Year 11 GCSE Physics Unit 1

Newton's Laws By the end of this section you should be able to: 1.1.6 recall and understand that forces arise between objects, that the

forces on these objects are equal and opposite and that friction is a force that always opposes motion;

1.1.7 calculate the resultant of two one-dimensional forces; 1.1.8 investigate experimentally Newton's f i r s t and second Laws, for

example using an air track and data logger, or a computer simulation, to study the e f fec t of balanced and unbalanced forces on an object, and through mathematical modelling derive the relationship between resultant force, mass and acceleration

1.1.9 recall and use the equation Resultant Force = mass x acceleration

Forces

When a force acts on a body it will cause one or more of the following to occur

Balanced Forces

Newton's 1 s t Law states that: ^ U - ^ . j f . „ \ i /*f>nn>*>

This means that if the forces acting on a body are balanced, the body will either

be sitting still (if it was still to start with), or moving with uniform velocity (if it

was moving to start with).

E.g. I f a car has a f rictional force acting

backwards and a driving force acting

forwards and the two forces are equal in

magnitude, it will be travelling with

uniform velocity. I n the diagram you can

also see that the vertical forces are equal.

81

Thrust Drag

Road

Weight

Year 11

Unbalanced Forces

GCSE Physics Unit 1

Resultant force - this means the result of two or more forces acting on a body

When a resultant force acts on a body it will either be accelerating or decelerating.

3N 3N

Resultant force = \ M -s> Resultant force =

Force. Mass and Acceleration

Newton's 2 n d Law states that: - H A J L ^krx^r r-ryc<z. r\ l -eH j

Newton's 2 law can be written as an equation.

m a Complete the memory triangle Where ZF = resultant force (N)

m = mass (kg)

a = acceleration (m/s 2)

IN can be defined as the force which gives a mass of 1kg an acceleration of

lm/s 2 .

Example:

A car accelerates at a rate of 5m/s 2 . I f it weighs 500kg how much driving force

is the engine applying? , ^ a

Physics for CCEA Questions 1 - 10, Pages 16 • 17

8 2

Year 11 GCSE Physics Unit 1

page 138 Force, mass, acceleration Extension

.Sheet

I f there is an unbalanced resultant force on an object, it accelerates.

Force = mass x acceleration (in N) (in kg) (in m/s2)

Acceleration = change in velocity

time taken

Example A car of mass 800 kg accelerates from rest to 10 m/s in 20 s. a) What is the acceleration? b) What is the force exerted?

Answer Acceleration

change in velocity 10 — 0 = 0.5 m/s2

time taken 20

Force = mass x acceleration = 800 x 0.5 — 400 N

Questions For each question show all your working clearly.

1. What force is needed to give a mass of 10 kg an acceleration of 2 m/s2. 2 0 A J

2. What acceleration is given to a mass o f 0.5 kg by a resultant force of 4 N? "cVvis"1

3. What is the mass of an object that accelerates at 3 m/s2 when a force o f 15 N is applied? <,

4. A car of mass 800 kg accelerates from 5 m/s to 25 m/s in 10 s. a) Calculate the acceleration. 2 r v r g ' b) Calculate the resultant force exerted.) (oC04^_

mass 800 kg

V acceleration

1»- Force

5. A car-driver of mass 60 kg is in a crash. He decelerates, from 20 m/s to rest, in 2 seconds. Calculate: a) His deceleration. ICv i ^ - * -b) The force exerted on him (by his seat and

seat-belt). b o o o

f--Fr - rWO.

6. A supermarket trolley has a mass o f 20 kg. When pushed by a force of 15 N it accelerates 12-- 1o<o.% at 0.5 m/s2. l 2 . * <0^ a) Calculate the resultant force on the trolley

that gives it this acceleration. b) What is the friction force on the trolley?

A car has a mass of 1000 kg and is travelling at 20 m/s. The brakes then exert a steady force of 5000 N. a) What is the deceleration? S ' vS " ' 1 -b) How long does it take to stop the car? a s

15 N

Friction

83>

Year 11

Proving F = mo Experimentally

GCSE Physics

^Trolley

Unit 1

F r i c t i o n Compensat ion Hanging

Masses " - - ^ S

Method

1. Allow the trolley to accelerate along the friction compensated track from

rest using one mass.

2. Measure the time taken for the trolley to reach the end of the track.

3. Use the equation s = ^(u + v) t to calculate the final velocity of the trolley,

where s = length of track, u = initial velocity, v = final velocity and t = time

taken to travel the track.

4. Use the equation v - u + at to calculate the acceleration of the trolley, where

u = initial velocity, v = final velocity, t = time taken and a = acceleration.

5. Repeat for different hanging masses, move masses from the trolley to the

hanger so that total mass of the system remains constant.

Results U=0 dkl/Sivi V = 2J3 OS u=o — — — l _ _

3

Mass (kg) Accelerating

Force (N) Time to reach

end of track (s) F nal Velocity

(m/s) Acceleration

(m/s 2)

0.1 1 \.&

0.2 2. i.io 2.2JU \p& 0.3 3 0.4 U O . p 2>PR 0.5 s o A s

0.6 la S 3 o

6. Draw a graph of force against acceleration. —

7. What does the gradient of this graph give us? p-s- (WO

f c r P^rv^O sV/OLCjUV

Year 11

Assessed Homework

GCSE Physics Unit 1

1. The diagrams below show several situations where two forces act on an

object. In each case work out the magnitude and direction of the resultant

force.

30N 20N

[3] Pull

200N

.300N -K\/Push

BeamX

Weight ofX

2. A magnet of weight 2.4N was hung from a Newton meter. An identical

magnet was hung from the f i rst magnet. The second magnet was pulled down

until it broke free from the f i rst magnet. The spring balance reading just

before the break was 9.6N. [1]

a) What was the reading on the meter when it supported the 1 s t magnet only? b) What was the reading when the 2 n d magnet was hung on? [1]

c) What was the force of attraction between the two magnets just before

they broke apart? [2]

3. An empty car has a weight of 5500N. I t s driver has a weight of 600N. The

weight on each wheel should not be greater than 2500N. In addition to the

driver, how much extra weight can the car support safely? [3]

4. A force of 800N is applied by a boy lifting a 20kg mass. How fast will he be

able to accelerate the mass if he lifts it straight up? [6]

5. A force of 3000N acts on a car to make it accelerate at 1.5m/s2. What is

the weight of the car? [4]

Total [20]

85

i o-\ t ^ Q ^ = l o r o t-<e*/ J (T.

C-Jj Ccr too. 1

Jn\ 0 U u H\ to) c

J c 6SCX3 ^fcj caCsO J c 6SCX3 ^fcj C D

= i n Q Q C A J O

I t • •

C D

: li \ LL \e w-*^ (T> P £ = rvACX (?) : li - j - — W T 2 O X : ( 0 G > B O O ' Z

wax ( 3 ) «P

I J L J = 9ncvO (7s CX-- £co» l o c

7 ^

- .

i

Year 11

Action-Reaction

GCSE Physics

Newton's 3 r d Law states that: ft/ eUeAj g&\cn p y c a HU&/e >S

;>r-> <scfuo-\ & epposi-o. r e a c h o s pcrca.

Unit 1

Example:

'Man on Earth

When a sprinter starts to run he exerts a force on the ground, moving the Earth

backwards. The Earth pushes back on the man with an equal force, causing him

to accelerate forward.

Since the Earth is so massive we don't notice the small acceleration gained by

the Earth that we give it. Since we are so small in comparison we gain a great

acceleration.

Example: iwosv-A /

The diagram shows a firework rocket.

As it flies through the air, there are

three forces acting on it.

a) Which 3 arrows show the 3 forces? ^ ^ , °

b) Copy the diagram with these 3 forces.

c) Label the 3 forces with these words: weight, thrust, air resistance (drag)

d) What can you Say about these forces when the rocket is just taking off? 2> ^ £ -lo r»cc_

e) Why does the rocket come back down?

• 3

8b

ecvtL.

Year 11

Friction

GCSE Physics

Friction is a (2rrce AUnY g y v ^ f - a r v n h ^

Unit 1

I t is measured in Newtons (N).

In general the rougher a surface is the more friction it will have.

Friction between two moving surfaces can cause a lot of energy to be wasted as

t I W W I .

How can we reduce friction between two surfaces?

W\\\ l ^ r i / i r l

Friction - Friend or Foe?

Friend - useful for staying upright and for brakes in a car

Foe - nuisance if air resistance to a racing driver

Physics for You Questions 35 - 38, Page 153 (omit 35 d + e and 36b)

87

Year 11 GCSE Physics Unit 1

page 99 Air resistance

o I f the forces on an object are balanced, there is no resultant force. The object stays sti l l , or carries on at a constant velocity.

• I f there is a resultant force, then the object w i l l accelerate or decelerate.

• A i r resistance (or friction) always opposes movement.

A sky-diver jumps out o f a helicopter. He uses air resistance to land safely.

1. F i l l in the missing words:

2. The sky-diver weighs 700 N. Label each o f the forces in all o f the diagrams wi th one o f these labels: 700 N less than 700 N more than 700 N

\0\fCyZ) 3. In your own words, rvLcy- resisVo^cx explain carefully

\ess s f*^ ^ a t h a P P e n s when 1l0Oecxs<2. his parachute opens.

4. Explain why a falling V O D . ^ raindrop does not go

*., fcf^V^a- faster and faster.

5. What two factors affect s - ^ c c the amount of air

c» ^ c x * speed or resistance experienced dro^ by the sky diver?

Draw a velocity t ime graph to represent the sky divers decent to Earth f rom the moment he exits the helicopter.

T e c >

"3<=CXJ

A t the start there is only one force on the sky-diver. This is his weu^kr "

This unbalanced or re.s^!Yr\r\Y force makes him a r c f ^ P n h c * •

As he travels faster, the friction (called 2L\/ r gSisVo^co. or drag) increases. Eventually the 2 forces are equal and bC\lQK-icaol. wifh no re-Sulrrmr- force, so he stops arg<s»l o jr .h . -y, and travels at a constant spggd . -' This speed is called his t e rmina l velocity.

When his parachute opens, the air resistance i r c / g p s g s . There is now a rp?a_)lr-p^ V force upwards. This makes him slow down, until .

the 2 forces are e g j . " again. Because the 2 forces are b^ ' .o^cgd (with no r e s J r o ^ r force) he now travels at a constant spee-e* . This is bis new t e v r v m n o i velocity.

When he hits the ground, it pushes up to make him d ^ I . p j r > b quickly. When he stands on the ground, the ground p_>.~ \c g up on his feet. The upward f OTCA. is equal to his w (so there is no r ).

88

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l e a p s pt>v> heUcqpfcaJ «*- \Cx/o^ a c c e J i e J o K l o ^ Q>r r^Sisrcince m o e c s e s a s s p e e d ir\c/aases CKCosU2ycxY\Cr\ r e d u c e s -jo o_y\sronh s p e e d

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i ppcrenV-Ga\i\eo dropeid 2 lead b^Us op (3 op HUe \eom^cj 4t>-oe/ cp p i s ^ , TT^cxj Sav^e +w\« , fyco./ig only g / ^ o L t y

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moss -to L > J = X C ^ V r xtvo s e w i e p e r

S O I s P ^ ^ r ™ * 3 3 r C ^ °

Year 11 GCSE Physics Unit 1

Circular Motion By the end of this section you should be able to 1.1.15 describe some examples of circular motion, explaining how the force

acting on an object causes this type of motion; 1.1.16 investigate qualitatively the factors affecting the centripetal force

for an object moving in a circle; 1.1.17 recall that:

- the direction of the centripetal force is towards the centre of the circle; - it increases with the mass and the speed of the object and decreases as the radius of the circle increases; and - if the force is removed, the object will move away at a tangent to the circle;

Many objects travel in a circle. List some examples,

1 1 y ^ m n z

rex ircinlnm

For an object to move in a circle a force is required. I n what direction does this

force act? h o o o ^ d s c p n k e r^. CUAJL

Draw an arrow to show the force acting on the object below, as it moves in a

circle, (label this F)

Draw an arrow to show the direction of the velocity of the object at the instant

shown, (label this v)

What direction object would the object move in if the force were suddenly

removed? i /

89

Year 11 GCSE Physic- Unit 1

I f the object moves at a steady speed of lOm/s around the circle. Compare the

velocity of the object at A and B.

I s the object accelerating?

- _ 9 ^ " ^ I c c i h .

3-

Factors affecting centripetal force

- As the mass of the object increases the centripetal force \nCJease~^.

- As the radius of the circle increases the centripetal force d e c / g p g ^ .

- As the speed of the object increases the centripetal force mc/gcne

The table shows examples of objects in circular motion, complete the force

column.

Example of object in circular Force provided by. .

motion

Earth orbiting the Sun cyc\\> VoYxcyy^ puS\ cf

Car going round a roundabout

-tujfes 4-rocxcJ.

Electron orbiting a nucleus

a . p re fe rs

Hammer thrower

Object swung on a strinc 1 S(Oi let SrVmCj

90