1.1. To revisit Newton’s 2To revisit Newton’s 2ndnd law in terms of law in terms of momentum.momentum.

2.2. To define the impulse of a force and connect To define the impulse of a force and connect it to the change in momentumit to the change in momentum

3.3. To understand the significance of the area To understand the significance of the area under a force v time graphunder a force v time graph

4.4. To be able to complete “impulse of a force” To be able to complete “impulse of a force” calculationscalculations

5.5. To revisit car safetyTo revisit car safety

Book Reference : Pages 4-17Book Reference : Pages 4-17

Newton’s 2Newton’s 2ndnd law : law :

The rate of change of momentum of an object The rate of change of momentum of an object is proportional to the resultant force on it. The is proportional to the resultant force on it. The resultant force is proportional to the change in resultant force is proportional to the change in momentum per secondmomentum per second

At AS we simply considered this to be F=maAt AS we simply considered this to be F=ma

We will now revisit this is terms of momentumWe will now revisit this is terms of momentum

Consider an object of constant mass Consider an object of constant mass mm acted on acted on by a constant force by a constant force FF..

The force causes an acceleration from the initial The force causes an acceleration from the initial speed of speed of uu to the final speed to the final speed vv..

Therefore initial momentum is Therefore initial momentum is mumu and the final and the final momentum is momentum is mvmv. So the change in momentum is . So the change in momentum is mv –mumv –mu

F F change in momentum change in momentum mv –mu mv –mu

time takentime taken tt

F F mv –mu mv –mu can be rewritten ascan be rewritten as

tt

F F m(v – u) m(v – u) From SUVAT a = (v-u)/tFrom SUVAT a = (v-u)/t

tt

F F ma ma after defining a suitable constant of after defining a suitable constant of proportionalityproportionality

F = kmaF = kma

We can make k=1 by defining the unit of force..... We can make k=1 by defining the unit of force.....

The Newton is the amount of force that gives an The Newton is the amount of force that gives an object of mass 1kg an acceleration of 1 msobject of mass 1kg an acceleration of 1 ms-2-2

We can write the 2We can write the 2ndnd law as: law as:

F = F = (mv)(mv)

tt

This can be used in two scenarios:This can be used in two scenarios:

Firstly if the mass is constant then Firstly if the mass is constant then (mv)/(mv)/t t becomes m becomes m v/v/t t

Change in velocity i.e. accelerationChange in velocity i.e. acceleration

Secondly if the mass changes at a constant rate Secondly if the mass changes at a constant rate then then (mv)/(mv)/t becomes v t becomes v m/m/tt

Where Where m/m/t is the change in mass per secondt is the change in mass per second

This could be applied to a rocket which is losing This could be applied to a rocket which is losing mass each second in the form of hot exhaust gasmass each second in the form of hot exhaust gas

Definition :Definition :

The impulse of a force is defined as the product of The impulse of a force is defined as the product of the force and the time which the force acts forthe force and the time which the force acts for

The impulse = FThe impulse = Ft = t = mvmv

The impulse of the force acting upon an object is The impulse of the force acting upon an object is equal to the change of momentum for the forceequal to the change of momentum for the force

An object of constant mass m is acted upon by a An object of constant mass m is acted upon by a constant force F which results in a change of constant force F which results in a change of velocity from u to vvelocity from u to v

From the 2From the 2ndnd law law F = (mv – mu )/tF = (mv – mu )/t

Rearranging : Ft = mv – mu Rearranging : Ft = mv – mu

Graphically.....Graphically.....

time

forc

e

F

t

Area under graph “Ft” = change of momentum

Units of momentum revisited :Units of momentum revisited :

From the area under the graph F x t we naturally From the area under the graph F x t we naturally arrive at units of “Ns” for change of momentum arrive at units of “Ns” for change of momentum and hence momentum itself.and hence momentum itself.

Ns is simply an alternative form of kgmsNs is simply an alternative form of kgms-1-1

During the Y11 course of study, it was discussed how During the Y11 course of study, it was discussed how many car safety features such as seatbelts, crumple many car safety features such as seatbelts, crumple zones and air bags increase safety by making the crash zones and air bags increase safety by making the crash “last longer”“last longer”

During our Y12 presentations, change in momentum was During our Y12 presentations, change in momentum was connected to car safety. Now taking it further and connected to car safety. Now taking it further and considering the impulse of a force :considering the impulse of a force :

The impulse = FThe impulse = Ft = t = mvmv

For a given crash the mass & velocity of the For a given crash the mass & velocity of the vehicle are predetermined. vehicle are predetermined. By By increasingincreasing t we t we decreasedecrease the force acting on the occupants the force acting on the occupants

A train of mass 24,000kg moving at a velocity of A train of mass 24,000kg moving at a velocity of 15ms15ms-1-1 is stopped by a braking force of 6000N. is stopped by a braking force of 6000N. Calculate :Calculate :

1.1. The initial momentum of the trainThe initial momentum of the train

2.2. The time taken for the train to stopThe time taken for the train to stop

An aircraft with total mass 45,000kg accelerates An aircraft with total mass 45,000kg accelerates on the runway from rest to 120mson the runway from rest to 120ms-1-1 at which at which point it takes off. The engines provide a constant point it takes off. The engines provide a constant driving force of 120kN. Calculate the gain in driving force of 120kN. Calculate the gain in momentum and the time to takeoffmomentum and the time to takeoff

The velocity of a car of mass 600kg was reduced The velocity of a car of mass 600kg was reduced from 15msfrom 15ms-1-1 by a constant force of 400N which by a constant force of 400N which acted for 20s and then by a constant force of 20N acted for 20s and then by a constant force of 20N for a further 20s.for a further 20s.

Sketch a force v time graphSketch a force v time graph

Calculate the initial momentum of the carCalculate the initial momentum of the car

Use your Force v time graph to establish the Use your Force v time graph to establish the change in momentumchange in momentum

Show the final velocity of the car is 1msShow the final velocity of the car is 1ms-1-1