Newton used the concept of momentum to
explain the results of collisions
Momentum
= mass x velocity
p
=mv
Units :
p (kg m/s)
= m (kg) x (m/s)
Note since velocity is a vector quantity, (both
magnitude and direction) then momentum is also
a vector quantity
When objects collide, assuming that there are no
external forces, then momentum is always
conserved.... Definition :
When two or more objects interact, the total
momentum remains constant provided that
there is no external resultant force
Mass 75 kg
Velocity 4m/s
Mass 50 kg
Velocity 0m/s
Mass 125 kg
Velocity ??? m/s
When objects hit each other the resulting
collision can be considered to be either elastic or
inelastic. Momentum and total energy are always
conserved in both cases.
Elastic :
momentum conserved, kinetic
energy conserved, total energy
conserved
Inelastic : momentum conserved, kinetic
energy NOT conserved, total energy
conserved
In an Inelastic collision some of the kinetic energy is
converted to other forms of energy (often heat & Sound)
Two trolleys on an air track are fitted with
repelling magnets. The masses are 0.1kg and
0.15kg respectively. When they are released the
lighter trolley moves to the left at 0.24m/s. What
is the velocity of the heavier trolley
A ball of 0.6kg moving at 5m/s collides with a
larger stationary ball of mass 2kg. The smaller ball
rebounds in the opposite direction at 2.4m/s
Calculate the velocity of the larger ball
Is the Collision elastic or inelastic. Explain your
answer
Definition :
The impulse of a force is defined as the product of
the force and the time which the force acts for
The impulse = Ft = mv
The impulse of the force acting upon an object is
equal to the change of momentum for the force
An object of constant mass m is acted upon by a
constant force F which results in a change of
velocity from u to v
From the 2nd law
F = (mv – mu )/t
Rearranging : Ft = mv – mu
F
force
Graphically.....
Area under graph
“Ft” = change of
momentum
time
t
A train of mass 24,000kg moving at a velocity of
15m/s is stopped by a braking force of 6000N.
Calculate :
1. The initial momentum of the train
2. The time taken for the train to stop
An aircraft with total mass 45,000kg accelerates
on the runway from rest to 120m/s at which point
it takes off. The engines provide a constant driving
force of 120kN. Calculate the gain in momentum
and the take of time
The velocity of a car of mass 600kg was reduced
from a speed of 15m/s by a constant force of
400N which acted for 20s and then by a constant
force of 20N for a further 20s.
Sketch a force v time graph
Calculate the initial momentum of the car
Use your Force v time graph to establish the
change in momentum
Show the final velocity of the car is 1m/s
During the Y11 course of study, it was discussed how
many car safety features such as seatbelts, crumple
zones and air bags increase safety by making the crash
“last longer”
During our Y12 presentations, change in momentum was
connected to car safety. Now taking it further and
considering the impulse of a force :
The impulse = Ft = mv
For a given crash the mass & velocity of the
vehicle are defined. By increasing t we decrease
the force acting on the occupants
We have seen that momentum is a vector
quantity since it’s related to velocity which is a
vector quantity.  direction is important and
therefore we need a “sign” convention to take
this into account.
If we consider a ball with mass m hitting a wall
and rebounding normally, (i.e. at 90°):
Towards the wall we
take as positive
Away from the wall we
take as negative
Initial velocity = +u
Initial momentum = +mu
Final velocity = -u
Final momentum = -mu
If we assume there is no loss of speed after the
impact then considering the change in
momentum...
Ft = final momentum – initial momentum
Ft = -mu – (+mu)
F = -2mu /t
When the impact is oblique, (i.e. At an angle, not
normally at 90°):


Initial velocity = +u
Initial momentum = +mu
In this case we use the normal components of the
velocity. Initially, this is +(u cos ). Similarly this
will give an overall change in momentum of :
Ft = -2mu cos 
A squash ball is released from rest above a flat
surface. Describe how the energy changes is i) it
rebounds to the same height, ii) It rebounds to a
lesser height
If the ball is released from a height of 1.20m and
rebounds to a height of 0.9m show that 25% of
the kinetic energy is lost upon impact
A shell of mass 2kg is fired at a speed of 140m/s
from a gun with mass 800kg. Calculate the recoil
velocity of the gun
A molecule of mass 5.0 x 10-26 kg moving at a
speed of 420m/s hits a surface at right angles and
rebounds at the opposite direction at the same
speed. The impact lasted 0.22ns. Calculate:
i) The change in momentum
ii) The force on the molecule
Repeat the last molecule question. This time the
molecule strikes the surface at 60° to the normal
and rebounds at 60° to the normal.
Angles can be measured in both degrees & radians :
Arc
length

r
The angle  in radians is defined as
the arc length / the radius
For a whole circle, (360°) the arc
length is the circumference, (2r)
 360° is 2 radians
Common values :
45° = /4 radians
90° = /2 radians
180° =  radians
Note. In S.I. Units we use “rad”
How many degrees is 1 radian?
Angular velocity, for circular motion, has
counterparts which can be compared with linear
speed s=d/t.
Time (t) remains unchanged, but linear distance (d)
is replaced with angular displacement  measured
in radians.
Angular displacement 
r

r
Angular displacement is the number of
radians moved
Consider an object moving along the arc of a circle
from A to P at a constant speed for time t:
P
Arc length
Definition : The rate of change of
angular displacement with time
A
“The angle, (in radians) an object
rotates through per second”
r

r
=/t
Where  is the angle turned through in radians, (rad),
yields units for  of rad/s
This is all very comparable with normal linear speed, (or velocity)
where we talk about distance/time
The period T of the rotational motion is the time
taken for one complete revolution (2 radians).
Substituting into :  = /t
 = 2 / T
T = 2 / 
From our earlier work on waves we know that the period (T) &
frequency (f) are related T = 1/f
f =  / 2
Considering the diagram below, we can see that
the linear distance travelled is the arc length
P
Arc length
r

r
A
Linear speed (v) = arc length (AP) / t
v = r /t
Substituting... ( =  / t)
v = r
A cyclist travels at a speed of 12m/s on a bike with
wheels which have a radius of 40cm. Calculate:
a. The frequency of rotation for the wheels
b. The angular velocity for the wheels
c. The angle the wheel turns through in 0.1s in
i radians ii degrees
The frequency of rotation for the wheels
Circumference of the wheel is 2r
= 2 x 0.4m = 2.5m
Time for one rotation, (the period) is found using
s =d/t rearranged for t
t = d/s = T = circumference / linear speed
T = 2.5/12 = 0.21s
f = 1/T = 1/0.21 = 4.8Hz
The angular velocity for the wheels
Using T = 2 / , rearranged for 
 = 2 /T
 = 2 /0.21
 = 30 rad/s
The angle the wheel turns through in 0.1s in
i radians ii degrees
Using  =  / t re-arranged for 
 = t
 = 30 x 0.1
 = 3 rad
= 3 x (360°/2) = 172°
Velocity v
If an object is moving in a circle
with a constant speed, it’s
velocity is constantly changing....
Because the direction is
constantly changing....
If the velocity is constantly
changing then by definition the
object is accelerating
If the object is accelerating, then
an unbalanced force must exist
We can substitute for angular velocity....
a = v2/r
From the last lesson we saw that:
v = r (substituting for v into above)
a = (r)2/r
a = r2
In exactly the same way as we can connect force
f and acceleration a using Newton’s 2nd law of
motion, we can arrive at the centripetal force
which is keeping the object moving in a circle
f = mv2/r
or
f = mr2
Any object moving in a circle is acted upon by a
single resultant force towards the centre of the
circle. We call this the centripetal force
The wheel of the London Eye has a diameter of
130m and takes 30mins for 1 revolution.
Calculate:
a. The speed of the capsule
b. The centripetal acceleration
c. The centripetal force on a person with a
mass of 65kg
The speed of the capsule :
Using v = r
we know that we do a full revolution (2 rad)
in 30mins (1800s)
v = (130/2) x (2 / 1800)
v = 0.23 m/s
The centripetal acceleration:
Using a = v2/r
a = (0.23)2 / (130/2)
a = 0.792 x 10-4 m/s2
The centripetal force:
Using f = ma
F = 65 x 0.792 x 10-4
F = 0.051 N
An object of mass 0.15kg moves around a circular
path which has a radius of 0.42m once every 5s at
a steady rate. Calculate:
a. The speed and acceleration of the object
b. The centripetal force on the object
During the last lesson we saw that an object
moving in a circle has a constantly changing
velocity, it is therefore experiencing acceleration
and hence a force towards the centre of rotation.
We called this the centripetal force: The force
required to keep the object moving in a circle. In
reality this force is provided by another force,
e.g. The tension in a string, friction or the force
of gravity.
Consider a car with mass m and speed v moving
over the top of a hill...
S
mg
r
At the top of the hill, the support force S, is in the
opposite direction to the weight (mg). It is the
resultant between these two forces which keep
the car moving in a circle
mg – S = mv2/r
If the speed of the car increases, there will
eventually be a speed v0 where the car will leave
the ground (the support force S is 0)
mg = mv02/r
v0 = (gr)½
Any faster and the car will leave the ground
A car with mass 1200kg passes over a bridge with a radius
of curvature of 15m at a speed of 10 m/s. Calculate:
a. The centripetal acceleration of the car on the bridge
b. The support force on the car when it is at the top
The maximum speed without skidding for a car with mass
750kg on a roundabout of radius 20m is 9m/s. Calculate:
a. The centripetal acceleration of the car on the
roundabout
b. The centripetal force at this speed
A car is racing on a track banked at 25° to the
horizontal on a bend with radius of curvature of
350m
a. Show that the maximum speed at which the
car can take the bend without sideways
friction is 40m/s
b. Explain what will happen if the car takes the
bend at ever increasing speeds
A car on a big dipper starts from rest and descends
though 45m into a dip which has a radius of
curvature of 78m. Assuming that air resistance &
friction are negligible. Calculate:
a. The speed of the car at the bottom of the dip
b. The centripetal acceleration at the bottom of
the dip
c. The extra force on a person with a weight of
600N in the train
A swing at a fair has a length of 32m. A passenger of
mass 69kg falls from the position where the swing is
horizontal. Calculate:
a. The speed of the person at the lowest point
b. The centripetal acceleration at the lowest
point
c. The support force on the person at the lowest
point
A wall of death ride at the fairground has a radius of
12m and rotates once every 6s. Calculate:
a. The speed of rotation at the perimeter of the
wheel
b. The centripetal acceleration of a person on
the perimeter
c. The support force on a person of mass 72kg at
the highest point