Motion and force

Physical Sciences

3C Motion and force (Criterion 5):

Measurement and units Kinematics Forces and Newton’s Laws Momentum

ROSNY COLLEGE SCIENCE DEPARTMENT 2014

Physical Sciences 3C

2 2014 Rosny College

Contents MOTION AND FORCE Page PART A: INTRODUCTION 3 Vectors 3 Scalars and vectors Addition and subtraction of Vectors Components of Vectors PART B: KINEMATICS 7 Introduction Graphs of Motion

Displacement~time graphs 10 Velocity~time graphs 12 Acceleration~time graphs 15 Linear Motion – Equations of Motion 17 Vertical Motion 20 Motion on an Inclined Plane 22 Projectile Motion 23 PART C: FORCES AND NEWTON’S LAWS

Forces 25 1st Law 30 2nd Law 32 3rd Law 36 Newton’s Laws of Motion summary Some Applications of Newton’s Laws

38

o Collisions & Safety 40 o Parachutes and Terminal Velocity 41 o Rocket Propulsion 42

PART D: MOMENTUM Momentum and Impulse 43 Conservation of Momentum 47



PART A: INTRODUCTION Vectors SCALARS & VECTORS If you were bush walking and asked someone for the air temperature, you would be satisfied to be told “5o C”. However if you asked how to get to the next hut and you were told to “walk for 5 km” you would be very unhappy if you then walked 5 km north and the hut turned out to be 5 km east!!. Similarly driving at 100 km/h north is quite different to driving at 100 km/h south. Sometimes direction can be very important in specifying physical quantities SCALAR quantities are specified by: (i) a numerical term e.g. Mass of object = 1.72 kg (ii) a unit term VECTOR quantities require a DIRECTION specification as well. i.e. they require (i) a numerical term (ii) a unit term e.g. Velocity of plane = 755 km h-1 (east) (iii) a direction Amongst the scalars and vectors you should become familiar with are:

SCALARS VECTORS energy distance displacement momentum temperature speed velocity power time acceleration volume force mass weight

ADDITION AND SUBTRACTION OF SCALARS This is simply achieved using the standard adding and subtracting process; e.g.

mass of beaker plus salt = 179.60 g - mass of empty beaker = 153.06 g

Thus: mass of salt = 26.54 g ADDITION OF VECTORS Any two vectors of the same type can be added together to give a single resultant vector. This resultant vector has the same overall effect as the original vectors combined. However, this process is made more difficult because of the direction term. We cannot just add the magnitudes together!! The method used is often described as the "head to tail" method. e.g. Consider the two force vectors A and B acting on an object X. (diagram 1) The combined effect of the two vectors is found by adding them. This is done by placing the tail of one vector (B) onto the head of the other vector (A) or vice versa. (diagram 2) The resultant can be found either by a scale diagram or by using standard trigonometric techniques. NOTE: The resultant vector needs two parts - magnitude (size and unit) and direction (E, W, N, S) - to state it completely. [N.B. Subtraction of vectors involves the addition of the 'negative' vector.] The sum of the two vectors, called the RESULTANT vector (R), is specified by the line going from the tail of the first to the head of the second. (diagram 3)



diagram 1 diagram 2 diagram 3 i.e A + B = R (where R is the resultant vector) N.B. Subtraction of vectors involves the addition of the 'negative' vector. The negative vector is a vector pointing in the opposite direction. The resultant can be found either by a scale diagram or by using standard trigonometric techniques. NOTE: The resultant vector needs two parts – magnitude (size and unit) and direction - to state it completely. Sample Problem (using a scale diagram; as a practice exercise, redraw to scale). A hiker goes 3km north and then 4km east. What is their total displacement from their starting point?

. 4 km . 4 km 3 km 3 km Resultant displacement= 5 km 53◦

N Scale 1cm = 1km

The hiker has walked a distance of 7 km, (3 km N + 4 km E)

but their displacement is 5 km N 53◦E of their starting point.

X

A B

X

A

B

X

A B

R



Examples 1 Draw a scale diagram to solve.

Two forces act on an object M. Force A is 6 newtons south while force B is 8 newtons S 60o W. Find the RESULTANT force on the object.

2 A yacht sails 200 km east then 400 km north. What is its resultant displacement?



COMPONENTS OF VECTORS Just as two vectors can be added to give a single resultant, a single vector can be broken down into two vectors which give the same effective result as the single vector. These component vectors allow us to see the effect of a vector in particular directions. We usually take the component vectors to be at right angles to each other (eg along the x- and y-axes).

θ θ

and

θ θ

Ax is the x-component, and Ay is the y- component of

Examples: 3 A rope is pulling a cart along a path. If the force in the rope is 300N and the rope is at 30o above the

horizontal, find the horizontal and vertical components of the force on the cart.

4 A bushwalker treks 25 km N 30o E. a) How far north of her starting position is she?

What is her easterly component of displacement?

θ



PART B: KINEMATICS Kinematics is the study of motion, it stems from the Greek word for motion kinema. Motion can be classified as one of two types: non-accelerating and accelerating. Accelerating motion is movement involving changes in speed or direction or both. Non-accelerating motion is movement in a straight line at a constant speed. We can describe the motion by way of words, graphs or equations. The quantities used to describe motion include: displacement distance velocity speed acceleration time It is very important to identify vector quantities and direction must be shown where appropriate. Time (t) is a scalar quantity and is measured in seconds (s) or hours (h). Displacement and Distance Displacement [s, or s ] = a vector quantity giving the resultant straight line distance travelled in a particular direction.

Distance [d] = a scalar quantity giving the total length of the path followed by a moving object, irrespective of changes in direction.

e.g. a person walks along a pathway from point A to point B. Displacement = = length and direction of straight line A-B

Distance = = Length of pathway

Both displacement and distance are measured in metres (m), kilometres (km) or centimetres (cm). However, displacement also requires a direction e.g. North, N37oW or 68o E of S. Velocity and Speed Velocity [v, or v ] = a vector quantity giving the rate of change of displacement.

Velocity =

Speed [v] = a scalar quantity giving the rate of change of distance.

Speed =

Both have the units m s-1 (metres per second) or km h-1 (kilometres per hour). Velocity also has a direction, which is the direction of the displacement.

Conversion between m s-1 and km h-1 : 1 km/h = = m s-1. s

Thus: km h-1 → m s-1 divide by 3.6 m s-1 → km h-1 multiply by 3.6

A

B



Average Velocity and Instantaneous Velocity Instantaneous velocity is the velocity of the object at a moment in time. If the velocity is varying, v can be found by reading the v~t graph or the slope of the s~t graph. Average velocity, vav, is the velocity averaged over the whole trip (or a part of it). It can only be found by dividing the displacement by time. Do not average the velocities. Acceleration Acceleration is a vector quantity giving the rate of change of velocity

a = = Units are m s-2 , m/s2 , m/s/s (metre per second squared). Direction is that of the change in velocity.

Example 1: An athlete runs the 400m around an oval track in 50s. Calculate his i) displacement ii) distance covered iii) average velocity iv) average speed Example 2: A car travels 50 km north at 100 km h-1 then 30 km south at 60 kmh-1. Calculate its average velocity and speed.



Example 3: A cyclist is travelling at 7.0 m s-1 east when she speeds up to 13.0 m s-1 east in 4.0s. Calculate her:

i) Change in velocity

ii) Acceleration.

Example 4: An Olympic sprinter accelerating at 1.4 m s-1 from rest reaches a speed of 12.0 m s-1. Calculate how long this will take him.



Displacement ~ time Graphs We will only consider motion in a straight line. Displacements and velocities northwards are taken as +ve.

This graph represents an object at rest, displaced …….. north of the zero point.

This graph represents an object moving with a fixed velocity, i.e. zero acceleration.

Velocity =

Represents an object moving at constant negative velocity (ie going southwards). It starts at ….. north and after 5.0 seconds it is ……………………… At 10s the object is …………………………………

Represents an object moving northwards with ………………. velocity, i.e. +ve acceleration. Velocity at 2 sec is less than at 4 sec.

Slope ( ) = ………………………… velocity.

Instantaneous velocity

= velocity at an moment in time (e.g at t = 2.0 s) = slope of s~t graph (slope of tangent). A car’s speedo gives an instantaneous reading.

Average velocity = velocity averaged over the whole (or part) of a journey.

= total displacement /time taken for journey

0 2 4 6 8

10 12

0 5 10

s (m)

time (s) 0 5

10 15 20 25

0 5 10 15

s (m)

time (s)

-15 -10

-5 0 5

10 15

0 5 10 15

s (m) time (s)

0 2 4 6 8

10

0 5 10

s (m)

time (s)



Examples 1 An object moves for 8.0 seconds as shown.

a) What is the object’s displacement at time

i) 3.0 s

ii) 7.0 s?

b) Calculate the object’s velocity at time i) 1.0 s

ii) 5.0 s

iii) 7.0 s

c) What was the object’s total displacement during the 8.0 s?

d) What was the total distance moved during the 8.0 s?

e) Find the object’s average velocity for the 8. 0 s

f) Calculate its average speed.

-15

-10

-5

0

5

10

15

0 5 10

s (m) time (s)



Velocity ~ time Graphs

Represents object moving with constant velocity of ………. m s-1 northwards i.e. ………………… acceleration. Displacement, s = area under the graph = s x t = x = 160 m north.

Describes an object with an initial velocity of ……… m s-1. Uniform acceleration given by a = …………… = 2.0 m s-2 North

Object has initial velocity ………… m s-1 , undergoing uniform negative acceleration

a = ………. = - 4.0 m s-2

= 4.0 m s-2 Southwards.

An object initially at rest, with high initial acceleration then undergoing decreasing acceleration. Displacement = area under curve ≈ ………………………………… m.

Summary

Displacement Graph Velocity Graph Displacement = direct reading Velocity = Slope Instantaneous velocity = slope of tangent to curve Av Vel = displacement/time

Velocity = direct reading Acceleration = Slope Instantaneous Accln. = slope of tangent Displacement = Area under graph Av Vel = displacement/time

Note: In a velocity~time graph, area below the x-axis represents a negative displacement.

0

10

20

30

0 2 4 6 8 10

v (m/s) North

time (s)

0

10

20

30

0 2 4 6 8 10 12

v (m/s) North

time (s)

-25

-15

-5

5

15

25

0 2 4 6 8 10 12

v (m/s)

(north)

time (s)

0

10

20

0 2 4 6 8

v (m/s)

time (s)



Examples 1 An object moves for 12 seconds as shown in this v ~ t graph

a) Calculate the object’s velocity at time:

i) 3.0 s

ii) 7.0 s.

b) Calculate its acceleration at: i) 2.0 s

ii) 5.0 s

iii) 8.0 s

c) Find its displacement after: i) 4.0 s

ii) 6.0 s

iii) 12 s

d) Calculate its average velocity over

i) 6.0 s

ii) 12.0 s

e) Find the distance travelled and the average speed over the 12.0 s

-8 -6 -4 -2 0 2 4 6 8

0 2 4 6 8 10 12 14

v (m/s) north

time (s)



2 This v ~ t graph describes the vertical motion for a ball thrown upwards.

a) What was the initial velocity?

b) What happened at 4.0 s?

c) What was the acceleration of the ball at: i) 2.0 s

ii) 4.0 s

iii) 6.0 s?

d) What happened at 8.0 s?

-40 -30 -20 -10

0 10 20 30 40

0 2 4 6 8 10

v (m/s)

time (s)



Acceleration ~ time Graphs Acceleration is found from the slope of the velocity~time graph.

a = slope of v~t graph =



0

10

20

30

0 2 4 6 8 10

v (m s-1)

time (s)

0

10

20

30

0 2 4 6 8 10 12

v (m s-1)

time (s)

accln (m s-2)

time (s)

accln (m s-2)

time (s)

-20

-10

0

10

20

0 2 4 6 8 10 12

v (m s-1)

time (s) 0

10

20

0 2 4 6 8 10

v (m s-1)

time (s)

accln (m s-2)

time (s)

accln (m s-2)

time (s)



Examples 1 An object moves for 12 seconds as shown in this v ~ t graph

Sketch an acceleration~time graph.

a) Draw the axes for the a~t graph (given here)

b) Calculate the slope (acceleration) of the four parts of the v~t graph.

i)

ii)

iii)

iv)

c) Draw the segments onto the acceleration graph above.

NOTE: If the relevant segment of the v~t graph is a straight line, the corresponding portion of the a~t graph is a horizontal straight line. If the v~t graph has a positive slope, the a~t graph is above the x-axis. If the v~t graph has a negative slope, the a~t graph is below the x-axis.

-8 -6 -4 -2 0 2 4 6 8

0 2 4 6 8 10 12 14

v (m s-1) north

time (s)

-4 -3 -2 -1 0 1 2 3 4

a (m s-2) north

time (s)



Linear Motion - Equations For the majority of motion problems encountered, you will be dealing with objects undergoing constant acceleration. Equations of Motion for Constant Acceleration

u = initial velocity s = displacement v = final velocity t = time a = acceleration

Acceleration = change in velocity/time

= (final velocity – initial velocity)/time a = (v – u)/t v = u + at …….. 1 [s u v a t]

Since the acceleration is constant,

vav = (u + v)/2 and s = vav.t

thus s = …….. 2 [s u v a t]

and by using v = u + at

and substituting v into equation 1, we get

s = …….. 3 [s u v a t]

Equating “t” from 1 & 2 gives: v2 = u2 + 2a. s …….. 4 [s u v a t] It is also possible to show that

s = …….. 5 [s u v a t]

however, this last equation is not often used. Note: that each equation uses 4 out of the 5 possible variables. It is often easier to identify the “missing” variable when choosing which equation to use to answer a question. Since 4 variables are involved, 3 must be “known” in order to find the one “unknown”. It helps to list the variables. Remember, these equations only apply if the acceleration is constant.



Examples 1 Convert:

a) a speed of 80 km h-1 to m s-1.

b) a velocity of 20 m s-1 southwest to km h-1.

2 Constant speed (i.e. no acceleration) a) What is the average speed of a runner who covers 42.195 km in 2 hrs 16 mins 40 seconds

i) in km h-1

ii) in m s-1?

b) How long will it take a cyclist travelling at 35 km h-1 to cover 2.5 km?

3 A roller skater decreases her velocity uniformly from 8.0 m s-1 northwards to 4.0 m s-1 northwards in 3.0 seconds. What is her acceleration?



4 A cheetah increases its velocity from 20 km h-1 eastwards to 80 km h-1 eastwards while accelerating at 5.0 m s-2. How far did it travel in that time?

5 An object travelling at 10 m s-1 southwards is accelerated at 4.0 m s-2 northwards for 5.0 seconds. a) What is the displacement of the object in that time?

b) Explain your answer.



Vertical Motion In vertical motion problems, note the following:

• The object is in “free fall” • unless otherwise noted, ignore all forces such as friction and air resistance • The acceleration is the acceleration due to gravity, ag = 9.80 m s-2 downwards (= g). • If an object is thrown upwards:

o at its highest point, the velocity is zero (but the a is still 9.80 m s-2 down). o by symmetry, the downward velocity is a mirror of the upward velocity.

• If an object is dropped, its initial velocity is zero.

Examples 1 A rock is dropped from a bridge 25 m above the water.

a) What was the rock’s initial velocity?

b) How long will it take to hit the water?

c) What is its impact velocity?



2 A ball is thrown upwards at 15 m s-1. Calculate: a) its initial velocity

b) the velocity of the ball at the highest point of its motion

c) the acceleration of the ball i) just after leaving the thrower’s hand

ii) at the highest point

d) the maximum height reached

e) the time taken to reach the maximum height

f) the velocity of the ball as it is caught by the thrower again.



Motion on an Inclined Plane A “plane” is simply a flat surface. “Inclined” implies that it is tilted to the horizontal. Assume that the plane is frictionless, unless told otherwise.

Any object placed in a gravitational field will experience a vertically downwards acceleration g (9.8 ms-2). However, if it is placed on an inclined plane, only a component of this acceleration will be experienced parallel to the plane. This is dependent upon the angle at which the plane is inclined as shown in the figure. The weight of the mass, W = mg vertically downwards, can be resolved into two components. Wd = W.Sin θ down the plane (parallel to plane) Wp = W.Cos θ perpendicular to the plane. The perpendicular component is balanced by a reaction force from the plane “pushing outwards”. The component down the plane accelerates the mass down the plane. If there is no friction, this acceleration is a = ag Sin θ down the plane = g Sin θ “ “

Examples: 1 What is the acceleration experienced by an object down a frictionless plane inclined at 16.5o to the

horizontal?

2 A skier starts from rest at the top of a hill and 5.0 s later is travelling at 7.5 m s-1 down the slope. What is the angle of inclination of the slope?

3 On the planet Poseidon a skier accelerates at 5.99 m s-2 down a slope inclined at 20.0o to the horizontal. What is the acceleration due to gravity on Poseidon?

θ θ

W Wp

Wd



Projectile Motion – Horizontal Projection The curved path of a ball thrown into the air is an example of a projectile path. If an object is projected horizontally from an elevated position (see diagram below), the time of falling along path 'A' is identical to the time the object would take to fall if dropped vertically along path 'B' (ignoring air-resistance). Throughout the flight path the object will have constant horizontal velocity (a = 0) and increasing vertical velocity (a = 9.80 m s-2 (down)) Consider an object projected horizontally at 35.0 m s-1 from the top of a 125 m cliff overlooking the sea. What is the object's horizontal range? i.e. How far out from the base of the cliff does the object hit the sea?

ANSWER: (i) VERTICAL MOTION: u = 0 m s -1 s = ut + at2/2 a = 9.80 m s-2 125 = 0 + 9.80t2/2 t = ? sec 9.81t2 = 250 s = 125 m t2 = 250/9.80 t = 5.05 sec

Thus the object falls for 5.05 seconds (ii) HORIZONTAL MOTION: u = 35.0 m s -1 s = ut + at2/2 a = 0 m s-2 s = (35.0 x 5.05) + 0t2/2 t = 5.05 sec s = 177 m. s = ? m Thus the object's horizontal range is 177 m

35 m s-1

HORIZONTAL RANGE

PATH 'A' PATH 'B'

Cliff 125 m



Examples: 1 A ball is dropped from a window 10 m above the ground.

a) How long will it take to hit the ground?

b) What will its speed be as it hits the ground?

At the same instant another ball was thrown horizontally at 10 m s-1 from the window. c) How long will it take for this second ball to hit the ground?

Just before it hits the ground, d) what will the vertical component of its velocity be and

e) what will the horizontal component of its velocity be?

2 A rock is thrown at 20 m s-1 horizontally from a cliff 25 m above the sea. Calculate: a) The time taken for the rock to reach sea level

b) The horizontal distance travelled in that time

c) The vertical component of the final velocity

d) The horizontal component of the final velocity

e) Its final velocity



Part C: FORCES and NEWTON’S LAWS OF MOTION Forces A force can be described in basic terms as a push or a pull. Forces are everywhere, we experience them continually. If there were no forces in our universe earth would not be in its orbit around the sun and we would not be able to walk, or even exist because objects need forces to keep their shape. A force exerted on an object can

• Start a stationary object moving • Stop an already moving object • Change its speed if already moving • Change its direction of motion

These four above are all aspects of acceleration • Change its shape [we will not deal with this aspect]

Some examples of forces

• Push • Pull • Tension • Reaction force • Friction • Air resistance • Weight (gravitational attraction)

Units The unit of force is the Newton (N). Large forces are expressed in kilonewtons (kN) [1kN = 1000 N = 103 N] Two or more forces acting on an object Forces are vector quantities and therefore both the magnitude and the direction must be taken into account when adding forces. The sum is the net force (or resultant). [Don’t forget to put the force vectors “nose-to-tail” to add them] Example 1 An object is subject to a 20 N force northwards and a 35 N force southwards. Find the resultant force.



Balanced and Unbalanced Forces If the resultant of two or more forces is zero, the forces are balanced If the resultant is not zero, i.e. the resultant has a magnitude and a direction, the forces are unbalanced. This unbalanced force is also known as the net force. Examples 2 An object is subject to a force of 2500 N northwards, a force of 1500 N southwards and a force of 1000

N southwards. Find the net force.

3 A mass is subject to a force of 200 N east and one of 200 N southwards. What is the resultant force?



Some important Forces Weight Force Weight is the gravitational force that the earth exerts on a body. It depends on: the mass of the body m and the strength of the gravitational field

= m where = 9.80 N kg-1 ( = 9.80 m s-2).

[Alternative representations of the weight vector are and .] Weight is a force (i.e. it is a vector) and its direction will be downwards (towards the centre of the earth). Because it is a force, the unit of weight is the Newton (N). It is important to remember that what we refer to as “weight” in everyday life is actually mass and has the units of kilograms (kg). Examples 4 What is the weight of a 65 kg student?

5 Calculate the mass of an object which has a weight of 4.9 N (downwards). Weight always acts on a body regardless of its position or motion. Unless we move a large distance from the surface of the earth, it is constant. The only time we can ignore weight is when an object is on a horizontal surface.

The downward weight force is balanced by the upward reaction force and we can usually ignore these two forces. In every other Dynamic case, we must consider the weight.



Some situations we will consider include: Inclined Plane Mass on a Rope Free Fall Parachute

Horizontal Projection Lift Satellite

Complete this table (don’t forget the weight force). Reaction Force When one object pushes against another, the second object “pushes back” – this is called a reaction force. For example a trolley is sitting on the ground. Its weight force is pushing the trolley downwards. The reaction force will be the road pushing up (and preventing the trolley from sinking to the centre of the earth!) Friction Force Friction always acts to resist motion. If an object is stationary, any frictional forces oppose the direction of an applied force. If the object is moving, the frictional force acts opposite to the direction of motion. Air resistance is a type of frictional force. It acts on a moving body opposite the direction of motion. It increases with the speed of the object. In some situations it is not immediately obvious that the object is stationary. Consider a car being driven; while the car itself is moving, at the point of contact with the road the tyres are stationary (if the tyres are not slipping). Similarly, when a person is walking or running their foot is stationary at the moment of contact with the ground.



Examples 6 A bicycle is coasting (the cyclist has stopped pedalling). Sketch a diagram showing the friction forces.

7 A 2.0 kg block is sitting at rest on a rough plane inclined at 200 to the horizontal. a) Draw a diagram showing the frictional force acting.

b) Calculate the frictional force acting.

8 An object is falling in the atmosphere (i.e. there is air resistance). Draw diagrams showing the forces on the body

c) At the start d) After a short time e) After some time.



NEWTON’S LAWS OF MOTION I NEWTON’S FIRST LAW (N1 or N I) “A body will remain at rest or continue with constant velocity unless acted upon by an unbalanced force”. Constant velocity means a constant speed in a straight line. SOME IMPLICATIONS OF N1:

(i) A body permanently at rest is being acted upon by balanced forces. (ii) A body with a constant velocity is being acted upon by balanced forces. (iii) An unbalanced force is needed to produce an acceleration (change in velocity). (iv) A change in speed or a change in direction requires an unbalanced force. (v) Safety features in cars e.g. seat belts and head-rests.

Examples 9 Draw, label and show the relative magnitude of the forces acting on the following: • Ball at rest on the table • Ball dropped • Ball initially pushed and allowed to roll along the table • Ball pushed along the table at a constant speed • Ball thrown into the air • Ball attached to a string and suspended at rest • Ball lifted at a constant speed • Ball attached to a string and lifted at increasing speed



10 In outer space where there is no air resistance, once a rocket had attained its set speed would it need to keep its engines on?

11 Why are you “thrown forward” in your seat as a car stops?

Is this why we need seat belts?

12 What happens to a passenger when a car turns a corner?

13 What causes “whiplash” injuries in a rear-end collision?



II NEWTON’S SECOND LAW (N2 or N II) “The acceleration (a) of an object is directly proportional to the size of the unbalanced force (F) applied to it and inversely proportional to the mass (m) of the object”.

NOTE: Force ( ) is a vector and is measured in newtons (N) Acceleration ( ) is a vector and is measured in m s-2 Mass (m) is a scalar and is measured in kilograms (kg) SOME IMPLICATIONS OF N2:

(i) An unbalanced force is needed to produce an acceleration. (ii) The acceleration is in the direction of the unbalanced force. (iii) In free-fall, a body’s weight is the unbalanced force exerted (down) on it and causes the

(Earth’s) gravitational acceleration: g = 9.80 N kg-1 (down) [= 9.80 m s-2 down] i.e. Fw = m.g

Examples 14 An 800 kg car accelerates from rest to 20 ms-1 in 6.0 sec. What is the average force being exerted on the

car during this period?

15 The brakes of a 3 tonne truck exert a stopping force of 4.5 kN. If the truck is initially travelling at 72 km/h, what will its acceleration be and how far will it travel before coming to rest?

16 A bucket of water with a total mass of 20.0 Kg is raised from a well. During the first part of the rise the bucket has an upward acceleration of 0.200ms-2. During this initial period:



a) what is the net force acting on the bucket of water? (4.00 N up)

b) what is the force exerted by the rope on the bucket of water? (200 N up)

17 A skydiver has a mass of 70.0kg. Find the average force exerted on her by her parachute when her vertical velocity is reduced from 13.0 ms-1 to 3.00 ms-1 in 2.00 s after it opens.(1040 N up)



18 The mass of a lift and its occupants is 600 kg. Calculate the tension at the bottom of the cable supporting the lift when it is moving:

a) upwards with a uniform velocity? (5880 N up)

b) with a uniform acceleration of 0.450ms-2 down? (5610 N up)

c) with a uniform acceleration of 0.450 ms-2 up? (6150 N up)

d) can you tell in b) or c) whether the lift is moving upwards or downwards?



19 A man of mass 100kg slides down a rope that can support a weight of only 755 N. a) how is this possible?

b) what is the least acceleration he can have without breaking the rope? (2.25 ms-2)

c) what will be his minimum speed after sliding down 8.00 m? (6.00 ms-1)



III NEWTON’S THIRD LAW (N3 or N III)

“To every force in nature, there is an equal force exerted in the opposite direction for the same period of time”. NOTE: The forces are sometimes referred to as the “action” force and the “reaction” force.

FAB = force of A on B FBA = force of B on A and FAB = - FBA

SOME IMPLICATIONS OF N3:

(i) Two bodies are required to create a force. No single body can create a force.

(ii) Forces always occur in pairs which are oppositely directed. (iii) The two forces of the pair act on different bodies. (iv) In an interaction between bodies A and B, the force exerted by A on B (FAB) is equal to but

opposite in direction to the force exerted by B on A (FBA) i.e. FAB = - FBA Examples

20 Someone on a stationary skateboard jumps forwards off it - what does the skate board do and why?

21 A bee collides with a truck travelling in the opposite direction. Compare: a) The force on the bee and the force on the truck

b) The acceleration of the bee and the acceleration of the truck.



22 During a tug of war between Red and Blue teams show forces on both teams: a) when both are taking the strain

b) when Red team is just about to win.

c) Explain this apparent paradox. (Don’t forget that there is also a force between each team and the ground.)

23 According to N III a caravan exerts an equal and opposite force on a car when it is towed. Why does the caravan move at all?



Newtons Laws of Motion Summary

Newton’s First Law (N I) The velocity of any object will remain constant unless the object is acted upon by an unbalanced force. N I implies: • a net force is required for acceleration to occur • the direction of acceleration is the same as the direction of the net force • an object at rest has zero acceleration and hence zero net force acting upon it. • an object travelling at constant velocity has zero acceleration and hence zero net force acting upon it. • by observing the motion of an object we can:

• determine the direction of acceleration (if any) • determine the direction of the net force (if any) • determine the relative magnitudes of forces that have been identified to contribute to the net force

Newton’s Second Law (N II)

The rate of change of momentum of a body is proportional to the total force acting on it and occurs in the direction of the force. N II has mainly mathematical implications such as: (commonly used formulae are in bold)

change in momentum F =

time impulse = change in momentum

F = Δp t

impulse = Δp

(mv – mu)

F = t impulse = (mv – mu)

m(v-u) F = t impulse = m(v-u)

m x (v-u) F = t

F = ma

Fg = mg = Fw this is often called force due to gravity (Fg) or the weight force (Fw). Some of these concepts will be covered later.

Things you need to take away from N II of motion: • The difference between weight and mass • How to calculate forces using information ascertained from NI and NIII • Force is a vector with units Newtons (or kg m s-2) • Working these problems out is no different to others we have used

• ie. do diagram, show and/or list the information we have, pick appropriate formula and away we go…

Newton’s Third Law (N III) If a body A exerts a force on body B then body B exerts an equal and opposite force on body A for the same time.



N III implies: • Mathematically we can then say FA = -FB • Wherever there is a force you should always ask: “Where is the opposing force?” • The transferral and magnitude of force through complex systems can be identified

Example 24 A car of mass 1.2 tonne is towing a 600 kg caravan.

If the car accelerates at 1.5 m s-2 westwards, find: a) the horizontal force exerted by the road on the car’s tyres,

b) the force that the car exerts on the caravan,

c) the force that the caravan exerts on the car,

d) the net force on the car.



Some Applications of Newton’s Laws. I. Collisions and Safety. Consider a collision of a moving object (e.g. your head) with a very large stationary object (e.g. the earth). The moving object is brought to rest by a force. The size of this force depends on:

o the mass of the object being brought to rest, and o its acceleration in coming to rest.

This is given by:

Since and

This has implications for human safety in collisions (e.g. car crashes, children falling from play equipment, design of safety helmets & seat belts etc). To reduce the force suffered during an impact, the stopping distance, s, (or, equivalently, the stopping time t) must be made as large as practicable. This is achieved by using materials which compress or stretch while slowing the body down. Some suitable materials are: grass (rather than concrete) for play areas, polystyrene in a helmet (rather than the tarmac of the road), stretchable fabric in a seat belt, crumple zones in a car, air bags (rather than the car dashboard). These are all designed to increase the stopping distance and hence reduce the force on the body. Example 25 A 20 kg boy falls 1.0m. a) What is his speed as he hits the ground?

b) He lands on his head, find the net force on it if the boy lands on: i) concrete with a stopping distance of 1 mm.

ii) grass with a stopping distance of 2.0 cm.



II. Parachute Jumping: The Changing Forces 1 Initially, the only force on the person is their weight (Fg) i.e. they accelerate downwards at 9.80 m s-2.

2 Their increasing downward velocity creates an increasing frictional force (FF). Thus downward acceleration is now less than 9.80 m s-2.

3 At “terminal velocity”, the weight force and the frictional forces are now balanced. Therefore acceleration = 0. i.e. velocity is now constant

4 Once the parachute is deployed, it makes a much greater FF. There is now a net upward force and hence an upward acceleration (deceleration down) causing a decrease in downwards velocity.

5 As the parachutist slows down, the frictional force FF decreases until at a relatively small downward speed the frictional force is once again in balance with the weight force. There is now no acceleration and the parachutist continues to descend at a (safe) constant velocity! (i.e. at a slower terminal velocity. Draw (to scale) the forces acting on the parachutist during her descent. Indicate also her speed and acceleration.



III. Rocket Propulsion Newton’s Third Law implies that for a body to have a force exerted on it there must be an equal but opposite force on another object. On earth we can achieve this in a car by pushing against the ground; an aeroplane does it by pushing some air backwards with its propellers, a boat pushes against the water. In outer space there is no ground, air or water to push against. To propel themselves, space travellers must carry their own object to push against. This can be compressed gas which is fired in a direction opposite to the desired direction of motion. The gas is pushed one way by the space ship and, by NIII, the space ship is pushed in the other direction by the gas. The gas is often not carried in a compressed form but is generated as needed by a rocket engine which burns on-board fuel to produce the hot, compressed gas. Examples: 26 Draw in the forces on the rocket and on the gases. 27 How does a space-walker manoeuvre about while floating in space?



Part D: MOMENTUM Momentum and Impulse THEORY: p = m.v F = m.a (Newton's 2nd Law) F = m(v-u)/t Impulse = I = F x t = mv – mu = pt - p0 = Δp (F x t) is called the 'impulse' and is equal to the change in momentum Units for 'change in momentum' are N.s or kg.m. s-1 Impulse (change of momentum) is in the direction of the resultant force.

Momentum The interaction of a moving body with other objects depends on both its mass and its speed. A large body moving slowly might have the same effect as a small object moving quickly. The product of a body’s mass (m) and velocity (v) is called its momentum (p).

p = m x v Momentum is a vector quantity and it is in the direction of the velocity of the body. The units for momentum are kg m s-1 (kilogram metres per second) Or, alternatively N s (Newton seconds)

Examples

1 What is the momentum of : a) A 0.75 tonne van moving at 15 m s-1 westwards?

b) A 60 g bullet travelling at 300 m s-1 to the north-east?



Impulse When an unbalanced force acts on a body for a time, it changes the momentum of the body. The product of the force (F) and time for which the force acts (t) is called the impulse of the force (I).

I = F.t Impulse is also a vector quantity and it is in the direction of the unbalanced force. The units for impulse are N s (Newton seconds). When a force acts on a body, the impulse of the force equals the change in momentum of the body.

I = Δp = pfinal – pinitial

= mv - mu Therefore, the change in momentum of a body is in the same direction as the impulse, i.e. in the same direction as the unbalanced force.

Alternative statement of Newton’s Second Law

F.t

=

I

= Δp F = Δp/t

i.e. the force on a body equals its rate of change of momentum

2 a) Find the impulse given to a 10 kg body by a force of 25 N eastwards for 2.5 s.

b) What is the change of momentum of the body?



3 A 3.0 tonne truck travelling at 20 m s-1 south comes to rest in 10 sec. a) Calculate the: i) Initial momentum

ii) Final momentum

b) What was the change in momentum (∆p) of the truck?

c) Calculate the change in velocity (∆v) of the truck.

d) By using (b), what average force was acting on the truck while it was slowing?

Consider a graph of force against time. The area under the graph gives the impulse of the force (since Area = F.t = I). This also gives the resulting change in momentum of the body in question.

The result is also true for non-constant forces.

0 0.5

1 1.5

2 2.5

0 0.5 1 1.5 2 2.5

Force N (North)

time (s)



4 This graph shows the force acting on a 1.5 kg object over a period of 4 seconds. The object was initially travelling at 2.5 m s-1 north.

a) What is the initial momentum of the object?

b) Find the impulse of the force.

c) What is the resulting change in momentum of the object?

d) Calculate the object’s final momentum.

e) Calculate the final velocity of the object.

0

2

4

6

0 1 2 3 4 5

Force (N) North

time (s)



Conservation of Momentum Theory: p = m.v L.C.M. ∑ (pinit) = ∑ (pfin) ∑ = "sum of" LCM = Law of Conservation of Momentum. Conservation of Momentum When two bodies collide or interact, their total momentum remains constant. This is a direct result of Newton’s Third Law. The forces during the interaction must be equal but opposite for the same period of time. Thus changes in momentum of the two bodies must be equal but opposite; i.e. the individual momentum changes balance out and the total momentum is unchanged. Remember that momentum is a vector quantity.

The total momentum of any system is constant and is not affected by the interaction of any parts of the system.

This assumes that there are no unbalanced external forces acting on the system (e.g. friction or weight). Examples 5 A car of mass 600 kg travelling northwards at 16.0 m s-1 collides with a stationary car of

mass 900 kg. If the cars coalesce (stick together) after the collision, calculate a) The total momentum before the collision

b) The total momentum after the collision

c) The velocity of the wreckage immediately after the collision.



6 Two ice skaters, Abigail of mass 60 kg and Bertrand of mass 90 kg are stationary and facing each other. Abigail then pushes Bertrand so that he moves at 1.5 m s-1 westwards. Calculate what happens to Abigail.

7 Two railway carriages approach each other. Carriage P (mass 50 tonne) is travelling at 3.0 m s-1 north and Q (mass 75 tonne) is travelling at 4.0 m s-1 south. If the carriages coalesce, what is their resultant velocity?



8 A 150 gram dynamics cart travelling at 0.75 m s-1 eastwards collides with a 300g cart travelling at 0.25 m s-1 eastwards. After the collision, the 150 g cart is travelling at 0.25 m s-1 westwards. Calculate the final velocity of the 300 g cart.

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Motion and force

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