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Momentum: Definition 2

Newton used the concept of momentum to explain the results of collisions Momentum = mass x velocity p = m v 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.

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Momentum: Definition 2

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  1. Newton used the concept of momentum to explain the results of collisions Momentum = mass x velocity p = m v 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 Momentum: Definition 2

  2. 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 Conservation of Momentum 1 Mass 75 kg Velocity 4m/s Mass 125 kg Velocity ??? m/s Mass 50 kg Velocity 0m/s

  3. 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. Types of Collision 1 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)

  4. 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 Problems 1

  5. 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 Impulse of a Force

  6. 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 Graphically..... Force v Time Graphs F Area under graph “Ft” = change of momentum force time t

  7. A train of mass 24,000kg moving at a velocity of 15m/s is stopped by a braking force of 6000N. Calculate : • The initial momentum of the train • 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 Problems 1...

  8. 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 Problems 2...

  9. 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 Car Safety

  10. 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°): Rebound Impacts 1 Towards the wall we take as positive Away from the wall we take as negative Initial velocity = +u Initial momentum = +mu

  11. Rebound Impacts 2 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

  12. When the impact is oblique, (i.e. At an angle, not normally at 90°): Rebound Impacts 3   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

  13. 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 Problems 1...

  14. 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: • The change in momentum • The force on the molecule Problems 2...

  15. Repeat the last molecule question. This time the molecule strikes the surface at 60° to the normal and rebounds at 60° to the normal. Problems 3...

  16. Angles can be measured in both degrees & radians : Arc length Radians & Degrees 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  r Common values : 45° = /4 radians 90° = /2 radians 180° =  radians Note. In S.I. Units we use “rad” How many degrees is 1 radian?

  17. 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 Angular displacement  r Angular displacement is the number of radians moved  r

  18. Consider an object moving along the arc of a circle from A to P at a constant speed for time t: Angular Velocity : Definition P Definition : The rate of change of angular displacement with time “The angle, (in radians) an object rotates through per second” Arc length r  A 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

  19. The period T of the rotational motion is the time taken for one complete revolution (2 radians). Angular Velocity : Period & Frequency • 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

  20. Considering the diagram below, we can see that the linear distance travelled is the arc length Angular Velocity : linear speed P Arc length r  A r Linear speed (v) = arc length (AP) / t v = r /t Substituting... ( =  / t) v = r

  21. A cyclist travels at a speed of 12m/s on a bike with wheels which have a radius of 40cm. Calculate: • The frequency of rotation for the wheels • The angular velocity for the wheels • The angle the wheel turns through in 0.1s in • i radians ii degrees Angular Velocity : Worked example

  22. 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 Angular Velocity : Worked example

  23. The angular velocity for the wheels • Using T = 2 / , rearranged for  •  = 2 /T •  = 2 /0.21 •  = 30 rad/s Angular Velocity : Worked example

  24. 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° Angular Velocity : Worked example

  25. 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 Velocity v Centripetal Acceleration : Introduction acceleration

  26. 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 Centripetal Acceleration : angular

  27. 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 Centripetal Force

  28. The wheel of the London Eye has a diameter of 130m and takes 30mins for 1 revolution. Calculate: • The speed of the capsule • The centripetal acceleration • The centripetal force on a person with a mass of 65kg Worked example 1

  29. 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 Worked example 1

  30. 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 Worked example 1

  31. 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: • The speed and acceleration of the object • The centripetal force on the object Problem 1

  32. 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. Centripetal Acceleration : Recap

  33. Consider a car with mass m and speed v moving over the top of a hill... Over the top 1 S mg r

  34. 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 Over the top 2

  35. A car with mass 1200kg passes over a bridge with a radius of curvature of 15m at a speed of 10 m/s. Calculate: • The centripetal acceleration of the car on the bridge • 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: • The centripetal acceleration of the car on the roundabout • The centripetal force at this speed Problems 1

  36. A car is racing on a track banked at 25°to the horizontal on a bend with radius of curvature of 350m • Show that the maximum speed at which the car can take the bend without sideways friction is 40m/s • Explain what will happen if the car takes the bend at ever increasing speeds Problems 2

  37. 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: • The speed of the car at the bottom of the dip • The centripetal acceleration at the bottom of the dip • The extra force on a person with a weight of 600N in the train Problems 1

  38. 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: • The speed of the person at the lowest point • The centripetal acceleration at the lowest point • The support force on the person at the lowest point Problems 2

  39. A wall of death ride at the fairground has a radius of 12m and rotates once every 6s. Calculate: • The speed of rotation at the perimeter of the wheel • The centripetal acceleration of a person on the perimeter • The support force on a person of mass 72kg at the highest point Problems 3

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