P4 P5 P6 Revision

1. P4 P5 P6 Revision P4 Explaining Motion P5 Electric Circuits P6 Radioactive materials

2. P4 Explaining Motion

3. speed (m/s) = distance travelled (m) / time taken (s) Usually when an object travels from ‘A’ to ‘B’ it’s velocity will vary so a calculation of it’s velocity is really an average velocity. An instantaneous velocity is the velocity at a given moment. Distances measured in one direction are positive, and in the other, negative. A negative velocity means moving in the opposite direction.

4. 10 9 8 7 6 5 4 3 2 1 0 D i s t a n c e (m) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Time (s) 1. What is the velocity of the object at first ? 9  3 = 3 m/s 2. For how long was the object stationary ? 6 s 3. What is the velocity in the last part ? 9  6 = - 1.5 m/s

5. 1. A ball is thrown and takes 4 seconds for its velocity to steadily increase to 4 m/s and then travels at a constant velocity for 5 seconds. It then hits a wall and rebounds at a constant velocity of 3 m/s for 5 s before it is caught. 5 4 3 2 1 0 -1 -2 -3 -4 -5 Velocity (m/s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Time (s)

6. 2. An object moves at a velocity of 2 m/s for 3 seconds and then accelerates at 1 m/s2 for 2 seconds. It then moves at a constant velocity for 3 seconds and then decelerates at 1 m/s2 until it is stationary. It remains stationary for 2 seconds and then accelerates backwards at 2 m/s2 for 1 second. It then takes 2 seconds to steadily decelerate till it stops. 5 4 3 2 1 0 -1 -2 -3 -4 -5 V e l o c i t y (m/s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Time (s)

7. A woman walks out onto the road A car is travelling at 30 km/hr X Will she survive ? 8 m The driver has a reaction time of 1 second 30 km = 30,000 m 1 hr = 3,600 s In 1 second the car would travel 30,000  3,600 = 8.33 m The woman is hit BEFORE the driver applies the brakes !!!!

8. steeper gradient - faster distance time For a distance-time graph a steeper gradient means a higher speed A tachometer continuously measures an objects speed and can be used to make a tachograph.

9. If the speed of an object is increasing, we say that it is accelerating. Acceleration (m/s2) = change of velocity, m/s time taken for the change (s)

10. box A force arises from an interaction between two objects. When one object exerts a force on another, it always experiences a force in return (a reaction force). A force and a reaction force are called an ‘interaction pair’. The two forces in an interaction pair are equal in size and opposite in direction and they act on different objects. the box acts downwards on the table due to gravity the table acts upwards on the box due to the reaction force the two forces are equal and opposite

11. Walking gravity reaction force of friction on acting on the feet force of the feet acting on the ground reaction force of the ground acting on the feet

12. The horizontal motion of objects (like cars and bicycles) can be analysed in terms of a driving force (produced by the engine or the cyclist), and a counter force (due to friction and air resistance). driving force greater than counter force – speeding up driving force equal to counter force – stationary driving force less than counter force – slowing down A resultant force takes into account all the acting forces. 30N 20N 50N resultant force = 20N

13. Friction is the interaction between two surfaces when they slide over each other There is a friction force on both objects involved Friction is caused by the roughness of the sliding surfaces Friction enables cars and people to get moving

14. momentum (kg m/s) = mass (kg) × velocity (m/s) A car has a mass of 5,000 kg and a velocity of 4 m/s. What is the car’s momentum ? 5,000 x 4 = 20,000 kg m/s A cyclist cycling at 10 m/s has a momentum of 540 kg m/s. The cyclist has a mass of 50 kg, what’s the mass of the bike ? total mass = 540 / 10 = 54 kg mass of bike = 54 – 50 = 4 kg If a resultant force acts on an object, it causes a change of momentum in the direction of the force If a resultant force on an object is zero then there is no change of momentum eg when the driving force = friction if it is stationary, it stays at rest if it is already moving, it continues at a steady speed in a straight line

15. Total momentum before = total momentum afterwards positive momentum = to the right negative momentum = to the left momentum before = m1v1 + m2v2 m1 = m2 v1 = -v2 = m1v1 + -m2v2 = 0 m1 m2 momentum after = m1v1 + m2v2 = -m1v1 + m2v2 = 0

16. m1 = m2 momentum before = m1v1 + m2v2 v2 = 0 = m1v1 + 0 = m1v1 m1 m2 momentum after = m1v1 + m2v2 = 0+ m2v2 = m2v2 therefore m1v1 =m2v2 ie all the momentum of the first ball is transferred to the second ball

17. m2 > m1 momentum before = m1v1 + m2v2 v2 = 0 = m1v1 + 0 = m1v1 m2 m1 momentum after = m1v1 + m2v2 = -m1v1 + m2v2 m1 rebounds of m2 and transfers some of it’s momentum to m2

18. momentum before = m1v1 + m2v2 m1 > m2 v2 = 0 = 0+ 0 v1 = 0 m2 pushes off m1 m1 m2 momentum after = m1v1 + m2v2 = -m1v1 + m2v2 = 0 therefore m1v1 = m2v2

19. When a force is applied to an object, its velocity increases The longer the force is applied, the greater the change in velocity The greater the force applied, the greater the change in velocity momentum = mass x velocity increasing the velocity increases momentum When a force is applied to an object, its momentum increases The longer the force is applied, the greater the change in momentum The greater the force applied, the greater the change in momentum change of momentum = resultant force x time during which it acts

20. change of momentum = resultant force x time for which it acts Increasing the time it takes for a change in momentum reduces the force that causes the change in momentum If the time from impact to the velocity becoming zero is increased then the impact force is reduced which means less injury Seat belts, air bags, crumple zones, cycle helmets etc increase the time during impact and therefore reduce the impact force. crumple zone

21. The energy of a moving object is called kinetic energy When a force moves an object, work is done work done (J) = force (N) × distance moved (m) A braking force of 1000N is applied by a driver to stop his car. The car covered 50m before it stopped. How much work did the brakes do ? 1,000 x 50 = 50,000 J

22. When an object is lifted to a higher position above the ground, work is done by the lifting force against the gravitational force acting on the object (its weight) As an object falls, its gravitational potential energy decreases as it is transferred into kinetic energy and heat (friction with the air)

23. When an object is lifted this increases the object’s gravitational potential energy (GPE) change in GPE (J) = weight (N) × height difference (m) A crane is lifting a 50kg load up into the air with a constant speed. If the load is raised by 20m how much work has the crane done ? remember that 1 kg has a weight of 10 N (on Earth) 50 kg = 500 N work done = 500 x 20 = 10,000 J

24. kinetic energy (J) = ½ × mass (kg) × [velocity]2 (m/s2) E = ½ m v2 A 70 kg boy runs at 10m/s. What is his kinetic energy ? kinetic energy = ½ x 70 x 102 = ½ x 70 x 100 = 3,500 J What is the kinetic energy of a 100g tennis ball being thrown at a speed of 5m/s ? 100g = 0.1 kg kinetic energy = ½ x 0.1 x 52 = ½ x 0.1 x 25 = 1.25 J

25. A parachutist with a total mass of 70 kg jumps from a helicopter at a height of 1,500 m. He pulls the cord of the parachute when he is 1,000 m above the ground. (a) Ignoring air resistance, what is the speed of the parachutist just as he pulls the cord ? You will need to use the formula E = ½ m v2. You are given the mass (70kg) in the question and you can work out E (energy) by using GPE = weight x height. Remember that 70kg = 700N. GPE = weight x difference in height GPE = 700 x (1,500 – 1,000) = 700 x 500 = 350,000 J E = ½ m v2 700,000 / 70 = v2 {divided both sides by 70} 350,000 = ½ m v2 v2 = 10,000 {replacing E with 350,000} 700,000 = m v2 v = √10,000 {multiplied both sides by 2} {getting the square root} 700,000 = 70 x v2 v = 100 m/s {replacing m with 70}

26. (b) Why doesn’t the parachutist actually reach the speed calculated in part (a) ? [2 marks] because of air resistance [1], some of the gravitational potential energy is dissipated as heat [1] (c) The parachutist actually reached the velocity of 40 m/s before the using the parachute. How much energy was dissipated ? total energy = 350,000 J velocity (without taking air resistance into account) = 100 m/s velocity (taking air resistance into account) = 40 m/s = 40% therefore 60% of the energy is dissipated 350,000 x 60 / 100 = 21,000 J (d) What principle is used to calculate part (c) ? the conservation of energy (all of the energy is accounted for)

27. {B to C = steady speed and C to D = fastest speed}

28. 1.65 hrs 144 – 112 = 32 gradient / slope 24.6 m/s 4.0 – 2.7 = 1.3 speed = 32 / 1.3 = 24.6 m/s Lance

29. outlier / anomoly /anomalous  (38 + 41 + 40 + 37) / 4 = 39

30. 

31. kinetic energy = ½ m v2  1875

32. no change / nothing / stays the same

33. P4 Explaining Motion P5 Electric Circuits P6 The Wave Model of Radiation

34. P5 Electric Circuits

35. Electric charge – objects become charged when electrons are transferred to or from them, for example, by rubbing Two types of charge are positive and negative (these names are just labels) Two objects with the same charge repel each other Two objects with different charges attract each other

36. voltage current = resistance Metal wire Normally the free electrons in a metal move around slowly at random. metal ions + electrons potential difference The potential difference (voltage) provides energy which makes the electrons move through the metal ie it generates a current. The electrons experience resistance when they flow through the metal. The symbol for voltage is V The symbol for current is I The symbol for resistance is R

37. r1 r2 R = + SERIES V I i2 v1 v2 i1 I = i1 = i2 The current is the same everywhere The sum of the voltages across each component equals the supply voltage V = v1 + v2 Resistance

38. Current PARALLEL I I3 i4 i1 i2 i5 I = I3 I1 = I4 I2 = I5 I = I1 + I2 Total current = the sum of the currents through each component Current does not get used up

39. PARALLEL V v1 v2 Voltage = energy per unit of charge V = v1 = v2 The voltage across each component is the same as the supply voltage.

40. If more bulbs are added in parallel to a circuit then they will all be as bright as normal and more current is drawn from the power supply The potential difference is largest across the component with the greatest resistance, because more energy is transferred by the charge flowing through a large resistance than through a small one The current is smallest through the component with the largest resistance, because the same battery voltage causes more current through a smaller resistance than a bigger one

41. SERIES 12 V 2A 4 V 5 V v3 1 ohm 3 ohm r3 i3 i3 = 2 A v3 = 3 V r3 = 2 ohm

42. PARALLEL 12 V i3 = 2 A 4 A 12 ohm v2 = 12 V 1 A v2 12 ohm 1 A i3 r3

43. Current is a flow of electrons Electrons have charge (negative) So current is a flow of charge How do we quantify current ? Current is the amount of charge flowing in a particular amount of time

44. Voltage provides energy to the electrons Electrons have charge (negative) So Voltage provides energy to the charge How do we quantify voltage ? Voltage is the amount of energy a particular amount of charge has

45. What about resistance ? All components will offer resistance to a flow of electrons How do we quantify resistance ? If a current of 1A flows through a component when the voltage across it is 1V then the component is said to have a resistance of 1 ohm [ 1 W ]

46. V I = R V I R = I I V V = R R = I I V I = R Multiply both sides by R R x x R R I = V Or V = I R Take V = I R and divide both sides by I or

47. V I = R V = I R V = I R V V R = I I R current = voltage / resistance voltage = current x resistance resistance = voltage / current Q. A current of 4 A flows through a circuit with resistance 3 W. What is the voltage ? use V = 4 x 3 Voltage = 12 V

48. 6 10 R = I = 2 5 V V V I R = = I R I R Q. A current of 5 A flows through a circuit with voltage 10 V. What is the resistance ? use resistance = 2 W Q. A circuit with voltage of 6 V has a resistance of 2 W . What current should flow ? use current = 3 A

49. V I = R 8 12 V = I R V = I R I = R = 4 2 V V V V R I R = = = R I I I R Q. A current of 4 A flows through a circuit with voltage 12 V. What is the resistance ? use resistance = 3 W Q. A circuit with voltage of 8 V has a resistance of 2 W . What current should flow ? use current = 4 A Q. A current of 60 A flows through a circuit with resistance 4 W. What is the voltage ? use V = 60 x 4 Voltage = 240 V

50. V I = R 16 V = I R V = I R R = 2 230 I = 5 V V V V I R R = = = R I I I R Q. A current of 2 A flows through a circuit with voltage 16 V. What is the resistance ? use resistance = 8 W Q. A circuit with voltage of 230 V has a resistance of 5 W . What current should flow ? use current = 46 A Q. A current of 25 A flows through a circuit with resistance 3 W. What is the voltage ? use V = 25 x 3 Voltage = 75 V