Download presentation

Presentation is loading. Please wait.

Published byBlaine Hoyland Modified over 2 years ago

1
P2.1.3 Terminal Velocity P2 Physics Ks4 Additional Science Mr D Powell

2
Connection Connect your learning to the content of the lesson Share the process by which the learning will actually take place Explore the outcomes of the learning, emphasising why this will be beneficial for the learner Demonstration Use formative feedback – Assessment for Learning Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex Offer different ways for the students to demonstrate their understanding Allow the students to “show off” their learning Consolidation Structure active reflection on the lesson content and the process of learning Seek transfer between “subjects” Review the learning from this lesson and preview the learning for the next Promote ways in which the students will remember A “news broadcast” approach to learning Activation Construct problem-solving challenges for the students Use a multi-sensory approach – VAK Promote a language of learning to enable the students to talk about their progress or obstacles to it Learning as an active process, so the students aren’t passive receptors

4
P2.1.4 Terminal Velocity a) The faster an object moves through a fluid the greater the frictional force that acts on it. b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). c) Draw and interpret velocity-time graphs for objects that reach terminal velocity, including a consideration of the forces acting on the object. d) Calculate the weight of an object using the force exerted on it by a gravitational force: W = mg (F = ma)

5
**a) The faster an object moves through a fluid the greater the frictional force that acts on it.**

6
**Copy this law in your own words...**

Newtons Law Number 1 Copy this law in your own words... If the resultant force is zero… The object will stay still OR, if already moving, move at a constant speed in a straight line. Force from muscles Air resistance There is a zero resultant force so the athlete runs at a constant speed. A

7
**Copy this law in your own words...**

Newtons Law Number 2 Copy this law in your own words... If there is a resultant force on an object it will accelerate (speed up, or slow down). mass The resultant forward force causes the puck to accelerate. Force = mass x acceleration A

8
**Copy this law in your own words...**

Newtons Law Number 3 Copy this law in your own words... For every ACTION there is an equal and opposite REACTION (think of an action as a sort of impact or hitting force) The tennis racket gives the ball a forward force (action) … but the ball also gives the racket a backward force (reaction). A

9
**Investigating TV with Oil.**

Pour 100ml of oil into the cylinder and get a spare cylinder to pour into. Get 2 stopwatches,3 dice and some paper to prevent spillage. Get two people down to eye level to time the fall between & 50&30ml. Each reading should be a paired reading. Record the times in your table. Repeat 5 times in total. Enter results in table

10
Example Results A 1ooMl Cylinder - Trying to Establish the Terminal Velocity of Oil 1st Timing /s 1st Distance /m Velocity m/s 2nd Timing /s 2nd Distance /m Velocity Difference m/s 1 0.53 0.034 0.06 2 0.56 0.69 0.05 0.01 3 0.55 0.57 4 0.52 0.07 -0.01 5 0.54 Dev 0.000

11
b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). A Why is this the case?

12
**? D c) Terminal Velocity Terminal Velocity Time (s) Velocity m/s 1 10**

1 10 2 20 3 30 4 36 5 40 6 44 7 48 8 50 9 52 54 11 55 12 13 ?

13
Progress Check…

14
Terminal Velocity Time (s) Velocity m/s 1 10 2 20 3 30 4 36 5 40 6 44 7 48 8 50 9 52 54 11 55 12 13

15
Summary Questions D

16
Multichoice... D

17
Match Up Matchup... D

18
d) What is the link? A

19
d) Investigation... A We can use the apparatus above to accelerate a trolley with a constant force. Use the newtonmeter to pull the trolley along with a constant force. You can double or treble the total moving mass by using double-deck and triple-deck trolleys. A motion sensor and a computer record the velocity of the trolley as it accelerates. Here is a typical set of results which show that for the same force the acceleration reduces as mass increases. Can you work out a simple formulae which relates force, mass & acceleration?

20
d) Weight or Gravity? A 7.36 × 1022 kilograms ¼ radius of Earth A lot of people slip into lazy English and try and use the ideas of Weight and Gravity interchangeably. However, they are not the same thing. Gravity is a special force which acts upon mass and is measured in Newtons per kilogram. Weight is measured in Newtons and is a unit of forces Gravity is on Earth is derived from the total mass of Earth. For this mass the Force of Gravity is 10N/kg. On the moon it is only 1.6N/kg as the Earth has 81.2 more mass and 4 times the radius. × 1024 kilograms

21
**Multiple of Earth gravity**

d) Comparison A Body Multiple of Earth gravity m/s² Sun 27.90 274.1 Mercury 0.3770 3.703 Venus 0.9032 8.872 Earth 1 (by definition) 10 Moon 0.1655 1.625 Mars 0.3895 3.728 Jupiter 2.640 25.93 Saturn 1.139 11.19 Uranus 0.917 9.01 Neptune 1.148 11.28 Pluto 0.0621 0.610

22
d) Gravity.... D So the force of gravity pulls down on masses accord to gravitational field strength. This varies with height but near to the Earth is a constant 10N/kg.So 1kg would weigh; Weight (N) = mass (kg) x Gravitational Field (g) (N/kg) (W=mg) W = mg W = 1kg x 10N/kg W = 10N The weight feels a force of 10N 800g would weigh; W = 0.800kg x 10N/kg W = 8N Using the above formula (on the earth) find the following; 1) W? if m = 0.5kg ) W? if m = 300g ) m? if W = 34N 4) m? if W = 280N ) m? if W = 0.1N 5N 3N 3.4kg 28kg 0.01kg

23
**Force = mass x acceleration**

d) Acceleration again.... D We can actually consider the gravity acting on objects as a form of acceleration. However, this time the units are different and we can quote the acceleration as the Force per kilogram or N/kg instead of m/s2. In fact both are the same; 10N/kg = 10m/s2 = 10ms-2, In which case we actually find that one Newton of force can be defined as the force required to give a mass of 1kg, an acceleration, of 1 m/s2 or 1ms-2 Force = mass x acceleration Work out the following; m= 50kg, a = 10N/kg, F = m= 100kg, a = 5N/kg, F = F= 50N, a = 10N/kg, m = F= 30N, m = 5kg, a = F= 28N, a = 2.5ms-2, m = 500N 5kg 6N/kg 11.2kg

24
**d) Applications to F=ma**

Complete the table below showing the resultant force, mass and acceleration of objects in different situations. Resultant force (in newtons) Mass (in kilograms) Acceleration (in m/s2 or N/kg) a) Athlete accelerating at start of 100 m race 70 8.0 b) Car accelerating 3000 1200 c) Lorry braking 16 000 0.8 d) Plane taking off 8000 5.0 Resultant force (in newtons) N Mass (in kilograms) kg Acceleration (in m/s2 or N/kg) a) Athlete accelerating at start of 100 m race 560 70 8.0 b) Car accelerating 3000 1200 2.5 c) Lorry braking 16 000 20000 0.8 d) Plane taking off 40,000 8000 5.0

25
**D d) Exam Questions... (0- 24)ms-1 /12s = -2.0 ms-2**

A vehicle of mass 1500 kg braked to a standstill from a velocity of 24 m/s in 12 s. 1) Show that the deceleration of the vehicle was 2.0 m/s2. 2) Calculate the resultant force on the vehicle. (0- 24)ms-1 /12s = -2.0 ms-2 F = ma = 1500kg x -2.0 ms-2 = 1500kg x -2.0 Nkg-1 = -3000N

26
**D d) Exam Questions... 12ms-1 /60s = 0.2 ms-2**

1) A cyclist accelerated along a flat road from a standstill to a velocity of 12 m/s in 60 seconds. The mass of the cyclist and the bicycle was 80 kg. Show that the acceleration of the cyclist was 0.2 m/s2. Calculate the resultant force on the cyclist and the bicycle. On reaching a velocity of 12 m/s, the cyclist in 1) stopped pedalling and slowed down to a velocity of 8 m/s in 10 s, when she started pedalling again. Calculate: 2) The deceleration of the cyclist when she slowed down. 3) The size and direction of the resultant force on the cyclist when she slowed down. 12ms-1 /60s = 0.2 ms-2 (8ms-1 -12ms-1) / 10s = -0.4 ms-2 F = ma = 80kg x -0.4 ms-2 = -32N

27
Summary Questions 1 D RF V M RF V RF

28
**D d) Summary Questions 2 kgm/s2 = kgms-2 = N**

a = F/m = 20N / 8 kg = 2.5 N/kg a = F/m = 0.2N / 500g = 0.2N / 0.5kg = 2 N/kg a = F/m = 5kN / 20 kg = 5000N / 20kg = 250 N/kg F = ma = 5kg x 5 m/s2 = 25 kgm/s2 or 25 kgms-2 or 25N F = ma = 15kg x 3 m/s2 = 45 kgm/s2 or 45 kgms-2 or 45N

29
**W = mg or F = ma W = mg or F = ma 70 8.0 3000 1200 16 000 0.8 8000 5.0**

Resultant force (in newtons) N Mass (in kilograms) kg Acceleration (in m/s2 or N/kg) a) Athlete accelerating at start of m race 70 8.0 b) Car accelerating 3000 1200 c) Lorry braking 16 000 0.8 d) Plane taking off 8000 5.0 Resultant force (in newtons) N Mass (in kilograms) kg Acceleration (in m/s2 or N/kg) a) Athlete accelerating at start of m race 70 8.0 b) Car accelerating 3000 1200 c) Lorry braking 16 000 0.8 d) Plane taking off 8000 5.0

30
**P2.1.4 Terminal Velocity P2.1.4 Terminal Velocity**

a) The faster an object moves through a fluid the greater the frictional force that acts on it. b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). c) Draw and interpret velocity-time graphs for objects that reach terminal velocity, including a consideration of the forces acting on the object. d) Calculate the weight of an object using the force exerted on it by a gravitational force: W = mg (F = ma) P2.1.4 Terminal Velocity a) The faster an object moves through a fluid the greater the frictional force that acts on it. b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). c) Draw and interpret velocity-time graphs for objects that reach terminal velocity, including a consideration of the forces acting on the object. d) Calculate the weight of an object using the force exerted on it by a gravitational force: W = mg (F = ma) P2.1.4 Terminal Velocity a) The faster an object moves through a fluid the greater the frictional force that acts on it. b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). c) Draw and interpret velocity-time graphs for objects that reach terminal velocity, including a consideration of the forces acting on the object. d) Calculate the weight of an object using the force exerted on it by a gravitational force: W = mg (F = ma) P2.1.4 Terminal Velocity a) The faster an object moves through a fluid the greater the frictional force that acts on it. b) An object falling through a fluid will initially accelerate due to the force of gravity. Eventually the resultant force will be zero and the object will move at its terminal velocity (steady speed). c) Draw and interpret velocity-time graphs for objects that reach terminal velocity, including a consideration of the forces acting on the object. d) Calculate the weight of an object using the force exerted on it by a gravitational force: W = mg (F = ma)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google