2 1, Acceleration (rate of change of Velocity) velocity is the vector version of speed, (speed in a straight line)v – ut= au = initial velocity (in m/s)v= final velocity (in m/s)t = time (in seconds (s))a = m/s2Can be rearranged tov = u + at
3 Sample questionA car increases it’s velocity from 10 m/s to 20m/s in 5 seconds. What is it’s acceleration?
4 Sample question (answer) v – ut= av = 20m/s u =10m/s t = 5 sec20 – 105= 2m/s2
5 Sample questionA car moving at 10 m/s accelerates at 2m/s2 for 4 seconds. What is it’s final velocity?
6 Sample question (answer) v = u + atCan be re-arranged tou = 10m/s a = 2m/s2 t = 4 secv = x 4 = 18m/s
7 Sample questionA stone block with a mass of 800Kg is lifted from rest with a uniform acceleration by a crane such that it reached a velocity of 14m/s. after 10 seconds. Calculate the tension in the lifting cable
8 Sample question To do this calculation we need to use another equation Force = mass x accelerationF = maWeight (force acting downwards) = mass x acceleration due to gravityWt = mgWt = m x 9.81 (acceleration due to gravity on earth)
9 Sample question (answer) First find the accelerationaccelerationForce due to accelerationF = maTensionWeight mgU = 0 from standing startv – u = a v = a = 1.4m/s2t t
10 Sample question (answer) F = ma F = 800 x 1.4 = NWt of box mg = 800 x 9.81 = N (7.8kN)Tension in cable == 8938N (8.9kN)accelerationForce due to accelerationF = maTensionWeight mg
11 Force due to acceleration Sample questionIf the stone is lowered at the same rate of acceleration the tension is less than the weight of the boxForce due to accelerationF = maTensionaccelerationWeight mgTension in cable == 6728N (6.7kN)
12 LiftYou get the same effect when you travel in a lift. You feel heavier when the lift begins and accelerates upwards and lighter when the lift accelerates downwards
13 A man standing on weighing scales in a lift has a mass of 60 Kg A man standing on weighing scales in a lift has a mass of 60 Kg. The lift accelerates uniformly at a rate of 2m/s2 Calculate the reading on the scales during the period of acceleration .
14 Force due to acceleration WeightmgForce due to accelerationF = ma
15 F = ma F = 60 x 2 = 120NWt of man mg = 60 x 9.81 = NForce acting on scale == 708.6NThe reading on the scale is ÷ 9.81= 72.23Kg
16 Displacement = average velocity x time 2, Displacement (s) is the vector version of distance (distance in a straight line)Displacement = average velocity x times = (v + u) x t2
17 Sample questionA car moving at 10 m/s increases it’s velocity to 20m/s in 4 seconds. How far will it have travelled during this time?
18 Sample question (answer) s = (v + u) x t2v =20m/s u = 10m/s t = 4 secs = ( ) x 42S = 60m
19 Sample questionHow long will it take for an athlete to accelerate from rest to 4 m/s over 8m ?
20 Sample question (answer) s = (v + u) x t2s = v x t2u =0 sot = 2svt = = 4 seconds4
21 3, Displacement related to acceleration With zero acceleration (constant velocity)a = s = uts = ut + at22From a standing start ut = 0s = at22
22 ExampleA car accelerates uniformly from rest and after 12 seconds has covered 40m. What are its acceleration and its final velocity ?
24 Finding v a = v – u a = v v = a x t t t v = 0.56 x 12 = 6.7 m/s From a standing start u = 0a = v – uta = vtv = a x tv = x 12 = 6.7 m/sv = a x t
25 Or alternatively s = (v + u) x t 2 s = v x t 2 v = 80 = 6.7m/s 12 U = 0 from standing starts = (v + u) x t2s = v x t2v = 80 = 6.7m/s12v = 2st
26 4, final velocity, initial velocity acceleration and displacement v2 = u2 + 2 as
27 ExampleIf a car travelling at 10m/s accelerates at a constant rate of 2m/s2, what is it’s final velocity after it has travelled 10m?
28 Sample Question (answer) v2 = u2 + 2 asV2 = (2 x 2 x 10)= = 140m/s2
29 D’Alembert’s Principle When an object is accelerating the applied force making it accelerate has to overcome the inertia. This is the force which resists the acceleration (or deceleration) and is equal and opposite to the applied force. This means that the total force acting on the body is zeroAcceleration (a)Applied Force (F)Inertia Force (Fi)F + Fi = 0 so F = -FiF = maso ma = - Fior Fi = - mamass
31 Free body diagram F (applied) 200N W (weight) Mg = 20 x 9.81 Fr (Friction)20NW(weight)Mg =20 x 9.81=196.2NF (applied)200NFs (spring)150NFree body diagram
32 D’Alembert’s Principle Downward forces are minus and upward forces are positive.From the diagram-200N – 196.2N +20N +150N= N (which is downward as to be expected)
33 Conservation of energy (Gravitational potential energy) GPE = mass x gravity x height(mgh)=5 x 9.81x10490.5joulesWork done (in lifting the mass) = force x distance49.05 x 10= 490.5joules(F = mg)10m5KgWork done = energy gained
34 Kinetic energyGPE = mgh = joulesNeglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom5Kg10mKE (at the bottom) =490.5J = mv22(v = velocity)
35 The velocity at the bottom v =√(2Ke÷m) Kinetic energyNeglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottomGPE = mgh = joules5KgThe velocity at the bottom v =√(2Ke÷m)= √ (2 x ÷ 5)√196.2= 14m/s10m
36 Work and energy Work done = energy used distance Work done (in joules) = Force x distance moved (in direction of force)Work done = energy usedForcedistance
37 ExampleA car of mass 800Kg is stood on a uniform 1 in 10 slope when it’s handbrake is suddenly released and it runs 30 metres to the bottom of the slope against a uniform frictional force of 50N. What is the car’s velocity at the bottom of the slope?
38 Tan of the angle = opposite/adjacent = 1/10= 0.1 ExampleFirst find the angle of the slopeTan of the angle = opposite/adjacent = 1/10= 0.1Tan = 5.7o110
39 Example Then find the height of the slope sin 5.7o = opposite/hypotenuseOpposite = hypotenuse x sine5.7o30 x sine 5.7o=2.98mht30m5.7o
40 Example 1 (conservation of energy method) Find the gravitational potential energy of the car at the top of the slope800 x 9.81 x 2.98= 23.4kNht30m5.7o
41 Example 1 (conservation of energy method) Work done against friction = force x distance = 50N x 30m = 1.5kjht30m5.7o
42 Example 1 (conservation of energy method) Kinetic energy of the car at the bottom of the slopemv2/2 = = 21.9kjv= √(2 x Ke/m)v= √(2 x 21900/800)v= √54.75v = 7.4 m/sht30m5.7o
43 Example 2 (Resolving forces method) Work out the forces involvedWeight of car (force acting vertically downward) = mass x gravity= 800x 9.817848Nht30m5.7o
44 Example 2 (Resolving forces method) Work out the forces involvedExample 2 (Resolving forces method)Wt = weight of car (7848N)Fn is the normal reaction force of the slope on the carFs is the force on the car down the slope5.7oFnFswt
45 Example 2 (Resolving forces method) Work out the forces involvedSine 5.7o = Fs ÷ wtFs = Wt x sin 5.7oFs = sin 5.7oFs = N (Force down the slope)5.7oFnFswt
46 Example 2 (Resolving forces method) Work out the forces involvedResultant force down the slope = Fs – friction force779.5 – 50=749.5N5.7oFnFswt
47 Example 2 (Resolving forces method) Acceleration down the slopea= F/m 749.5/800= 0.94m/s25.7oFnFswt
48 Example 2 (Resolving forces method) Velocity at the bottom of the slopev2 = u2 + 2as (u2 = 0) sov2 = 2asv2 = 2 x 0.94 x 30v =√56.4v = 7.5m/s5.7oFnFswt
49 Linear Momentum and Collisions Conservationof Energy49
50 Momentum = mass x velocity The total momentum remains the same before and after a collisionMomentum is a vector quantity
51 Linear Momentum and Collisions A railway coach of mass 25t is moving along a level track 36km/hr when it collides with and couples up to another coach of mass 20t moving in the same direction at 6km/hr. Both of the coaches continue in the same direction after coupling. What is the combined velocity of the two coaches?
52 Linear Momentum and Collisions Let the mass of the first coach be M1 and the mass of the second coach be M2 and the velocity of the first coach be V1 and the velocity of the second be V2
53 Linear Momentum and Collisions Before coupling the momentum of the first coach is 25 x 36 = 900tkm/hr and the momentum of the second is 20 x 6 =120 tkm/hrWhich is a total of 1020tkm/hr
54 Linear Momentum and Collisions After the coupling the momentum of both is the same as before the coupling which is tkm/hrAnd the combined mass is 45t
55 Linear Momentum and Collisions Velocity after coupling is momentum divided by mass1020÷4522.6 km/hr
56 ExampleA hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and drives it 100mm into the grounda) Calculate the loss of energy on the impact.b) Calculate the work done by the resistance of the ground.c) calculate the average resistance to penetration.
57 ExampleBefore falling the GPE of the hammer is 200kg x 9.81 x 5m = 9810jKinetic energy of the hammer just before impact = 9810j (0.5 x m x v2)Velocity of the hammer just before impact v2 = (9810)/0.5 x m(9810)/0.5 X 200 =98.1v = √98.1 = 9.9m/s
58 thus 500kg x velocity after collision = 1980 kgm/s ExampleMomentum of hammer just before impact = mass x velocity= 200kg x 9.9m/s = 1980 kgm/sMomentum before collision = momentum after collisionmass of hammer plus pile after collision = 200kg + 300kg = 500kgthus 500kg x velocity after collision = 1980 kgm/sVelocity after collision = 1980 ÷ 500 = 3.96m/s
59 ExampleKinetic energy after collision (0.5 x mass x v2) = x 500 x 3.962= 3920jLoss of energy = 9810 – = 5890jTo find deceleration of pile hammerv2 = u2 +2as (v2 =0)u2 = - 2as
60 Examplea = - u2/2s3.962/0.2- 78.4m/s2The minus sign means deceleration, Resistive force of the ground = mass x deceleration500 x78.439.2kN
61 Work done by ground = Force x distance ExampleWork done by ground = Force x distance39.2kN x 0.1 m3.92kj(this agrees with the fact that fact that the kinetic energy of the hammer and pile after impact was 3.92kj and zero when the pile stopped moving in the ground