Heinemann Physics 11 3rd Edition Worked Solutions Chapter 5

Heinemann Physics 11 (3e) Copyright © Pearson Education Australia 2008 Teacher’s Resource and Assessment Disk (a division of Pearson Australia Group Pty Ltd) ISBN 9781442501249 1

Worked solutions

Chapter 5 Newton’s laws

5.1 Force as a vector

1 a Velocity and force are vectors so B, D. b i 150 N up

ii 40 N west

iii zero

iv ∑F = √(102 + 102) = 14 N NW

2 a F ≈ 10 N

b F ≈ 10 N

c Assuming the mass is 5 to 10 kg, F ≈ 50–100 N

d Assuming your mass is 40 to 80 kg, F ≈ 400– 800 N

Worked solutions Chapter 5 Newton’s laws


3 Pulling on an anchored rope: F = 1.0 kN Tug-of-war team: F = 10 × 1.0 kN = 1.0 × 104 N

4 The directions are identical in alternatives B, C and D. True north is taken as 0°T. All angles greater than 0°T are taken in the clockwise direction from north. Then 40°T would correspond to N40°E.

5 True north is taken as 0°T. All angles greater than 0°T are taken in the clockwise direction from north. a N60°E = 60°T b N40°W = 320°T c S60°W = 240°T d SE = 135°T e NNE = 22.5°T

6 a Length of vector ≈3 cm, 1 cm ≈20 N; magnitude of vector = 3 × 30 = 60 N i.e. 60 N right

b 50 N up (vector ≈2.5 cm)

c 70 N down (vector ≈3.5 cm)

7 a 1 N west b

Magnitude of resultant force = √(602 + 802) = 100 N tan θ = 80/60 and θ = 53° F = 100 N 143°T

8

∑F = 2 × 400 cos 40° (or use sine rule) = 610 N in a direction that bisects the two ropes

9 a FS = 100 cos 60° = 50 N south FE = 100 cos 30° = 87 N east b FN = 60 N north c FS = 300 cos 20° = 280 N south FE = 300 cos 70° = 103 N east



d Fv = 3.0 × 105 cos 60° = 1.5 × 105 N up

Fh = 3.0 × 105 cos 30° = 2.6 × 105 N horizontal

10 Horizontal component Fh = 300 cos 60° = 150 N Vertical component Fv = 300 sin 60° = 260 N



5.2 Newton’s first law of motion

1 Aristotle felt that the natural state for any object was at rest in its natural place. This meant that any moving body would come to rest of its own accord. Galileo introduced the idea that friction was a force that could be added to other forces that act on a moving body, but it was Newton who explained that the moving object should continue to travel with constant velocity unless a net force is acting.

2 a According to Aristotle, a constant force is needed to maintain a constant velocity so A is correct.

b According to Newton, an object moving with a constant velocity has zero net force acting on it. In this situation there are no driving forces or retarding forces, so D is correct.

c Newton is correct, so D.

3 No horizontal force acts on the person. In accordance with Newton’s first law of motion, the bus slows, and the standing passenger will continue to move with constant velocity unless acted on by an unbalanced force; usually the passenger will lose his or her footing and fall forward.

4 Constant speed, so ∑F = 0, then frictional force = applied force = 20 N.

5 Constant speed, so ∑F = 0 in both vertical and horizontal directions. To exactly balance the other forces, lift = 50 kN up, and drag = 12 kN west.

6 a Constant speed, so ∑F = 0, then frictional force = applied force = 25 N.

b 25 N c F cos 30° = 25 N

F = 29 N at an angle of 30° to the horizontal.

7 When the car or aircraft slows suddenly, a passenger will continue to travel with the same velocity as before, until being acted on by an unbalanced force. The purpose of the seat belt is to supply this force, but evenly spread across the body to reduce the effects of the force. Without the seatbelt, the stopping force may be provided by the windscreen or steering wheel.

8 a gravitational force of attraction between the two masses b electrical force of attraction between the negative electron and the positive nucleus c friction between the tyres and the road d tension in the wire

9 a If the cloth is pulled quickly, the force on the glass acts for a short time only. This force does not overcome the tendency of the glass to stay where they is, i.e. its inertia.

b Using a full glass makes the trick easier because the force will have less effect on the glass due to its greater mass. The inertia of the full glass is greater than that of an empty glass.

10 The fully laden semitrailer will have most difficulty pulling up. Its large mass means that more force is required to bring it to a stop.



5.3 Newton’s second law of motion

1 a The netball is accelerating, so forces are unbalanced: U. b Bus is stationary with acceleration of zero, so forces are balanced: B. c Constant speed means acceleration of zero, so forces are balanced: B d Cyclist is accelerating, so forces are unbalanced: U.

2 a a = Δv/Δt = +30/0.0070 = 4300 = 4.3 × 103m s–2

b ΣF = ma = 0.060 × 4300 = 260 N

3 The 1.5 kg shot-put is the larger of the two masses, and so for the same applied force, its acceleration will be lower. This means a lower speed on leaving the athlete’s arm, and so it cannot be thrown as far as the lighter 1.0 kg ball.

4 a ΣF = (1002 + 1252)1/2 = 160 N

b tan θ = 125/100 = 1.25 θ = 51° south of east = 141°T

c a = ΣF/m = 160/0.75 = 210 m s–2 at 141°T

5 Net force acting on the car = 900 × 2.0 = 1800 N ΣF = force applied by motor – 800 N Force applied by motor = 2.6 × 103 N in the direction of travel

6 a 50 kg b Fg = mg = 50 × 9.8 = 490 N

c ΣFh = 45 N south + 25 N north = 20 N south

d ΣF = ma 20 = 65a a = 0.31 m s–2

7 The mass of the hammer remains constant at 1.5 kg. The weight of the hammer on Mars is Fg = mg = 1.5 × 3.6 = 5.4 N.

8 60 km h–1 = 60/3.6 = 16.7 m s–1 v2 = u2 + 2ax 0 = 16.72 + 2 × a × 150 a = –0.93 m s–2 ΣF = FB = ma = 1200 × –0.93 = 1.1 × 103 N opposing the motion



9

As the parachutist leaves the aircraft, his or her weight will be the net force acting, accelerating at 9.8 m s–2. But as the speed increases, the drag force (air resistance) opposing the motion also increases until it equals the weight. At this point in time, the net force will be zero and the parachutist will travel with a constant speed.

10 a ΣF = Fg = mg = 0.50 × 9.8 = 4.9 N

a = )5.05.2(

9.4

+=

!

!

m

F = 1.6 m s–2

b v = u + at = 0 + 1.6 × 0.5 = 0.8 m s–1

c ΣF = ma = Fg – Ff so ΣF = 4.9 – 4.3 = 0.6 N

and 20.03

6.0==

!=m

Fa m s–2



5.4 Newton’s third law of motion

1 a Action: force from the racquet on to the ball Reaction: ball exerts a force on the racquet in the opposite direction

b Action: weight of pine cone, i.e. gravity pulling it to the Earth Reaction: Earth pulled toward the pine cone with the same force

c Action: downwards force that the pine cone exerts on the ground. Reaction: upwards force that ground exerts on the pine cone.

d Action: force of attraction (gravity) that the Sun exerts on the Earth. Reaction: force of attraction (gravity) that the Earth exerts on the Sun.

2 a The boat exerts an equal and opposite reaction force, ie 140 N in the opposite direction to the leaping fisherman.

b 5.340

140==

!=m

Fa m s–2 in the opposite direction to the fisherman

c Acceleration of the fisherman: 0.270

140==

!=m

Fa m s–2

Speed of the fisherman: v = 0 + 2 × 0.5 = 1 m s–1 Speed of the boat: v = 0 + 3.5 × 0.5 = 1.8 m s–1

3 a If no other forces act, there will be an action–reaction force pair in which the action force is the force on the tool kit, and the reaction force will act on the astronaut in the opposite direction. If the kit is thrown directly away from the ship, hopefully the reaction force will propel her back to the craft.

b Calculate force that acts on tool kit: ΣFT = mT × aT =2.5 × 8.0 = 20 N . An equal 20 N force will also act on the astronaut.

c Acceleration of astronaut: a = ∑F/m = 20/100 = 0.20 m s–2 v = u + at = 0 + 0.20 × 0.50 = 0.10 m s–1

d The astronaut will travel with a constant speed of 0.10 m s–1 so: v = d/t t = d/v = 10/0.10 = 100 s or 1 min 40 s

4 Force exerted on jack by bowl = mJaJ = 1.0 × 25 = 25 N north Therefore force exerted on bowl by jack = 25 N south Acceleration of bowl a = ΣFB/mB =25/2.0 = 12.5 m s–2 south

5 a The speed is increasing but acceleration is constant along a uniform incline, so B is correct.

b The speed is increasing. The angle of inclination becomes smaller as the ball travels down the ramp, so the acceleration will also become smaller (a = g sin θ). C is correct.

6 a Weight vertically down, normal force up and perpendicular to the incline, slight friction up the incline.

b Weight Fg = mg = 65 × 9.8 = 640 N downwards Normal reaction force, FN = Fg

cos 50° = 410 N



c Net force = mg sin θ = 640 sin 50° = 490 N down the incline

d Acceleration = 490/65 = 7.5 m s–2 down the incline

7 The braking force must balance the component of the gravitational force that is acting down the incline. Braking force FB = mg sin θ = 110 × 9.8 sin 15° = 280 N

8 The normal force will be smallest at A and will become progressively larger as he travels through point B and C.

9 Tania is correct. While the forces are equal in size but opposite in direction, they are both acting on the lunchbox and so cannot be an action/reaction pair in the sense of Newton’s third law.

10 a Net force on rope = (50 × 9.8) – (30 × 9.8) = 196 N down on the side of the student b Acceleration = ΣF/Σm = 196/80 = 2.45 m s–2 down on the side of the student

c Consider the student only: ΣFs = ms × as = 50 × 2.45 = 50 × 9.8 – FT FT = 370 N



Chapter review

1 C, as described by Newton’s third law.

2 Frictional force = 30 N ΣF = ma = 35 × 0.50 = 17.5 N ΣF = pushing force – 30 N Pushing force = 17.5 + 30 = 48 N

3 a = ΣF/m = 25/15 = 1.67 m s–2 v = u + at = 0 + 1.67 × 4.0 = 6.7 m s–1

4 Three examples are: • Kicking the tyre of a car hurts because you apply a force to the tyre (action) and the

tyre will apply a reaction force to your foot (reaction). • Hot gases are forced out of a jet engine (action) and the gases push the engine

forward (reaction). • The weight of any object can be considered an action force—the Earth pulls on the

body. The reaction force acts on the Earth—the object pulls on the Earth.

5 a v = u + at = 0 + 2.0 × 2.5 = 5.0 m s–1

b i Fg = mg = 55 × 9.8 = 540 N down ii Fg = mg = 55 × 9.8 = 540 N down

6 a Mass = 85 kg b Mass = 85 kg c Fg = mg = 85 × 3.6 = 306 N down

7 A block is struck with a sharp blow so that it overcomes the grip of the blocks within which it is in contact, and so it is ejected from the pile. The other blocks experience no horizontal net force, and so stay in the vertical stack.

8 a Fh = 120 cos 20° = 110 N north Fv = 120 sin 20° = 41 N down b Forces are balanced, so frictional force is opposite to horizontal pushing force

= 110 N south c FN = mg + Fp(v) = (20 × 9.8) + 41 = 240 N up d When a cart is pulled, the vertical component of the applied force is upwards rather

than downwards, and so a smaller (upwards) normal force is needed—helping the wheels to rotate more freely.

9 Using Newton’s second law, the glider will accelerate at 5.0 m s–2 for 1 s and then at 2.0 m s–2 for 1 s. Its final speed will be 7.0 m s–1.

10 The weight and normal forces balance. There is no driving force or drag force acting, so B is correct.

11 a a = ΣF/Σm = 120/(30 + 50) = 120/80 = 1.5 m s–2 in the direction of the force



b Consider 50 kg trolley. The only horizontal force acting on it is the pushing force from the 30 kg trolley. ΣF(50 kg) = ma = 50 × 1.5 = 75 N

12 a These forces are an action–reaction pair and are therefore equal in magnitude. Consequently, their ratio = 1:1.

b Acceleration of girl = F/40 Acceleration of skateboard = F/10 Acceleration of girl/acceleration of skateboard = 1:4

c Ratio of final velocities = ratio of accelerations = 1:4 (using v = u + at).

13 To maintain a tension of less than 100 N in the rope, the bucket must be allowed to accelerate downwards as it is lowered. If it ever travels with a constant speed or stops, the tension in the rope will be 120 N and the rope will snap.

14 a No b Both the weight and normal forces are acting on the water tank, so they are not an

action/reaction pair as used in Newton’s third law. Action/reaction forces each act on different objects.

15 The slope of the road is: θ = sin–1 0.2 = 11.5° The car will have an acceleration of: a = g sin θ = 9.8 × 0.2 = 1.96 m s–2 v2 = u2 + 2ax = 0 + 2 × 1.96 × 100 = 392 v = 19.8 m s–1 = 19.8 × 3.6 = 71 km h–1

16 a = g sin θ 4.9 = 9.8 sin θ sin θ = 0.50 θ = 30°

17 Component of weight force along incline = mg sin θ = 70 × 9.8 sin 45°= 485 N Net force along incline = 485 – 250 = 235 N a = ΣF/m = 235/70 = 3.4 m s–2

18



19 The net force on the system is: ΣF = m1g – m2g = 98 – 49 = 49 N a = ΣF/Σm = 49/15.0 = 3.27 m s–2 Consider the 10.0 kg mass: ΣF(10 kg) = m × a = 98 – FT 10 × 3.27 = 98 – FT FT = 98 – 32.7 = 65 N

20 a Therese is correct. The marble and billiard ball exert equal forces on each other as described by Newton’s third law.

b Therese is correct. The floor and basketball exert equal forces on each other as described by Newton’s third law.

Documents

Heinemann Physics 11 3rd Edition Worked Solutions Chapter 5