Chapter 9: Force, Mass and Momentum

Force: Anything that causes the velocity of an object to change (i.e. to speed
up, to slow down or change direction) is called a force.
 Unit: Newton (N) or 1kg m/s/s.
 Vector Quantity
 Symbol is F
A force causes the velocity of a body to change.
i.e. force causes acceleration.
»The acceleration a is directly proportional to the force F causing it.
aαF
 Frictional force:
A force that can cause an object to slow down and stop. To keep an object moving at
a constant speed we must negate friction.
* Air exerts a friction-called air resistance-on any object moving through it.
*when the size of the air resistance is equal to the weight of the man.

Mass: The mass of a body is a measure of how difficult it is to accelerate the
body.
 Unit: Kilogram (kg)
 Scalar quantity
 Symbol is m
The greater the amount of matter being accelerated by a force, the less acceleration it
gets.
»For a given body the acceleration produced by a given force is inversely proportional
to the mass of the body.
a α 1/m
 Newton: 1 Newton is the force that gives a mass of 1 kilogram an
acceleration of 1 m/s.
 aαF
 a α 1/m
Therefore: a α F/m → F = kma where k is a constant.
F = ma i.e. Force = Mass × Acceleration.

Weight and Mass: The weight of an object is the force of the Earth’s
gravity acting on it.
 Unit: Newtons or kg m/s/s
 Symbol is W
W = mg i.e. Weight = Mass ×Acceleration due to gravity.
 Momentum = Mass × Velocity
Unit – kg m/s
 Newton’s laws of motion
1. Every body will remain in a state of rest or travelling at constant velocity
unless an unbalanced external force acts on it.
2. When an unbalanced force acts on a body the rate of change of the body’s
momentum is directly proportional to the force and takes place in the direction
of this force.
Special case
F= ma or F = mv – mu
t
3. If a body A exerts a force on body B then body B exerts an equal but opposite
force on body A
 Principle of Conservation of momentum
States that in any interaction between two or more bodies the total
momentum of the bodies before the collision is equal to the total
momentum after collision PROVIDED NO EXTERNAL
FORCES ACT ON THE SYSTEM OF BODIES.
m1u1 + m2u2= m1v1 + m2v2
Important Questions:
Problem: A gun of mass 3kg fires a bullet of mass 10g with a velocity of
500ms-1. Calculate the recoil velocity of the gun.
Solution: Before the explosion the system had no momentum after the
explosion both the gun and the bullet have momentum – but in the opposite
direction to each other. The sum of these moments must be 0.
Let ‘v’ metres per second be the recoil velocity of the gun. Then by the
principle conservation of momentum we have:
Momentum before = Momentum after.
 (3)(0) + (.01)(0) = 3v + (.01)(500)
 0 = 3v + 5
 v = -1.67ms-1.
Problem: A spacecraft of mass 400kg moving at 1000m/s ejects an object of
mass 20kg at a speed of 2000m/s at right angles to the direction in which the
craft is moving. Calculate the resultant velocity of the craft in magnitude and
direction.
Solution: Since no force act in the original direction of motion the velocity of
the craft in this direction remains 1000m/s. Let x be the recoil velocity acquired
by the craft at right angles to the 1000m/s. Apply conservation of momentum
perpendicular to the original direction of motion:
 (400)(0) = (380)(x) + (20)(-2000)
 0 = 380x – 40000
 x= 105.26 m/s
Resultant velocity v, of spacecraft is found as follows:
 v2 = (105.26)2 + (1000)2
 v = 1005.5 m/s
 Tan θ = 105.26/1000
 θ = 60
I.e. v = 1005.5m/s at 60 to the original direction of motion.
Describe an experiment to verify the principle of conservation of momentum.
Apparatus – two trolleys with velcros, track on incline, ticker tape timer, balance,
ruler.
Procedure –
(insert diagram here)
1)
2)
3)
4)
5)
6)
Attach velcro to each of the trolleys and then using a balance, measure
the mass of each of the trolleys m1 and m2
The apparatus is arranged on a slope as in the diagram. The first trolley
is placed near the top of the track with the ticker tape timer attached to
it. The slope of the track is adjusted until the trolley moves at a
constant speed when given a slight push. This is indicated by regular
intervals between dots on the ticker tape. The second trolley is placed
further down the track at rest.
The first trolley is then given a slight push. The speed of the first
trolley, u1 , and the speed of the combined trolleys after the collision,
v , are then measured using the ticker tape timer.
The speed of the first trolley, u1 , is found by measuring the distance
between the first dot and the eleventh (i.e. ten time intervals) and
dividing this by 10 x (0.02) or 0.2 s.
(0.02 seconds is the time interval between successive dots on our ticker
tape.)
Repeat using the last eleven dots before impact, to determine the speed
of the combined masses, v.
Repeat the experiment with various masses on each trolley.
Result –
To verify the principle of conservation of momentum it must be shown that, in
each case, the total momentum before the collision is equal to the total
momentum after the collision.
m1u1 + m2u2 = m1v1 + m2v2
or in this experiment
m1u1 + 0 = (m1 + m2)v