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Module 2

Module 2. Mechanics. Year 11 recap. Distance – A length between 2 defined points measured in meters Displacement – A distance travelled in a defined direction from a specific point measured in meters

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Module 2

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  1. Module 2 Mechanics

  2. Year 11 recap • Distance – A length between 2 defined points measured in meters • Displacement – A distance travelled in a defined direction from a specific point measured in meters • Speed – is the distance covered by an object per unit time measured in meters per second (ms-1) • Average speed = distance/time • Instantaneous speed is different and can be worked out by decreasing the time interval to approaching zero using the gradient of a distance-time graph. • Velocity – is the displacement per unit time • Average velocity = displacement/time • The distinction between speed and velocity plays a vital role in areas such as circular motion. • Q1. A hockey player runs from one end of a 100m hockey pitch to the other end and back to the half way line in 19s. What was his average speed and average velocity?

  3. Acceleration • An acceleration occurs when an object changes velocity. The magnitude of an acceleration is the rate at which this change in velocity occurs. • Acceleration = change in velocity/time • Units are m/s/s written as ms-2 • As acceleration is a change in velocity and not just speed and object with constant speed can accelerate if it changes direction. • Q2. A car accelerates from 10 ms-1 to 40 ms-1 in a time of 6 seconds. What is the acceleration?

  4. S.U.V.A.T Equations • Only apply to uniformly accelerating bodies s = displacement (m) u = initial velocity (ms-1) v = final velocity (ms-1) a = acceleration (ms-2) t = time (s) Q3. A ball is thrown upwards with a velocity of 25 ms-1. What is a) the time taken to return to his hand? And b) the maximum height reached by the ball? Q4. A stone is thrown from a cliff that is 50 m high. How long and at what speed does it hit the sea?

  5. Motion Graphs, distance-time • Summary – (see text book for graphs) • The gradient of a graph is Δy/Δx, so the gradient of a distance-time graph is Δdistance/Δtime which gives the average speed • The gradient at any given point gives you the instantaneous speed • (If we are dealing with a displacement-time graph then the gradient gives average velocity, also as displacement is a vector quantity the graph is able to have negativevalues and gradients) • The area under the graph has no meaning

  6. Motion graphs, velocity-time • The gradient of a graph is Δy/Δx, so the gradient of a velocity-time graph is Δvelocity/Δtime which gives the average acceleration • The gradient at any given point gives you the instantaneous acceleration • The area under the line of any graph gives you the quantities of each axis multiplied together so in this case velocity x time, which equates to displacement. • (If the shape of the graph changes the total area = the sum of any smaller areas such as rectangles/triangles you can find added together.)

  7. Motion graphs, acceleration-time • Acceleration-time graphs that deal with uniform acceleration will only ever consist of vertical and horizontal lines. • The gradient of an acceleration time graph has no meaning for uniform acceleration • The area under the graph gives the acceleration x time = change in velocity.

  8. Mass and Weight • The mass of an object is how much of it there is. (A slightly weak definition but nobody really knows what mass is at this stage!) • The mass of an object does not depend on its location and is (generally although not in nuclear reactions) a conserved scalar quantity.(although we know mass-energy is conserved!) • Weight is a vector quantity as it is a force due to a gravitational field measured in Newtons. • The issue is slightly confused because without a reaction force there is no way to experience weight (astronauts being ‘weightless’) so weight can also be defined as the reaction force provided by a supporting scale. • Weight (N) = mass (kg) x gravitational field strength (N/kg) • Therefore weight depends on the location of the object.

  9. Newton’s Laws • 1st law – A body will remain at rest or continue with constant velocity unless a resultant force acts (ie- no resultant force, no acceleration) • 2nd law – The rate of change of momentum is proportional to and in the direction of the resultant force (ie- F = m x a as we shall see later) • 3rd law – If body A exerts a force on body B, then body B exerts and equal and opposite reaction force on body A.

  10. Parachutist Q5. Using Newton’s laws describe the motion of the parachutist in as much detail as possible.

  11. What role does air resistance play? • Watch the following video • Using Newton’s Laws can you explain why the hammer and feather fall at the same rate on the moon? • Why is this not the case on the Earth?

  12. Static free body diagrams • From Newton’s 1st law we know that any static body must be in a state of equilibrium with all forces balanced. • We can represent this by a free body diagram where we just consider the forces about the object’s centre of mass. • Balanced forces • W x sinθ = friction and W x cosθ = Normal reaction force • If W x sinθ > friction then the block will accelerate down the slope • Q6. Using the diagram above. The block hass a mass fo50kg and the slope is 25°. • A) write out two equations that are true if the block is in equilibrium • B) Work out the size of the friction force and the size of the normal reaction force using the 2 equations.

  13. Work, energy and power • In physics, WORK DONE is the amount of ENERGY TRANSFORMED when a force moves (in the direction of the force) • The amount of work done (measured in Joules) is equal to the force used (Newtons) multiplied by the distance the force has moved (metres in the direction of the force). • Work Done (J) = Force (N) x distance (m) • This means no work is done if no distance is moved so holding a weight is not work in physics! • The work done by or against a force is also INDEPENDENT of the PATH taken. If you lift a weight 5m straight up you do the same work against gravity as if you carried it up a flight of stairs that went up 5m but along 5m as well.

  14. What if the force is at an angle to the distance moved? We need to resolve the force to find the size of the component that is parallel with the distance moved so the equation becomesWork = Force xdistance x cosθqu 7 – A 10N force pulls a block at an angle of 30˚ to the horizontal 50m along the floor. How much work is done by the force? F θ s

  15. Work done by a varying force • When you stretch a spring the longer it gets the harder it is to stretch it further so more work is required. • We can used force-distance graphs to work out the Work Done by a non-constant force • The area under any graph is the product of the quantities of each axis so • force x distance = W.D. = area under a F-d graph.

  16. Work done in stretching a spring Work done in strectching spring = area under graph F/N x/m

  17. Power • Power relates to the RATE at which ENERGY is TRANSFORMED • Power is measured in Watts (W) where; 1 W = 1Js-1 • Mechanical Power can be calculated by using • Power = Force x velocity • As P = W.D/time • P = (F X s)/time • P = F x v

  18. G.P.E and K.E • Energy cannot be created or destroyed, only transformed. • This means that at times it may be easier to use the conservation of energy law to solve a problem rather than force diagrams • Kinetic energy = ½ x m x v2 • G.P.E = m x g x Δh • Q8 - A dog of mass 12 kg falls from an aeroplane at a height of 3.4 km. How much gravitational energy does the dog lose as it falls to the ground

  19. Momentum • Momentum is a property of moving objects just as kinetic energy is. • The faster an object moves the more momentum it has. • More massive objects have more momentum at the same speed than less massive objects Momentum (kg.m.s-1) = Mass (kg) x Velocity (m.s-1) p = m x v

  20. Law of Conservation of Momentum • In any collision or explosion the total momentum before will equal the total momentum after, PROVIDING NO EXTERNAL FORCE ACTS • This means that we can use the simplified formula • Total Pbefore = Total Pafter (m1.u1) + (m2.u2) = (m1.v1) + (m2.v2) Where; u = initial velocity v = final velocity • or if the objects stick together (m1.u1) + (m2.u2) = (m1+m2) x v

  21. An example of using momentum Before 2000kg 1000kg 5 m.s-1 Momentum before = 1000 x 5 + 2000 x 0 = 5000 kg.m.s-1 After Combined mass = 3000 kg V m.s-1 Momentum after = 3000 x v v = 5000/3000 v = 1.67 ms-1

  22. What about direction • Momentum involves the vector quantity of velocity, therefore momentum is also a VECTOR QUANTITY so direction is important • The convention is that anything moving to the right has positive momentum and anything moving to the left has negative momentum

  23. Snoopy (mass 10kg) running at 2.5 m.s-1 jumps onto a skateboard of mass 4 kg travelling in the opposite direction at 7 m.s-1. What is the velocity of Snoopy and skateboard after Snoopy has jumped on? Because they are in opposite directions, we make one velocity negative 10kg 2.5 m.s-1 4kg -7 m.s-1 Momentum before = 10 x 2.5 + (4 x –7) = 25 - 28 = -3 Momentum before = Momentum after -3 = 14v V = -3/14 = -0.21 m.s-1 14kg v m.s-1 Momentum after = 14v

  24. Explosions and recoil • The law of conservation of momentum applies equally to explosions • However the total momentum before the explosion must be 0 as nothing is moving • Therefore we get; 0 = Total Pafter 0 = (m1.v1) + (m2.v2) (m1.v1) = -(m2.v2) Therefore the two (or more fragments) must move off in opposite directions to conserve momentum

  25. Impulse • Newton’s second law is commonly written as • F = m x a • From SUVAT eq a = (v-u)/t • So F = m x(v-u)/t F = (m x v) – (m x u)/t F x t = Δp 2 things come out of this manipulation; 1) Force is equal to the rate of change of momentum 2)Impulse is defined as the change of momentum

  26. Force = rate of change of momentum • This is really what Newton’s 2nd law states and is the principle behind airbags and crash mats and crumple zones of cars • If an object comes to rest with an impact it has to lose momentum • If it loses momentum quickly the force must be large • However if the momentum is lost over a longer time then the force is smaller an therefore the risk of injury is less.

  27. Impulse = F x time = Δp • Impulse can be measured in either • N.s (from force x time) • Or kgms-1 as a change of momentum • We can calculate the impulse of a collision from a Force-time graph

  28. Example Szymon punches Eerik in the face. If Eerik’s head (mass 10 kg) was initially at rest and moves away from Szymon’s fist at 3 m/s, what impulse was delivered to Eerik’s head? If the fist was in contact with the face for 0.2 seconds, what was the force of the punch? m = 10kg, t = 0.2, u = 0, v = 3 Impulse = Ft = mv – mu = 10x3 – 10x0 = 30 Ns Impulse = Ft = 30 Fx0.2 = 30 F = 30/0.2 = 150 N

  29. Area under a force-time graph = impulse Area = impulse

  30. Elastic/inelastic • During a collision momentum is conserved, however K.E is not always conserved as work can be done in changing the shape of a ball for example and lost as heat etc • If all K.E is conserved then this is referred to as an ELASTIC COLLISION eg- snooker balls • If energy is lost during the collision then this is INELASTIC. We are only concerned with a TOTALLY inelastic collision which is where the objects stick and move off together as one object.

  31. Circular motion • When describing motion in a circle or rotation it is better to describe angles rather than linear distances so we have some definions to LEARN • Time Period (T) – Time for 1 complete circle • Angular displacement (θ) – The angle moved through by a line joining the object to the centre of the circle • Angular velocity (ω) – The angle moved through per second - ω(rads s-1) = 2∏(rads) / T • Frequency (f) – Num of complete circles per sec– f(Hz or s-1) = 1/T

  32. Not uniform velocity • It is important to remember that though the speed is constant, the direction is changing all the time, so the velocity is changing. • A change in velocity is an acceleration! • Even at constant speed an object travelling in a circle is constantly accelerating!!!! Uniform speed ≠ uniform velocity

  33. Direction of centripetal acceleration/force Change in velocity VB VA VB VA VA + change in velocity =VB The change in velocity (and thus the acceleration) is directed towards the centre of the circle.

  34. Uniform circular motion The centripetal acceleration/force is always directed towards the centre of the circle Centripetal force/acceleration velocity

  35. How big is the centripetal acceleration? a = v2 = 4π2r r T2 where a is the centripetal acceleration (m.s-2), r is the radius of the circle (m), and v is the constant speed (m.s-1).

  36. How big is the centripetal force? F = mv2 r Or F = mω2r from F = ma (Newton’s 2nd law) Centripetal Force - The Real Force

  37. Work done? • None! Because the force is always perpendicular to the motion, no work is done by the centripetal force.

  38. Examples of centripetal force • Centripetal force describes the force towards the centre of the circle but this must be provided by a force eg- • Gravity in orbits • Tension in a string • Electrostatic attraction in an orbiting electron • Friction between tyres and the road • Normal reaction force from the wall in “wall of death”

  39. Answers to practice Questions • Q1. 7.9 ms-1 and 2.6 ms-1 • Q2. 5 ms-2 • Q3. 2.5 s and 25 m • Q4. 31.6 ms-1 and 3.16 s • Q5 • Initially the air resistance is zero so the sky diver accelerates due to his weight, Newton’s 2nd law. • As his velocity increases the air resistance increases, so the resultant force decreases providing a decreasing rate of acceleration, Newton's 2nd law. • Eventually at terminal velocity his weight balances the air resistance and Newton's 1st law • When the parachute is opened the force of air resistance is larger than his weight so the resultant force is in the opposite direction to his motion so he decelerates. • He decelerates until the forces are balanced at a new lower terminal velocity. • Q6. 500 x sin (25) = friction and 500 x cos(25) = Normal reaction force • B) Fr = 211N and Nr = 453N • Q7. W.D = 10 x 50 x cos(30) = 433N • Q8 . GPE lost by dog = mgh = 12 x 10 x 3400 = 408 000 J

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