Download A2_Unit4_03_Momentum_02

Book Reference : Pages 4-17
1.
To revisit Newton’s 2nd law in terms of
momentum.
2.
To define the impulse of a force and connect
it to the change in momentum
3.
To understand the significance of the area
under a force v time graph
4.
To be able to complete “impulse of a force”
calculations
5.
To revisit car safety
Newton’s 2nd law :
The rate of change of momentum of an object
is proportional to the resultant force on it. The
resultant force is proportional to the change in
momentum per second
At AS we simply considered this to be F=ma
We will now revisit this is terms of momentum
Consider an object of constant mass m acted on
by a constant force F.
The force causes an acceleration from the initial
speed of u to the final speed v.
Therefore initial momentum is mu and the final
momentum is mv. So the change in momentum is
mv –mu
F  change in momentum
time taken
mv –mu
t
F  mv –mu
can be rewritten as
t
F  m(v – u)
From SUVAT a = (v-u)/t
t
F  ma after defining a suitable constant
of proportionality
F = kma
We can make k=1 by defining the unit of
force.....
The Newton is the amount of force that gives an
object of mass 1kg and acceleration of 1 ms-2
We can write the 2nd law as:
F = (mv)
t
This can be used in two scenarios:
Firstly if the mass is constant then (mv)/t
becomes m v/t
Change in velocity i.e. acceleration
Secondly if the mass changes at a constant rate
then (mv)/t becomes v m/t
Where m/t is the change in mass per second
This could be applied to a rocket which is losing
mass each second in the form of hot exhaust gas
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
Units of momentum revisited :
From the area under the graph F x t we naturally
arrive at units of “Ns” for change of momentum
and hence momentum itself.
Ns is simply an alternative form of kgms-1
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 predetermined. By increasing t we
decrease the force acting on the occupants
A train of mass 24,000kg moving at a velocity of
15ms-1 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 120ms-1 at which
point it takes off. The engines provide a constant
driving force of 120kN. Calculate the gain in
momentum and the time to takeoff
The velocity of a car of mass 600kg was reduced
from 15ms-1 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 1ms-1