Hf s Worked Examples

Worked Examples

WJH

5th July 2010

CONTENTS CONTENTS

Contents

0.1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1 Questions 3

1.1 HFS - Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 SUVAT . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.3 Motion Graphs . . . . . . . . . . . . . . . . . . . . . . . . 41.1.4 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.5 Energy and Power . . . . . . . . . . . . . . . . . . . . . . 61.1.6 Projectiles . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Numerical Answers 8


3 Worked Solutions 10


1

CONTENTS 0.1. PREFACE

0.1 Preface

This is a set of worked examples. To get the most out of the worked examples,attempt the questions yourself �rst.

Feedback and corrections to [email protected] an ellipses . . . is seen in calculations, it is implied that the preceding

number is carried through to a higher precision in following calculations than isshown to avoid rounding errors.

Unless otherwise indicated, ignore air Resistance at all times.The acceleration due to gravity g should be taken to be 9.81ms−2

This text uses inverse trigonometric functions in the following form:

y = tanx =⇒ x = arctan y

y = sinx =⇒ x = arcsin y

y = cosx =⇒ x = arccos y

Instead of the alternative form occasionally seen:

y = tanx =⇒ x = tan−1 y

y = sinx =⇒ x = sin−1 y

y = cosx =⇒ x = cos−1 y

2

CHAPTER 1. QUESTIONS

Chapter 1

Questions

1.1 HFS - Mechanics

1.1.1 Vectors

1. De�ne the word `Vector' and give three example of a vector.

2. De�ne the word `Scalar' and give three examples of a scalar.

3. An airplane travels west at a speed of 400kmph into a wind blowing eastat 50kmph. What is the velocity of the airplane?

4. A racing car travels 5 laps around a circular circuit with a radius of 200m.The start line is at the north end of the track. What is:

(a) The total distance the car travels?

(b) The maximum displacement of the car?

(c) The �nal displacement of the car?

5. During a hike, a group walk 4km north to the base of a hill, then 3kmeast to the 500m high summit? The entire hike takes 3 hours. Calculate:

(a) The magnitude of the total displacement.

(b) The average speed.

(c) The average velocity.

1.1.2 SUVAT

1. Identify what each of the letters in the `SUVAT' acronym stand for.

2. State three common forms of the SUVAT equations.

3. Rearrange any appropriate equations from question 2 to make time thesubject.

4. In a drag race, two cars accelerate uniformly from rest along a straighttrack with a length of 400m. Car A crosses the �nish line in 7.34 seconds,and Car B crosses the �nish line with a speed of 350kph.

3

CHAPTER 1. QUESTIONS 1.1. HFS - MECHANICS

(a) Which car won the race?

(b) How far from the end of the race was the looser when the winnercrossed the �nish line?

(c) Car C is released from a height of 400m above the �nish line; does itbeat the other cars if it travels in free-fall?

1.1.3 Motion Graphs

Figure 1.1:

0

2

4

6

8

10

12

0 1 2 3 4 5

Vec

ocity

v/m

s-1

Time t/s

AB

1. Figure 1.1 shows a velocity-time graph for two objects, A and B whichmove with a constant acceleration but begin their motion at di�erenttimes.

(a) What is the Velocity of A at time t = 2s?

(b) What is the acceleration of B?

(c) What distance does A cover by t = 4s?

(d) At what time will the velocities of A and B be equal?

4


Figure 1.2:

-2

-1

0

1

2

3

0 2 4 6 8 10

Vel

ocity

v/m

s-1

Time t/s

2. Figure 1.2 shows the velocity-time graph for an object. Using the graph,estimate the following within the displayed range:

(a) The maximum velocity of the object.

(b) The acceleration of the object at its maximum velocity.

(c) The time at which the object is furthest from it's starting position.

(d) The displacement of the object at the time identi�ed in (b).

(e) The average velocity of the object to the time identi�ed in (b)

(f) The acceleration of the object at time t = 6s.

1.1.4 Forces

1. State Newtons three laws.

2. A 1500kg car travels along a road propelled by its engine with a forceof 10kN . Air resistance applies a force of 7.5kN against the direction ofmotion of the car. What is the acceleration of the car?

3. 20s after jumping from a plane, a 60kg skydiver has travelled 0.5km.

(a) What is the average drag force acting on the diver in the �rst 20s oftheir jump?

(b) What is the drag force acting on the sky diver at terminal velocity?

4. A 1kg mass hangs at the end of a string. Air is blown over the stringapplying a horizontal force of 3N to the mass. What angle is the stringat from vertical once it settles to equilibrium?

5. A 65Gg cruise ship travels at a constant velocity of 10ms−1 due westacross an ocean. The current applies an eastward force on the boat of5.5MN .

5


(a) If the engine of the ship were to fail, how long would the ship taketo stop.

(b) If, in addition to the current, a wind blows south applying a forceof 3MN on the ship, what force must the ships engines supply tomaintain it's original velocity?

(c) At this speed, the ships engines can supply a maximum force of 6MN .If this force is applied in the direction determined in (b), what is theacceleration of the ship?

Figure 1.3:

θ

6. Figure 1.3 shows an object of mass 1kg suspended between walls by ahorizontal rod and a rope which is at an angle of 30◦ above the horizontal.What is the tension in the horizontal rod?

1.1.5 Energy and Power

1. Identify and de�ne the acronyms K.E and G.P.E.

2. Show that the equations for calculating G.P.E and G.P.E are of the formEnergy = Force× distance

3. An object is dropped from 10m above the ground.

(a) By considering its energy, calculate the speed of the object as itreaches the ground.

(b) After colliding with the ground, the object bounces back upwardswith half the speed it had before the impact. What is the maximumheight the object reaches?

4. A train is travels 1km due north on a �xed track. The wind blows southeast applying a force in this direction of 10kN to the train.

(a) What work is done by the train over this distance?

(b) The train travels the 1km in 2 minutes. What is the power outputof the train engine?

5. During a jump an 80kg sky diver reaches equilibrium at a speed of 180kmph.What power is supplied to the sky diver by the earth to maintain thisspeed?

6


6. A 75kg athlete climbs a rope at 0.5ms−1; assuming the human body is20% e�cient, what is the power generation of the athletes body?

7. At a fun-fair, a 2kg mallet is swung at a `test your strength' game, collidingwith the target at a speed of 5ms−1. The 500g indicator hits the bell whichis 2m above the ground with a speed of 0.5ms−1. What is the e�ciencyof the system that transfers energy from the mallet to the indicator?

1.1.6 Projectiles

1. A ball is thrown upwards from 1m above the ground at a speed of 3ms−1.How long after being released does the ball hit the ground?

2. A bullet is �red horizontally from eye level (1.5m above ground), leavingthe muzzle at 500ms−1.

(a) What horizontal distance does the bullet travel before hitting theground?

(b) What is the velocity of the bullet when it hits the ground?

3. An archer launches an arrow from ground level at 50ms−1 at an angle of30◦ above the horizontal.

(a) How far away does the arrow land?

(b) Will the arrow be able to pass over a wall 5m high 40m away?

4. A space probe lands on an extra-terrestrial planet. In an experiment todetermine the strength of gravity on the planet, the probe launches a ballbearing at 40◦ above the horizontal at a speed of 5ms−1. The bearinglands 50m away. What is the acceleration due to gravity on the planet?

7

CHAPTER 2. NUMERICAL ANSWERS

Chapter 2

Numerical Answers

2.1 HFS - Mechanics

2.1.1 Vectors

Q3 350kmph west

Q4a 6280m

Q4b 400m south

Q4c 0m

Q5a 5.02km

Q5b 2.35kmph

Q5c 1.67kmph at 36.9◦ east of north and 5.71◦ above the horizontal

2.1.2 SUVAT

Q4a Car A

Q4b 81.7m

Q4c No

2.1.3 Motion Graphs

Q1a 6ms−1

Q1b 6ms−2

Q1c 24m

Q1d 6s

Q2a ≈ 2.75ms−1

Q2b 0ms−2

8

CHAPTER 2. NUMERICAL ANSWERS 2.1. HFS - MECHANICS

Q2c ≈ 9.5s

Q2d ≈ 14.4m

Q2e ≈ 1.51ms−1

Q2f ≈ 1.06ms−2

2.1.4 Forces

Q2 1.67ms−2

Q3a 439N

Q3b 589N

Q4 17.0◦

Q5a 118ks

Q5b 6.26MN at 28.6◦ north of west

Q5c 4.08µms−2 at 28.6◦ south of east

Q6 17.0N


Q3a 14.0ms−1

Q3b 2.5m

Q4a 7.07MJ

Q4b 58.9kW

Q5 39.2kW

Q6 1.84kW

Q7 39.5%

2.1.6 Projectiles

Q1 0.851s

Q2a 277m

Q2b 500ms−1 at 0.622◦ below the horizontal

Q3a 221m

Q3b Yes

Q4 −0.492ms−2

9

CHAPTER 3. WORKED SOLUTIONS

Chapter 3

Worked Solutions

3.1 HFS - Mechanics

3.1.1 Vectors

Q1 De�ne the word `Vector' and give three example of a vector.

A1 A vector quantity is one which has both magnitude and a direction.Examples include Displacement, Velocity, Acceleration and Force.

Q2 De�ne the word `Scalar' and give three examples of a scalar.

A2 A scalar quantity is one which has magnitude, but no associated direction.Examples include distance, speed, energy and temperature.

Q3 An airplane travels west at a speed of 400kmph into a wind blowing eastat 50kmph. What is the velocity of the airplane?

A3 The velocity of the airplane is given by the net of the two providedvelocities. As the velocity of the airplane and the wind are in opposite directions,one must take a negative value. If we take west to be the positive direction,then the wind is blowing with a velocity of −50kmph. The net velocity v istherefore:

v = 400kmph− 50kmph = 350kmph

That this is a positive value means that the net motion of the airplane is totravel west at at a velocity of 350kmph. Alternatively we could say that theairplane is travelling east at a velocity of −350kmph

Q4 A racing car travels 5 laps around a circular circuit with a radius of 200m.The start line is at the north end of the track What is:

Q4a The total distance the car travels?

10

CHAPTER 3. WORKED SOLUTIONS 3.1. HFS - MECHANICS

A4a The distance s of travelled by the car is given by the circumference ofthe circular track multiplied by the number of laps n the car completes.

s = n2πr

Where r is the radius of the circle which is given.. Substituting in the knownvalues for the variables gives a value for s of:

s = 6280m

Q4b The maximum displacement of the car?

A4b Displacement will be a maximum when the car has moved from the startline at the north end of the track to the south end of the track. At this pointthe displacement of the car is south of the start line with a magnitude equal tothe diameter of the track. Therefore the maximum displacement of the car is:

400m south

Q4c The �nal displacement of the car?

A4c Because the car travels a complete number of laps, i.e. 5, the �nal positionof the car is at back at the start line, so the displacement of the car is 0m. Asthe magnitude is zero, it is not necessary to include a direction.

Q5 During a hike, a group walk 4km north to the base of a hill, then 3kmeast to the 500m high summit? The entire hike takes 3 hours. Calculate:

Q5a The magnitude of the total displacement.

A5a This is calculated by using Pythagoras's theorem twice. First use thedistance traveled north sn then east se to determine the magnitude of the dis-placement in the horizontal plane sh. Then combine that with the height of thehill svto �nd the magnitude of the displacement s. In the horizontal plane:

sh =√s2n + s2e =

√16 + 9 = 5km

Which combined with sv gives:

s =√s2h + s2v =

√25 + 0.25 = 5.02km

Alternatively we can do the entire calculation in one step, i.e. Pythagoras'stheorem in 3D:

s =√s2n + s2e + s2v =

√16 + 9 + 0.25 = 5.02km

Q5b The average speed.

11


A5b To �nd the average speed we divide the total distance travelled by thetime taken. The distance travelled in the �rst leg of the hike d1 to the base ofthe hill is 4km. The second leg going up the hill d2 requires that the distance becalculated using Pythagoras's theorem on the distance east de and the verticaldistance dv.

d2 =√d2e + d2v =

√9 + 0.25 = 3.04 . . . km

To give a total distance d of:

d = d1 + d2 = 7.04 . . . km

Which is divided by the time taken t to provide the average speed v:

v =d

t=

7.04 . . .

3= 2.35kmph

Q5c The average velocity.

A5c The calculation to determine the magnitude of the average velocity vis simpli�ed as we have already determined the magnitude of the displacement.So this is given by:

v =s

t=

5.02 . . . km

3hr= 1.67kmph

But as velocity is a vector we must also give the direction. Due to the natureof this question, i.e. that it is in 3 dimensions, we need to identify a directionin each of the horizontal and vertical planes. The direction in the horizontalplane θh can be given by the angle away from north. To calculate this we usetrigonometry, using the inverse tan function on the north and east legs of thehike.

θh = arctansesn

= arctan3

4= 36.9◦ east of north

The direction of the velocity in the vertical plane θv can be identi�ed by givingthe angle above the horizontal using similar trigonometry.

θv = arctansvsh

= arctan0.5

5= 5.71◦ above the horizontal

Combining the magnitude and the two directions, the average velocity is givenby:

v = 1.67kmph at 36.9◦ east of north and 5.71◦ above the horizontal

3.1.2 SUVAT

Q1 Identify what each of the letters in the `SUVAT' acronym stand for.

A1 The SUVAT equations are a set of formulae which can be used to describelinear motion with uniform acceleration in one dimension. Each of the lettersstand for:

S Displacement

12


U Initial velocity

V Final velocity

A Acceleration

T Time

Q2 State three common forms of the SUVAT equations.

A2 Any of the following, or rearranged versions there-of would be valid

s = ut+1

2at2 (3.1)

s = vt− 1

2at2 (3.2)

v = u+ at (3.3)

v2 = u2 + 2as (3.4)

s =u+ v

2t (3.5)

Q3 Rearrange any appropriate equations from question 2 to make time thesubject.

A3 Four of the �ve equations listed above contain t and so can be rearranged.These are 3.1, 3.2, 3.3 and 3.5. Starting with equation 3.1:

s = ut+1

2at2

Which is a quadratic equation in t. Putting this into the usual form for aquadratic equation, i.e. ax2 + bx+ c = 0, gives:

1

2at2 + ut− s = 0

so:

a =1

2a, b = u, c = −s

Which can be substituted into the solution for a quadratic equation to give thedesired solution:

x =−b±

√b2 − 4ac

2a=⇒ t =

−u±√u2 + 2as

a

The method for equation 3.2 is very similar to equation 3.1 but with some minorchanges, namely u→ v and 1

2a→ −12a, so the solution changed slightly to

t =−v ±

√v2 − 2as

−2a

13


Equations 3.3 and 3.5 are much simpler; starting with 3.3:

v = u+ at (−v)

v − u = at (÷a)v − ua

= t

And �nally equation 3.5:

s =u+ v

2t (×2)

2s = (u+ v)t (÷(u+ v))

2s

u+ v= t

Q4 In a drag race, two cars accelerate uniformly from rest along a straighttrack with a length of 400m. Car A crosses the �nish line in 7.34 seconds, andCar B crosses the �nish line with a speed of 350kph.

Q4a Which car won the race?

A4a We already know that car A �nished the race in 7.34s, so to �nd out ifit won, we need to determine the time that car B took to �nish the race. Weknow three of the suvat variables for car B:

s = 400m, u = 0ms−1, v = 350kmph = 97.2 . . .ms−1

Note that the �nal speed has been converted from kmph into the SI units ofms−1. The suvat equation to �nd t from these quantities is:

s =v + u

2t

Which when rearranged to make t the subject becomes:

t =2s

v + u

Substituting in the values from above gives:

t = 8.23s

Which is greater than 7.34s, so car A won the race.

Q4b How far from the end of the race was the looser when the winner crossedthe �nish line?

14


A4b To �nd the answer to this question, we must �nd the distance that carB travels in the time it takes car A to �nish. The di�erence between this andthe course distance, i.e. 400m, gives the remaining distance to the �nish line.To �nd this we only have the initial speed u = 0 and the time t = 7.34s. The�nal velocity v = 350kmph does not apply as that is for the complete 400minstead of this partial distance. With just these two pieces of information, wedo not know enough to determine the distance that car B travels in this time.We therefore must �rst determine a third variable to �nd the distance travelled.The acceleration rather than the �nal velocity will be easier to �nd. To �nd theacceleration we know the following:

u = 0ms−1, v = 97.2 . . .ms−1, s = 400m, t = 8.23 . . . s

The easiest of the suvat equations to �nd acceleration from these is:

v = u+ at

Which solved for a becomes:

a =v − ut

To give:a = 11.8 . . .ms−2

Now we know the acceleration, the distance can be found using the relevantequation containing s, u, a and t, which is:

s = ut+1

2at2

In this form this does not require rearranging to �nd s, and due to the value ofu simpli�es to:

s =1

2at2

By substituting in the appropriate values, taking care to use the correct valuefor t, we reach a value of:

s = 318. . . .m

So taking the di�erence between this and the track length, car B was behindcar A by 81.7m at the end of the race.

Q4c Car C is released from a height of 400m above the �nish line; does itbeat the other cars if it travels in free-fall?

A4c.1 The simple answer to this question is to observe that the acceleration infree-fall is 9.81ms−2. This is less than the acceleration of car B, the slowest car,so the dropped car must take longer to complete the same distance. Thereforethe dropped car �nishes last.

15


A4c.2 For the more rigorous approach, we should �nd the time it takes thecar to cover the distance. We know the following values:

s = 400m, u = 0ms−1, a = 9.81ms−2

Taking the downwards direction to be positive for ease. The appropriate equa-tion to �nd the time t from these values is:

s = ut+1

2at2

Which when rearranged for t becomes:

t =−u±

√u2 + 2as

a

We can recognise than because u = 0ms−1 the original equation simpli�es to:

s =1

2at2

Rearranging this simpli�ed version, or simplifying the quadratic solution leadsus to:

t = ±√

2s

a

For a value of t once substituted of:

t = 9.03s

Which is greater than the time for either car A or B, so the dropped car �crosses�the �nish line last.

3.1.3 Motion Graphs

Figure 3.1:

0

2

4

6

8

10

12

0 1 2 3 4 5

Vec

ocity

v/m

s-1

Time t/s

AB

Q1 Figure 3.1 shows a velocity-time graph for two objects, A and B whichmove with a constant acceleration but begin their motion at di�erent times.

16


Q1a What is the Velocity of A at time t = 2s?

A1a Reading o� the graph on the line for A at 2s gives a velocity of 6ms−1.

Q1b What is the acceleration of B?

A1b Acceleration a on a velocity-time graph is given by the gradient.

gradient =∆y

∆x=

∆v

∆t= a

B has reached 6ms−1 at 1s after starting its motion, and 12ms−1 after 2s.Using either of these �gures to substitute into the above equation produces avalue for a of:

a = 6ms−2

Q1c What distance does A cover by t = 4s?

A1c On a velocity-time graph, the distance s is given by the area beneath thegraph. On a straight line graph such as this, the area can be calculated using:

area =1

2∆y∆x =

1

2∆v∆t = s

I.e. the area of a triangle. After 4s, object A reaches a velocity of 12ms−1 soafter substitutions, the distance travelled is:

s = 24m

Q1d At what time will the velocities of A and B be equal?

A1d.1 Having calculated the acceleration of B we can produce a formula forhow its velocity vB changes with time t based on the suvat equation:

vB = uB + aBt

However uB = 0ms−1 and B only begins it's motion after 3s have passed.Because of the time o�set of B, we substitute t → t − 3. After these twoadjustments, this becomes:

vB(t) = aB(t− 3)

For object A we �rst need to calculate the acceleration of A. As A reaches avelocity of 12ms−1, in 4s the acceleration is aA = 3ms−2. Putting this into asimilar equation to B, but without the time o�set, we reach:

vA(t) = aAt

For the two objects to be travelling at the same velocity, these two equationsmust be equal:

vA(t) = vB(t)

17


∴ aAt = aB(t− 3)

As we wish to �nd the time, this needs to be rearranged to make t the subject:

aAt = aB(t− 3) (expand the braket)

aAt = aBt− 3aB (−aBt)

aAt− aBt = −3aB (factorise t)

t(aA − aB) = −3aB (÷(aA − aB))

t =−3aBaA − aB

Which after substituting in the values calculated earlier for the accelerations ofthe objects, gives a time of:

t = 6s

A1d.2 Alternatively, sketch out the graph to extend the axis, and extrapolatethe two lines for A and B to see the time at which they intersect to reach thesame answer.

Figure 3.2:

-2

-1

0

1

2

3

0 2 4 6 8 10

Vel

ocity

v/m

s-1

Time t/s

Q2 Figure 3.2 shows the velocity-time graph for an object. Using the graph,estimate the following within the displayed range:

Q2a The maximum velocity of the object.

A2a The maximum velocity v on a velocity-time graph is given by the highestpoint on the graph. In this case that is at the top of the right-hand peak.Reading from the y-axis at this point gives an approximate value:

v ≈ 2.75ms−1

18


Q2b The acceleration of the object at its maximum velocity.

A2b The acceleration a is given by the slope of the graph at this point onthe graph. At the very peak of the velocity, the tangent to the curve is exactlyhorizontal. This therefore provides a value for the acceleration of:

a = 0ms−2

In fact, at the maximum or minimum of any curve, the gradient will be exactlyzero.

Q2c The time at which the object is furthest from it's starting position.

A2c Displacement s on a velocity-time graph is given by the area beneath thecurve. The maximum will at the time t where velocity �rst becomes negative.At this point the object begins to return towards the starting point, decreasingits displacement. This can also be perceived by the negative area between thecurve and the v = 0ms−1 line. Once this becomes negative, the total areabeneath the graph is decreasing. The point where the curve passes throughv = 0ms−1 can be read from the graph to give:

t ≈ 9.5s

Q2d The displacement of the object at the time identi�ed in (b).

A2d To determine the value of the displacement of the object, we must some-how measure the area beneath the curve up to the time identi�ed in part (c).Measuring the area could be done by splitting the area into shapes or rectanglesand summing the area of each shape. However as a grid of moderately sizedsquares is already positioned on the graph, an alternative method of countingsquares can be used. The method of counting squares counts any square whichis at least half within the curve. Counting up the number of squares n betweenthe curve and the v = 0ms−1 line by this method yields a value in the range of:

n ≈ 115

However, depending on who is counting, this �gure could vary in by 10% or so.Now we know the number of squares, we need to know the area, or equivalentlythe displacement, that each square represents Sn. The sides of each square are0.25ms−1× 0.5s so each square represents a displacement of 0.125m. The totaldisplacement after the 9.5s is therefore given by the product of the displacementrepresented by each square and the number of squares:

Snn = s ≈ 14.4m

Q2e The average velocity of the object to the time identi�ed in (b)

A2e We already know the time and displacement the object covers to thispoint, so the average velocity is simply given by the ratio:

v =s

t≈ 1.51ms−1

19


Q2f The acceleration of the object at time t = 6s.

A2f Acceleration is given by the gradient on a velocity-time graph.

gradient =∆y

∆x=

∆v

∆t= a

To measure the gradient we draw a tangent to the curve at the appropriatepoint. Once this is done we can measure from the tangent a value for ∆v and∆t. Doing so provides values such as:

∆v ≈ 4.25ms−1, ∆t ≈ 4s

To give a value for the acceleration of:

a ≈ 1.06ms−1

3.1.4 Forces

Q1 State Newtons three laws.

A1 The three fundamental laws of motion developed by Newton paved theway for developments in the �eld of classical mechanics.

Newtons First Law A body will remain at rest or move with a constantvelocity unless a non-zero net force is applied to it.

Newtons Second Law The acceleration a of a body has a linear rela-tionship with the net force ΣF applied to it, and an inverse relationship withthe mass m of the body. The body is accelerated in the same direction as theaction of the force.

a =ΣF

m=⇒ ΣF = ma

Newtons Third Law When one body exerts a force F1 upon another,the second body will exert a second force F2 upon the �rst that is equal andmagnitude and opposite in direction.

F1 = −F2

One of the implications from this law is that there will be no net force insidea closed system. An object or collection of objects cannot move their center ofmass without an external force.

Q2 A 1500kg car travels along a road propelled by its engine with a force of10kN . Air resistance applies a force of 7.5kN against the direction of motionof the car. What is the acceleration of the car?

20


A2 Newtons second law tells us that the acceleration a of the car can be foundby �nding the net of the forces acting upon the car and dividing that by themass m of the car. The net force F on the car can be found by combining theforce from the engine and the air resistance. As the air resistance acts againstthe direction of motion it takes a negative sign:

F = ΣFn = 10kN − 7.5kN = 2.5kN

Substituting this value into the equation provided by Newtons second law withthe value for the mass gives:

a =F

m= 1.67ms−2

Q3 20s after jumping from a plane, a 60kg skydiver has travelled 0.5km.

Q3a What is the average drag force acting on the diver in the �rst 20s of theirjump?

A3a To determine the nature of a particular force acting on a body, we mustknow the net force acting on the body and all other individual forces acting onit. I.e one force is given by the net force minus all other forces. In this case theonly force other than the drag force Fd acting on the diver is the weight forceFw. To determine the net force F acting on the body we must apply the suvatequations to the motion of the diver to calculate the average acceleration a. Forthis we know the following variables:

u = 0ms−1, s = −500m, t = 20s

Where the downwards direction is taken as negative. The appropriate equationto �nd a from this data is:

s = ut+1

2at2

Which when rearranged and simpli�ed due to the value of u becomes:

a =2s

t2

To give a value for a of:a = −2.5ms−2

So using the value for mass m given in the question, the net force on the diveris:

F = ma = −150N

Net force is given by:F = Fd + Fw

So rearranging for Fd and substituting Fw = mg gives:

Fd = F −mg

Substituting in appropriate values, taking care with signs gives a value of:

Fd = 439N

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Q3b What is the drag force acting on the sky diver at terminal velocity?

A3b At terminal velocity, the sky diver is travelling with a constant velocity.Newtons �rst law therefore tells us that the net force on the diver is zero.

F = Fd + Fw = 0

Which implies that Fd is equal in magnitude and opposite in direction to Fw sowe �nd the solution to be:

Fd = −Fw = −mg = 589N

Q4 A 1kg mass hangs at the end of a string. Air is blown over the stringapplying a horizontal force of 3N to the mass. What angle is the string at fromvertical once it settles to equilibrium?

A4.1 The condition of the mass being at equilibrium tells us that all theforces acting on the ball will be at balanced, i.e the net force on the ball willbe zero. There are three forces acting on the ball to balance: the weight forceFw, the given force from the air Fa, and the tension in the string Ft. Therewill be a single angle θ for the string at which these forces balance, and so bybalancing these forces we can �nd the solution. One way to balance these forcesis by resolving them into their horizontal and vertical components and balancingthese independently. Starting with the horizontal direction, the components are:

Fwh = 0, Fah = Fa, Fth = Ft sin θ

Note that as θ is the angle to the vertical, the horizontal component of thetension is given by using the sine of this angle. The tension and force from theair will act in opposite directions, so balancing them provides:

Fa = Ft sin θ

Repeating in the vertical direction:

Fwv = Fw = mg, Fav = 0, Ftv = Ft cos θ

Which balance to give:Fw = Ft cos θ

We now have a pair of equations each containing a di�erent known force, plusthe unknown tension and angle. As we are not interested in the tension inthe string, we can rearrange one of the equations to make Ft the subject, thensubstitute this into the other. Choosing the �rst equation to rearrange andsubstitute:

Ft =Fa

sin θ

=⇒ Fw =Fa cos θ

sin θ

Now we need to use a trigonometric identity to simplify this, knowing that:

sin θ =O

H, cos θ =

A

H, tan θ =

O

A

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Suggests that:sin θ

cos θ=O

H

H

A=O

A= tan θ

Which can be substituted into the equation we devised to get to:

Fw =Fa

tan θ

Which rearranged for θ becomes:

θ = arctanFa

Fw

Substituting in appropriate values gives:

θ = 17.0◦

A4.2 Alternatively, we can recognise that three balanced forces will form aclosed vector triangle. As Fw and Fa are perpendicular, this will be a rightangled triangle with Ft as the hypotenuse. From θ, Fa will be the opposite sideand Fw the adjacent. This allows us to state that for the angle from the vertical:

tan θ =O

A=Fa

Fw

Which rearranges in the same way as in the �rst method, to provide an identicalvalue of:

θ = arctanFa

Fw= 17.0◦

Q5 A 65Gg cruise ship travels at a constant velocity of 10ms−1 due westacross an ocean. The current applies an eastward force on the boat of 5.5MN .

Q5a If the engine of the ship were to fail, how long would the ship take tostop.

A5a The engine failing implies that the engine is no longer providing the forcerequired to maintain a constant velocity. The calculation to determine the timeit takes the ship to stop is a suvat calculation, for which we know an initialvelocity u = 10ms−1 and a �nal velocity v = 0ms−1 but not the acceleration.We now use Newtons second law to determine the acceleration of the ship fromits mass m and the net force on the ship. Without the engine running the netforce is just that from the current, so from:

F = −5.5MN, m = 65Gg

We reach:

a =F

m= −84.6 . . .mms−2

Now that a is know, we choose the appropriate suvat equation:

v = u+ at

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Rearranged and simpli�ed gives:

t =−ua

Which after appropriate substitutions gives an answer:

t = 118ks

Q5b If, in addition to the current, a wind blows south applying a force of3MN on the ship, what force must the ships engines supply to maintain it'soriginal velocity?

A5b This part of the question is very similar to question 4. Due to the simi-larity, we shall just work through the simpler solution. To maintain the originalvelocity, the forces acting on the ship must balance. As there are three forces,they will form a closed triangle in a vector diagram. We recognise that the forcefrom the wind Fw and the force from the current Fc are perpendicular, so theclosed vector triangle will be a right angled triangle with the force from theengines Fe as the hypotenuse. The magnitude of the force is therefore foundusing Pythagoras's theorem:

Fe =√F 2w + F 2

c

Which after substituting in appropriate values:

Fc = 5.5MN, Fw = 3MN

Gives a value of:Fe = 6.26MN

To �nd the direction of the force θ, we use trigonometry on the two knownforces. If we take Fw to be the side of the vector triangle opposite to θ, andFc to be adjacent, then θ will give us the direction of the vector as a directionnorth of west. The appropriate trigonometric function linking the opposite andadjacent sites is the tangent. Working through this provides a value of:

θ = arctanFw

Fc= 28.6◦

So the force required by the engines is:

Fe = 6.26MN at 28.6◦ north of west

Q5c At this speed, the ships engines can supply a maximum force of 6MN . Ifthis force is applied in the direction determined in (b), what is the accelerationof the ship?

A5c To determine the acceleration of the ship, we must �rst �nd the net forceacting on it. We shall do this by resolving each force into its north/south andeast/west components, and �nding the net force in each direction. This netforce in each direction can then be combined to give the magnitude and thedirection of the acceleration. Starting with the north/south direction, taking

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north to be positive, and taking the direction of the new maximum engine forceFm from the previous part of the question:

Fwn = −3MN, Fcn = 0MN Fmn = Fm sin θ = 2.87 . . .MN

Which gives the net northward force Fn as:

Fn = −0.127 . . .MN

In the east/west direction, taking west as positive:

Fwn = 0MN, Fcn = −5.5MN Fmn = Fm cos θ = 5.27 . . .MN

For a net westward force Fw of:

Fw = −0.233MN

The magnitude of the net acceleration is given by combining Fn and Fw throughPythagoras theorem and applying newtons second law:

a =

√Fn + Fw

m

Which after substituting appropriate values provides a value of:

a = 4.08µms−2

The direction θ is found using trigonometry. Treating Fw as the adjacent to θand Fn as opposite gives the direction in degrees south of east:

θ = arctanFn

Fw= 28.6◦

So the net acceleration of the ship is:

a = 4.08µms−2 at 28.6◦ south of east

Note that this is the opposite direction to that determined in the prior part ofthe question. This could have been taken without performing the calculation,but it is good to prove it by being thorough.

Figure 3.3:

θ

Q6 Figure 3.3 shows an object of mass 1kg suspended between walls by ahorizontal rod and a rope which is at an angle of 30◦ above the horizontal.What is the tension in the horizontal rod?

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A6.1 As with the earlier questions, this question of balanced forces can beapproached by balancing resolved forces, or using vector triangles. The threeforces to balance are the weight of the mass Fw, the tension in the rod or barFb and the tension in the rope Fr which is at an angle of θ above the horizontal.Finding the vertical components of each of these, taking upwards as positivegives:

Fwv = mg = −9.81N Fbv = 0N, Frv = Fr sin θ

Which balances to give:mg = Fr sin θ

In the horizontal direction, taking left to be positive:

Fwh = 0N Fbv = Fb, Frv = −Fr cos θ

Which balances to give:Fb = Fr cos θ

As θ is known, we can rearrange the equation from the vertical plane for Fr

and substitute that new equation into the equation for the horizontal plane toprovide an equation for Fb in terms of known quantities:

Fr =mg

sin θ

=⇒ Fb =mg cos θ

sin θSubstituting in appropriate values then provides a value of:

Fb = 17.0N

A6.2 After recognising that Fw and Fb are perpendicular, studying the rightangled vector triangle for this problem shows that relative to the known angleθ, Fr is the hypotenuse, Fb is the adjacent side and Fw is the opposite. As Fw

is known and Fb is desired, the appropriate trigonometric function is:

tan θ =O

A=Fw

Fb

Which rearranged for Fb gives:

Fb =Fw

tan θ

Substituting appropriate values gives a �gure which agrees with our earlier me-thod:

Fb = 17.0N


Q1 Identify and de�ne the acronyms K.E and G.P.E.

A1 K.E stands for Kinetic Energy. Kinetic energy is the energy stored in themotion of an object.G.P.E stands for Gravitational Potential Energy. Gravita-tional potential energy is the energy stored in an object that is associated withits position in a gravitational �eld.

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Q2 Show that the equations for calculating K.E and G.P.E are of the formEnergy = Force× distance

A2 The equation for Kinetic energy is:

K.E =1

2mv2 (3.6)

The suvat equations tell us:v2 = u2 + 2as

As we measure kinetic energy relative to a stationary object, u = 0ms−1 so theabove equation becomes:

v2 = 2as

Which substituted into equation 3.6 gives:

K.E =1

2m(2as) = mas

And as we know that F = ma, this equation is equivalent to a force multipliedby a distance.

The equation for Gravitational potential energy is:

G.P.E = mgh

g is the acceleration of an object in free fall, which multiplied by m gives thegravitational force acting on an object. As h is a distance, this equation is ofthe form of force multiplied by distance.

Q3 An object is dropped from 10m above the ground.

Q3a By considering its energy, calculate the speed of the object as it reachesthe ground.

A3a This question makes use of the conservation of energy. As the objectfalls, it's gravitational potential energy decreases, so the kinetic energy mustincrease. More quantitatively:

∆G.P.E = −∆K.E

Which after substituting in the relevant equations this becomes:

mgh = −1

2mv2

As we are after the �nal speed of the object, this equation must be rearrangedto make v the subject:

v = ±√−2gh (3.7)

Appropriate values for substitution are g = 9.81ms−2 and h = −10m. h isnegative as the height of the object in the gravitational �eld decreases. These�gures when substituted give a value for the speed of:

v1 = 14.0ms−1

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Q3b After colliding with the ground, the object bounces back upwards withhalf the speed it had before the impact. What is the maximum height the objectreaches?

A3b.1 There are two approaches to this question. The �rst is to recognisethat K.E is proportional to the square of the speed, and G.P.E is proportionalto the height. Therefore if the speed reduces by a factor of two, the K.E willreduce by a factor of four. Conservation of energy tells us that the �nal G.P.Eand therefore the �nal height will also decrease by a factor of four, so the �nalheight will be a quarter of the original height, i.e. 2.5m.

A3b.2 The alternative method uses the value v1 from previous part question.This can be done by rearranging equation 3.7 for h:

h =v2

2g

Which with a value for the speed of v12 gives a value for the height of 2.5m

Q4 A train is travels 1km due north on a �xed track. The wind blows southeast applying a force in this direction of 10kN to the train.

Q4a What work is done by the train over this distance?

A4a Work done w is given by the product of the force F and the distance stravelled in the direction of the force; or equivalently, the distance and the forcein the direction of travel. In this question, there is an angle θ of 45◦ betweenthe direction of travel, and the direction of the force. The relevant equation istherefore:

w = Fs cos θ

Which after substitution of the appropriate values given in the question providesa value of:

w = 7.07MJ

Q4b The train travels the 1km in 2 minutes. What is the power output ofthe train engine?

A4b Power is given by the time t taken to provide some amount of energy ordo some work w. In this case the work done is the 7.07 . . .MJ calculated inthe previous part of this question. Taking care with the conversion of 2 minutesinto SI units, we �nd the power to be:

P =w

t=

7.07 . . .× 106

120= 58.9kW

Q5 During a jump an 80kg sky diver reaches equilibrium at a speed of180kmph. What power is supplied to the sky diver by the earth to maintainthis speed?

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A5 Power is given by the rate of work, i.e. the amount of work done divided bythe time it took to perform that work. As work is given by a force multiplied bythe distance through which that force was applied, we can re-write the equationfor power as follows:

P =w

t=Fs

t= Fv

I.e. power is given by the product of the velocity of an object and the forceapplied to it. In this case we are given the velocity of the sky diver, and fromthe given mass can determine the weight force acting on the skydiver. Thisleads to the following equation:

P = Fv = mgv

Which with appropriate substitutions provides a value of:

P = 39.2kW

Q6 A 75kg athlete climbs a rope at 0.5ms−1; assuming the human body is20% e�cient, what is the power generation of the athletes body?

A6 Power is given by the product of force and velocity. In this case the climberacts against gravity, so:

P = Fv = mgv = 368. . . .W

This is the useful work power output of the climber, however we are told that theclimber is only 20% e�cient, so this �gure must be multiplied by 5 to determinethe raw power generated by the climber, which gives us a value of:

P = 1.84kW

Q7 At a fun-fair, a 2kg mallet is swung at a `test your strength' game,colliding with the target at a speed of 5ms−1. The 500g indicator hits the bellwhich is 2m above the ground with a speed of 0.5ms−1. What is the e�ciencyof the system that transfers energy from the mallet to the indicator?

A7 E�ciency ε is given by the ratio of the useful energy leaving a system tothe energy input to the system . In this case that corresponds to the ratio ofthe energy in the indicator Ei to that in the mallet Em.

ε =Ei

Em

In this example, we need only consider K.E and G.P.E. To determine the e�-ciency we can therefore calculate the total K.E and G.P.E of the indicator inratio to that of the mallet:

ε =K.Ei +G.P.Ei

K.Em +G.P.Em=

12miv

2i +mighi

12mmv2m +mmghm

Substituting in values as follows:

mi = 0.5kg, mm = 2kg, vi = 0.5ms−1, vm = 5ms−1, hi = 2m, hm = 0m

Gives a value for the e�ciency of:

ε = 0.395 = 39.5%

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3.1.6 Projectiles

Q1 A ball is thrown upwards from 1m above the ground at a speed of 3ms−1.How long after being released does the ball hit the ground

A1.1 As the ball moves only in the vertical plane, this is simple linear motionin one dimension, so the SUVAT equations can be used. We start by listing theknown variables:

s = −1m, u = 3ms−1, a = −9.81ms−2

AThe �nal displacement of the ball will be 1m below the starting position,

and as the ball moves in free-fall, the acceleration is that due to gravity. Theupwards direction is taken to be positive to give s and a a negative sign. Theequation containing these three variables, plus time is:

s = ut+1

2at2

This must be re-arranged to make t the subject as we wish to �nd the time of�ight. As the equation is quadratic in t, we can use the solution for a quadraticequation:

x =−b±

√b2 − 4ac

2a

Where we substitute:

x = t, a =1

2a, b = u, c = −s

To get:

t =−u±

√u2 + 2as

a

Substituting the values determined earlier with careful consideration of sign, weget an answer of:

t = −0.240s or t = 0.851s

The negative value of t corresponds to the time the ball would have been atground level had it been travelling freely. The positive value is the value ofinterest as this represents the time of �ight for the ball to travel upwards thenback to ground level, so:

t = 0.851s

A1.2 The second method for this question still uses a SUVAT equation,but avoids using quadratic equation by �nding the �nal velocity of the ball �rst.The �nal velocity can be found by two methods which turn out to be equivalent.Conservation of energy tells us that the G.P.E lost due to have descended 1mwill be converted into K.E.

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K,Ev = K.Eu +G.P.E

1

2mv2 =

1

2mu2 +mgh

v2 = 9u2 + 2gh

Alternatively, we can use the suvat equation

v2 = u2 + 2as

Which is equivalent to the energy solution with a = g = −9.81ms−2and s =h = −1m. To make v the subject of this formula, take the square root of bothsides to reach:

v = ±√u2 + 2as

By substituting in the values for u, aand s we determined earlier, we obtain avalue for v of:

v = ±5.35 . . .ms−1

We use common sense to choose the negative value as the ball will be travellingdownwards at the end of its path. As we now additionally have the �nal velocityto use in our calculations, we can choose a di�erent SUVAT equation to avoidthe use of a quadratic equation, and rearrange it to make t the subject of theformula:

v = u+ at

To give:

t =v − ua

When we substitute in the values for u and a in addition to the newly calculatedvalue of v we get a value for t of:

t = 0.851s

Which agrees with the value obtained through the �rst method.

Q2 A bullet is �red horizontally from eye level (1.5m above ground), leavingthe muzzle at 500ms−1.

Q2a What horizontal distance does the bullet travel before hitting the ground?

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A2a This is a 2 dimensional problem. To determine the distance travelled,we need the horizontal velocity of the bullet which we are given, plus the timeof travel which we are not given so must �rst calculate. The time of travel isthe the time it takes for the bullet to fall the 1.5m distance. The known suvatvariables are:

s = −1.5m, u = 0ms−1, a = −9.81ms−2

And the appropriate equation to �nd the time t is:

s = ut+1

2at2

Which simpli�es and rearranges to:

t =

√2s

a

To give:t = 0.553 . . . s

So the horizontal distance the bullet travels is:

s = ut = 500ms−1 × 0.553 . . . s = 277m

Q2b What is the velocity of the bullet when it hits the ground?

A2b This velocity calculation requires we �nd the magnitude v and directionθof the bullet. We have been given the horizontal component of the velocity vhbut must still calculate the vertical component of the velocity vv. To �nd thiswe know u, a, and s so the appropriate equation to �nd the �nal velocity is:

v2v = u2v + 2asv

Which once simpli�ed and rearranged becomes:

vv = ±√

2as

To give a value of:vv = ±5.42 . . .ms−1

Using Pythagoras this is combined with vhto give v as:

v =√v2h + v2v = 500ms−1

Note how the square dependence results in the vertical velocity having a negli-gible e�ect on the total velocity magnitude. For the direction, we use trigono-metry to determine the angle below the horizontal of the bullet using the inversetangent function:

θ = arctanvvvh

= 0.622◦

So the velocity of the bullet as it hits the ground is:

500ms−1 at 0.622◦ below the horizontal

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Q3 An archer launches an arrow from ground level at 50ms−1 at an angle of30◦ above the horizontal.

Q3a How far away does the arrow land?

A3a This is very similar to Q2 in that to �nd the distance traveled we must�rst �nd the time of �ight. However, this time the projectile is not launchedhorizontally, but at an angle θ. Therefore we must �rst resolve the initial velocityvector u into it's horizontal uh and vertical uv components, and then proceedwith the calculation as before.

uh = u cos θ = 43.3 . . .ms−1, uv = u sin θ = 25ms−1

Which in the vertical dimension provides the following values for suvat variables:

s = 0m, a = −9.81ms−2, u = 25ms−1

The appropriate equation is, as in Q2:

s = ut+1

2at2

However as u is non-zero, solving this equation requires �nding the solution toa quadratic. To avoid this we can use the symmetry of the problem to recognisethat the �nal vertical velocity vv will be −25ms−1. By recognising this, we canuse an equation which is simpler to solve for t such as:

v = u+ at, or s =v + u

2t

The later breaks due to the zero value of the displacement, so the former shallbe used, which solved for t becomes:

t =v − ua

To give a value for t of:t = 5.10 . . . s

So the distance the arrow travels is:

s = uht = 221m

Q3b Will the arrow be able to pass over a wall 5m high 40m away?

A3b The solution to this problem requires that we �nd the height of thearrow at a distance of sh = 40m. To do this we use the horizontal dimension tocalculate the travel time t to this point, then move into the vertical dimensionto determine the height s of the arrow at this time. From the �rst part of thisquestion, we already know the vertical and horizontal components of the initialvelocity:

uh = 43.3 . . .ms−1, uv = 25ms−1

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So the time to travel 40m horizontally is:

t =shuh

= 0.924 . . . s

To calculate the height of the arrow at this time, we know he following variables:

uv = 25ms−1, a = −9.81ms−2, t = 0.924 . . . s

The appropriate equation for �nding the height of the arrow using these variablesis:

s = ut+1

2at2

Which in this form does not need to be rearranged, and provides a value for sof:

s = 18.9m

So the arrow will be able to clear a wall of this height at this distance.

Q4 A space probe lands on an extra-terrestrial planet. In an experimentto determine the strength of gravity on the planet, the probe launches a ballbearing at 40◦ above the horizontal at a speed of 5ms−1. The bearing lands50m away. What is the acceleration due to gravity on the planet?

A4 Acceleration a due to gravity acts in the vertical direction. After resolvingthe vertical component uv of the initial velocity vector u we know the followinginformation in the vertical direction:

sv = 0m, uv = 3.21 . . .ms−1

But we need a third variable, either time t or the �nal velocity vv to be ableto determine the acceleration. Of these two, the easiest to calculate would bethe time of �ight from the horizontal component of the initial velocity uh andthe distance s traveled as there will be no horizontal acceleration.Choosing theappropriate formula gives us:

t =shuh

=50m

5ms−1 × cos 40◦= 13.1 . . . s

Which can now be used with our values of sv and uv to �nd a using the appro-priate equation:

sv = uvt+1

2at2

Which when rearranged and simpli�ed gives:

a =−2uvt

= −0.492ms−2

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