1. 1.To revisit Newton’s 2 nd 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 Book Reference : Pages 4-17

2. Newton’s 2 nd 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

3. 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 mv –mu time takent

4. F  mv –mucan be rewritten as t F  m(v – u)From SUVAT a = (v-u)/t t F  maafter defining a suitable constant of proportionality F = kma We can make k=1 by defining the unit of force.....

5. The Newton is the amount of force that gives an object of mass 1kg an acceleration of 1 ms -2 We can write the 2 nd 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

6. 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

7. 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

8. 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 2 nd lawF = (mv – mu )/t Rearranging : Ft = mv – mu Graphically..... time force F t Area under graph “Ft” = change of momentum

9. 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

10. 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

11. 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

12. 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