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Worksheet

Worked examplesPractical 1: Using the triangle of forces to determine the

weight of an object (hammer)

Practical 2: Projectile motion

End-of-chapter testMarking scheme: Worksheet

Marking scheme: End-of-chapter test

Chapter 4

Working with vectors

OCR (A) specifications: 5.1.1b,c,d,e; 5.1.2e

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Higher level5 A child of mass 35 kg on a swing is pulled to one side.

The diagram shows the forces acting on the seat of

the swing when it is in equilibrium.

a What is the net force on the seat? [1]

b Draw a vector triangle of forces. Hence determine:

i the tension Tin the rope; [4]

ii the angle made by the rope with the

vertical. [2]

32 Cambridge University Press 2005 4 Working with vectors

Worksheetacceleration of free fallg= 9.81ms2

Intermediate level1 A small aeroplane travels 30km due north and then 40km due east.

a Draw a vector triangle for the final displacement. [2]

b Determine the magnitude of the final displacement. [2]

2 Calculate the magnitude of the resultant force in each case below.

a b c

[2] [2] [3]

3 The diagram shows a swimmer attemptingto swim across a river.

The swimmer swims at a velocity of

2.5ms1 normal to the riverbank and the

velocity of the river water is 3.0ms1

parallel to the riverbank. Calculate:

a the magnitude of the actual velocity

of the swimmer; [3]b the direction of the final velocity relative to the riverbank. [2]

4 In each case below, resolve the vector into two perpendicular components inthe x andy directions.

a b c

5.0N

7.0N

20N

10N

10N

20N

40N

80N

bank

river

3.0

2.5ms1ms1

10N

45

x

y

85N

20x

y

40N

70

x

y

T

180N

weight

[2] [2] [2]

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4 Working with vectors Cambridge University Press 2005 33

6 A gardener pulls a 50kg roller along levelground, as shown in the diagram.

The roller moves at a steady speed along the

level ground when the handle makes an

angle of 30 to the horizontal ground and

the gardener pulls with a force of 300N

along the handle.

a Calculate the horizontal component

of the force 300N. [2]

b What is the net force in the horizontal direction? Hence determine the

magnitude of the resistive force acting on the roller. [2]

c Determine the vertical contact force acting on the roller due to the ground. [3]

7 A marble is flicked off the edge of a platform. The marble has a horizontalvelocity of 2.5ms1. The marble hits the ground after travelling a vertical

distance of 2.0m. You may assume that air resistance has a negligible effect

on the motion of the marble.

a How long does it take for the marble to travel from the edge of the platform

to the ground? [2]

b Determine the range of the marble the horizontal distance travelled by

the marble before it hits the ground. [2]

8 A stone is thrown horizontally at a velocity of 15m s1 from a 120m tall tower.You may assume that air resistance has a negligible effect on the motion of the

stone. Calculate:

a how long it remains in flight; [2]

b the horizontal distance travelled; [2]

c the magnitude of its impact velocity. [4]

Extension9 The diagram shows a stunt person of mass

82kg holding on to a rope.

The rope on either side of the person makes

an angle of 5.0 to the horizontal

a Determine the tension Tin the rope. [3]

b What would be the consequence of

making the angle between the rope

and the horizontal equal to zero? [2]

10 The trajectory of a water-jet from a garden hose is as shown in the diagram.

30

300N

T T

5.05.0

82 kg

You may assume that air resistance has a negligible effect on the motion of

the water-jet. Use the information provided above to determine the speed V

of the water emerging from the pipe and the rangeR. [6]

2.5 m

30

V

R

water-jet

garden hose

Total: 59

Score: %

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Example 2The diagram shows a pencil rubber resting on an

inclined plastic ruler. The rubber remains in a state of

equilibrium when the ruler is tilted to an angle of 62

to the horizontal. The weight of the rubber is 0.50N.

Determine the magnitude of the frictionFacting on

the rubber.

friction force = component of weight down the ruler

F = Wsin 62

F= 0.50 sin 62 0.44N

rubber

ruler

62W(0.50 N)

F

contactforce R

The netforce is zero.

Tip

The rubber is in equilibrium. We can therefore draw a triangle of forces torepresent the three forces acting on the rubber.

From the triangle, we have:

Fsin 62 =

0.50

F = sin 62 0.50 0.44N

(You will get the same answer.)

F

0.50 N

62R

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Practical 1

Using the triangle of forces to determine the weight ofan object (hammer)Safety

There are not likely to be any major hazards in carrying out this experiment. However,

teachers and technicians should always refer to the departmental risk assessment before

carrying out any practical work.

Apparatus

clamp stands hammer

G clamps string

two newtonmeters protractor

Introduction

In this experiment you will determine the weight of an object (a hammer in our case) by

constructing a triangle of forces. When three forces acting at a point are in equilibrium,

we can construct a triangle of forces to represent the forces. This is described on page 40

ofPhysics 1.

Procedure

1 Cut three 0.60m lengths of string andtie them together at one end.

2 Tie the other ends of two of the stringsto newtonmeters and the end of the

third string to the hammer, as shown in

the diagram.

3 By either raising or lowering thenewtonmeters or by moving the clamp

stands, adjust the angle between the

vertical and horizontal strings to be 90.

4 Measure the angle with a protractor.

5 Measure the tensions TandFfrom thenewtonmeters.

6 Repeat the experiment for different angles . Record your results in a table.

7 For each experiment, construct a triangle of forces. Determine the weight Wof thehammer. (This may either be done by a scale drawing or using trigonometry.)

8 What is the average value for the weight Wof the hammer from your experiment?How does your value compare with the actual weight of the hammer when

measured using a newtonmeter?

Guidance for teachersInstead of using newtonmeters, you can use known weights to represent the forces T

andF. The weights can be hung over two smooth pulleys.

W

T

F

F

T

W hammer

triangle offorces

T(N)

F(N)

(degrees)

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Practical 2Projectile motionSafety

There are not likely to be any major hazards in carrying out this experiment, but it is abit messy! However, teachers and technicians should always refer to the departmental

risk assessment before carrying out any practical work.

Apparatus

A3 paper taped onto a

wooden board (or drawing board)

clamp stand

marble

Introduction

In this experiment you have the opportunity to investigate the projectile motion of

a marble.

Procedure

This is a simulation of projectile motion. The wooden board is tilted slightly. The

acceleration of the marble down the wooden board will be a component of the

acceleration of free fallg.

marble

wooden board wooden board

ramp

marble

paper

x

y

1 Raise one end of the wooden board and tilt it to an angle of about 10.

2 Dip the marble in ink.

3 Place the marble on the ramp and release it so that it travels parallel to the topedge of the paper. The marble will leave a trace of its path on the paper.

4 The path of the marble should be parabolic. When you double the horizontal

distance x, then the vertical dropy should increase by a factor of 4. Is this true?

5 Try releasing the marble at different initial speeds. The rangeR of the marbleshould increase when the initial speed of the marble is increased.

6 Increase the angle of tilt of the wooden board. This is equivalent to increasingthe vertical acceleration of the marble. How does this affect the path described

by the marble?

Guidance for teachers

An excellent alternative to the arrangement above would be to use a stroboscope and a

camera to record the motion of a metal ball bearing. The advantage of this is that:

it clearly shows accelerated motion in the vertical (y) direction;

you can determine the speed of the ball at different points in its path.

If a stroboscope is used, then you must check if anyone in the class suffers from epilepsy.

non-permanent ink

disposable gloves

ramp

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End-of-chapter testAnswer all questions.

acceleration due to gravityg= 9.81ms2

1 The diagram shows an aeroplane flying at a constantvelocity of 80kmh1 in a north-westerly direction.

The aeroplane is in flight for 1.5 hours. For this

aeroplane, determine in km:

a its total displacement; [1]

b its displacement in the northerly direction. [2]

2 The diagram below shows a stage light of mass 5.2 kg attached by two supportingcablesAand B.

The stage light is in equilibrium.

a What is the net force on the stage light? [1]

b Calculate the vertical component of the tension T1 in cableA. [2]

c Use your answer tob to determine the tension T1. [2]

d Calculate the tension T2 in cable B. [2]

3 The diagram shows forces acting on a ball moving horizontally through the air.

a Draw a vector diagram to show the resultant forceF

acting on the ball. [2]

b The ball has a mass of 330g. Use the vector diagram from a

to determine:

i the magnitude of the total forceFacting on the ball; [2]

ii the magnitude of the acceleration of the ball. [2]

4 A bullet is shot from a rifle at a horizontal velocity of 150ms1. It hits a targetat a horizontal distance of 30 m. You may assume that the effect of air resistance

is negligible.

a Determine the flight time of the bullet. [1]

b Explain why the impact speed of the bullet is not 150ms1. [1]

c Determine the vertical displacement of the bullet. [2]

N

45

80kmh1

ceiling

wall

28A

B

T1

T2

weight

drag

1.2N

weight

3.2N

Total: 20

Score: %

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3.0

v

2.5

ms1

ms1


Marking schemeWorksheet1 a Correct diagram [1]

Correct arrows on the vectors [1]

b s2 = 302 + 402 [1]; s = (302 + 402) = 50km [1]

2 a R2 = 7.02 + 5.02 [1]; R = 49 + 25 = 8.6N [1]

b R2 = 102 + 202 [1]; R = 100 + 400 = 22.4 22 N [1]

c Force vertically = 40 10 = 30N and force horizontally = 80 20 = 60N [1]

R2 = 302 + 602 [1]; R = 900 + 3600 67N [1]

3 a

[1]

b tan

=

2.5

= 0.833 [1]3.0

= tan1 (0.833) = 39.8 40 [1]

4 a Fx =Fcos = 10 cos 45,Fx = 7.07 7.1 N [1]

Fy =Fsin = 10 sin 45,Fy = 7.07 7.1 N [1]

b Fx =Fcos = 85 cos 20,Fx = 79.9 80 N [1]

Fy =Fsin = 85 sin 20,Fy = 29.1 29N [1]

c Fx =Fcos = 40 cos (90 70),Fx = 37.6 38 N [1]

Fy =Fsin = 40 sin (90 70),Fy = 13.7 14 N [1]

5 a The net force is zero because the seat is in equilibrium. [1]b i

[1]

Weight = mg= 35 9.81 343N [1]

T2 = 1802 + 3432 [1]

T= 1802 + 3432 390N [1]

ii tan =180

= 0.525 [1]; = tan1 (0.525) 28 [1]343

N 40 km

30 kms

180N

mg

T

v2 = 2.52 + 3.02 [1]

v2 = 6.25 + 9.00 3.9m s1 [1]

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6 a Fx =Fcos = 300 cos 30 [1]; Fx 260 N [1]

b The net force is zero, because the roller is moving at constant velocity. [1]

Resistive force 260N to the left. [1]

c Fy =Fsin = 300 sin 30 = 150N [1]

The net vertical force is zero.

150 + contact force = mg[1]

contact force = (50 9.81) 150 340N [1]

7 a Vertically

s = 2.0m u = 0 a = 9.81ms2 t= ?

s = ut+1

at2 (u = 0) [1]; t=2s

=2 2.0

0.64s [1]2 a z9.81

b Horizontally

u = 2.5ms1 t= 0.64 s (from a)

range = ut[1]; range = 2.5 0.64 = 1.6m [1]

8 a Vertically

s = 120m u = 0 a = 9.81ms2 t= ?

s = ut+1

at2 u = 0 [1]2

t=2s

=2 120

= 4.946s 4.9s [1]

a 9.81b Horizontally

u = 15ms1 t= 4.95s (from a and retaining 3 s.f.)

distance = ut[1]; distance = 15 4.95 74m [1]

c Vertically

u = 0 s = 120m a = 9.81ms2 v = ?

v2 = u2 + 2as [1]

v = 2 9.81 120 48.5ms1 [1]

Final velocity = V (see diagram)

V2 = 48.52 + 152 [1]; V 51m s1 [1]

9 a The net vertical force = 0

Tsin 5.0 + Tsin 5.0 = mg[1]

2Tsin 5.0 = 82 9.81 [1]

T=82 9.81

4.6 103 N (4.6kN) [1]2 sin 5.0

b The tension Tin the rope is given by: T=82 9.81

[1]2 sin

As the angle becomes equal to zero, the tension Tbecomes infinite. The rope

will snap. [1]

150N

260N

mg

contact force

componentsof 300 N force

48.5

15

Vms1

ms1

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10 Vertically for motion to the highest point

u = Vsin 30 s = 2.5m a = 9.81 m s2 v = 0 [1]

v2 = u2 + 2as

0 = (Vsin 30)2 (2 9.81 2.5) [1]

(0.5V)2 = 2 9.81 2.5

V=2 9.81 2.5

14m s1 [1]0.52

Vertically for motion back down to the ground

u = 14 sin 30 = 7.0m s1 v = 7.0m s1 a = 9.81ms2 t= ?

v = u + at where t= total time of flight

t=7.0 7.0

= 1.43s [1]9.81

Horizontally

range = (Vcos 30)t= 14 cos 30 1.43 [1]; range = 17.3m 17m [1]

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Marking schemeEnd-of-chapter test1 a Displacement = 80 1.5 = 120km [1]

b Displacement = s cos

, displacement = 120

cos 45 [1]displacement 85km [1]

2 a The net force is zero because the stage light is in equilibrium. [1]

b Vertical component of T1 = weight [1]

vertical component of T1 = mg= 5.2 9.81 51N [1]

c T1cos 28 = 51 [1]

51T1 = cos 28

= 57.8 N 58N [1]

d T2 = T1 sin 28 (the net horizontal force = 0) [1]

T2 = 57.8 sin 28 27 N [1]

(You can also calculate T1 and T2by drawing the triangle of forces.)

3 a Correct labels shown on the diagram [1]

Correct direction for the resultant [1]

b i F2 = 3.22 + 1.22 [1]

F= 3.22 + 1.22 = 3.42 N 3.4N [1]

F 3.42ii a =

m=

0.330[1]; a 10.4m s2 [1]

4 a30

Time =150

= 0.20s [1]

b Apart from the horizontal component 150ms1 the bullet also has a vertical

component because of the acceleration due to gravity. The resultant velocity is

greater than 150ms1. [1]

c Vertically

s = ? u = 0 a = 9.81ms2 t= 0.20s (from a)

s = ut + 1 at2 (u = 0) [1]2

s =1 9.81 0.202 = 0.196 0.20m [1]

2

1.2N

3.2NF

mg= 51NT

1

T2

28

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