetic energy
1
∆𝐸 = 𝑚𝑣 2 − 0
2
𝑊 = ∆𝐸 = 𝐹𝑠
1
𝐹𝑠 = 𝑚𝑣 2
2
𝑚𝑣 2
𝑠=
2𝐹
Worked Example
A goalkeeper catches a football of m=0.45 kg travelling at u=10 ms-1, while a cricketer catches a
cricket ball of m=0.16 kg travelling at u=25 ms-1. Both players exert a stopping force
𝐹 = 1.0 × 103 N on their respective ball. Calculate the momentum, kinetic energy and the stopping
time and distance of both balls using the equations found above.
Football
mass / kg
Speed / m s-1
momentum / kg m s-1
Kinetic energy / J
Force / N
Stopping time / s
Stopping distance / m
Cricket Ball
−2
4.5 × 10
10
4.5
23
1.0 × 103
4.5 × 10−3
2.3 × 10−2
1.6 × 10−2
25
4.0
50
1.0 × 103
4.0 × 10−3
5.0 × 10−2
The heavier, slower football has a larger momentum than the cricket ball and takes a longer time to
stop. The higher kinetic energy of the cricket ball means that it takes less time to stop, but it travels
further.
Worked Example
The impulse can be determined from the area underneath a Force-time diagram, by counting the
number of squares underneath the curve (for more detailed guidance, see the Maths Skills
Handbook). This technique is especially useful for highly varying forces such as experienced during a
collision.
By counting the number of rectangles in the graph below the change in momentum can be
determined. The area of each rectangle is 0.1 × 1 = 0.1 kg m s-1. The change in momentum is then
∆𝑠 = 𝑛𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠 × 0.1 = 303 × 0.1 = 30.3 kg m s-1
Momentum
CPD Resource for First Teaching
Force on an Object
16
14
12
10
F/N 8
6
4
2
0
0
0.5
1
1.5
2
t/s
Worked Example Collision
A small car with mass 1000kg is travelling at 𝑢𝑐𝑎𝑟 = 25 ms-1 when it collides with a van with mass
3000kg travelling in the opposite direction at 𝑢𝑣𝑎𝑛 = −15 ms-1. Directly after the collision, assume
that both vehicles travel as a combined object and that no momentum is exchanged with any other
objects including the road. Calculate the velocity immediately after impact.
(𝑚𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 )𝑣 = 𝑚𝑐𝑎𝑟 𝑢𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 𝑢𝑣𝑎𝑛
𝑣 = (𝑚𝑐𝑎𝑟 𝑢𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 𝑢𝑣𝑎𝑛 )/(𝑚𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 )
𝑣 = 1000 × 25 + 3000 × (−15))/(1000 + 3000) = −5.0 ms-1
The combined velocity is -5 m s-1
Calculate the change in total kinetic energy.
1
1
2
2
𝐸𝑠𝑡𝑎𝑟𝑡 = 𝑚𝑐𝑎𝑟 𝑢𝑐𝑎𝑟
+ 𝑚𝑣𝑎𝑛 𝑢𝑣𝑎𝑛
2
2
1
𝐸𝑒𝑛𝑑 = (𝑚𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 )𝑣 2
2
1
1
1
2
2
∆𝐸 = (𝑚𝑐𝑎𝑟 + 𝑚𝑣𝑎𝑛 )𝑣 2 − 𝑚𝑐𝑎𝑟 𝑢𝑐𝑎𝑟
+ 𝑚𝑣𝑎𝑛 𝑢𝑣𝑎𝑛
2
2
2
1
1
1
∆𝐸 = (1000 + 3000)(−5)2 − × 1000 × 252 + × 3000 × (−15)2 = 6.0 × 105
2
2
2
From this example, it can be seen that while momentum is conserved, the kinetic energy is not. The
kinetic energy “lost” is converted to allow the deformation of the vehicles and the generation of
heat.
Momentum
CPD Resource for First Teaching
Perfectly Elastic and Inelastic Collisions
The example above showed that the kinetic energy was not conserved. This is an example of an
inelastic collision. Perfectly elastic collisions, such as the collisions between atoms in a gas, do fully
conserve kinetic energy (and momentum).
Summary
Momentum and kinetic energy are both ways to relate the combine the velocity and mass of an
object. There are some important differences an overview is given in the table below.
Basic Equation [units]
Change
Used to calculate
In collisions
Momentum
Kinetic Energy
𝑝 = 𝑚𝑣 [kg ms-1 ]
𝐸 = 2 𝑚𝑣 2 [J]
∆𝑝 = 𝐹∆𝑡
∆𝐸 = 𝐹∆𝑠
stopping time
stopping distance
conserved
not-conserved except in fully
elastic collisions
1
Exercises

A student drops a bouncy ball from height h=2 m with zero initial velocity. The ball bounces
a number of times before coming to rest.
o Calculate the momentum and kinetic energy of the ball just before hitting the
ground the first time. Ignore air-resistance.
o The student argues, since the momentum of the ball continuously changes during
the experiment, that momentum is not conserved in this case. Do you agree with
the student? Explain your reasoning.

The two GE90 jet engines on the Boeing 777 aircraft are some of the largest ever built, with
a thrust rating of F=500kN each. When on the runway the pilot increases the engines to
maximum thrust before releasing the brakes. The mass of the aircraft is 𝑚 = 3.0 × 10−5 kg.
o What is the acceleration immediately after releasing the brakes?
o The pilot takes off at 80 ms-1
 What is the momentum of the aeroplane?
 What is the kinetic energy?
o Ignoring air resistance, calculate the minimum time and distance needed to take off.
Momentum
CPD Resource for First Teaching

Determine the change in momentum on the object from the graph below
Force on an Object
20
15
10
5
F/N
0
-5
0
0.5
1
1.5
2
-10
-15
t/s

A fireman is testing a new hose by spraying water against a wall. The hose has a diameter of
5 cm and the flow rate is 20 litres per second. What is the force experienced by the wall?
Ignore the effects of gravity or air resistance.

Watch the following video about the Galilean Cannon on Youtube. Discuss why the small ball
bounces very high. https://www.youtube.com/watch?v=2UHS883_P60