← Grade 9 Science

Force and Motion

Unit 7 · CDC textbook pages 97–126

नेपालीमा पढ्नुहोस्

In this unit 9 sections
  1. The bus journey of Batuli
  2. Speed and velocity
  3. Acceleration
  4. Equations of linear motion
  5. Graphs of linear motion
  6. Inertia
  7. Newton’s laws of motion
  8. Elasticity and plasticity
  9. Exercise

Batuli gets on a bus. It sits still, then the engine roars and it creeps forward, faster and faster. On the bends of the hill road the driver turns the wheel; when he brakes, Batuli’s body jerks forward. At the end of the day she has traveled 74 km along the road, yet the map says her home is only 25.5 km away in a straight line.

Every part of that ride is in this unit. How do we describe motion with numbers (speed, velocity, acceleration)? How do we predict it (the equations of motion) and draw it (motion graphs)? And why does anything start, stop or turn at all? The answer to the last question is Isaac Newton’s: force. You will finish with his three laws of motion, and with what force does to the shape of things: elasticity and plasticity.

The bus journey of Batuli

Batuli travels by bus from place A to place B. The bus is at rest when she gets on. The engine pushes it, so it starts to move slowly; then it speeds up (it accelerates). Each time the driver brakes, her body lurches forward. The road bends, and the driver turns the steering wheel to change the direction of the bus. (What would happen if the steering broke on a bend?)

The roadside milestones tell her how far the bus has come:

Time (minutes) 10 25 30 40 50 …
Distance (km) 5 10 bus stopped (10) 15 20 …

On a straight stretch, the speedometer in front of the driver steadily reads 50 km/h: the bus is in uniform motion. At the end, the distance along the road from A to B is 74 km. Batuli divides the total distance by the total time to find the average speed. Then she opens a map on a phone and finds that B is 25.5 km to the south of A in a straight line. That straight-line distance, with its direction, is called the displacement.

Speed and velocity

Distance and displacement

Distance is the total length of the path an object has traveled, whatever its shape. It has a size (a magnitude) but no direction, so it is a scalar quantity.

Displacement is the shortest, straight-line distance from the starting point to the end point, in a stated direction. It has both a magnitude and a direction, so it is a vector quantity.

On the map below, a bus goes from A along the winding road through B, C, D and E to F. The distance is the whole length of road ABCDEF. The displacement is the straight arrow AF, towards the north-east. However the road twists, the displacement only cares where you started and where you are now.

The bus on a winding road

Press Drive, or drag the slider. Watch the red dashed arrow (the displacement) and compare it with the distance along the road. The blue arrow is the bus’s velocity: see how it turns on the bends even when the speed stays the same. (Directions are short: NE is north-east, ESE east-south-east.)

NSEWABCDEFdisplacement from Avelocity of the bus
Distance along the road
0 m
Displacement from A
0 m
Velocity now
10 m/s N
Average speed
–
Average velocity
–
Speed now
10 m/s

A boy walks 3 m east and then 4 m north. What is his displacement?

Pick an answer to see why.

Speed and velocity

The distance an object covers in one unit of time is its speed. The distance it covers in a particular direction in one unit of time (its displacement per unit time) is its velocity. The SI unit of both is the meter per second (m/s); in daily life we also use km/h.

At B on that map, the bus covers 10 m every second towards the east: its velocity is 10 m/s east. At C it is going faster, 20 m/s, and in a new direction; between D and F it does 15 m/s towards the north-east.

Speed is a scalar; velocity is a vector. That means velocity can change even when the speed does not: a bus going round a bend at a steady 40 km/h has a constant speed but a changing velocity, because its direction changes.

Average speed and average velocity

A real bus goes fast and slow and stops: in equal intervals of time it covers different distances. This is variable (non-uniform) motion. For a whole journey we use averages:

Average speed=total distance coveredtotal time taken \text{Average speed} = \frac{\text{total distance covered}}{\text{total time taken}}

Average velocity=total displacementtotal time taken(with the direction of the displacement) \text{Average velocity} = \frac{\text{total displacement}}{\text{total time taken}} \quad \text{(with the direction of the displacement)}

Positive and negative velocity

Along a straight line, we choose a reference point (call it 0) and a positive direction. Motion one way is then given a + sign and motion the other way a − sign.

0 (reference point)− side+ sidespeed = 25 m/svelocity = +25 m/sspeed = 25 m/svelocity = −25 m/sSame speed; opposite directions, so opposite signs of velocity.
Figure 7.1 Two cars move away from a child at the reference point, both at 25 m/s. The one going right has a velocity of +25 m/s; the one going left, −25 m/s. Their speeds are the same: speed is always positive.

So a negative velocity is not “slower than zero”: it just means moving in the opposite direction. Speed, which has no direction, is never negative.

A ball rolls 6 m to the right and then 2 m back to the left, taking 4 s altogether. Which is right?

Pick an answer to see why.

Acceleration

pile of bookssmooth plank, 3 m0 s1 s2 s3 sthe gaps grow: the marble speeds up
Figure 7.2 Marks every second on a sloping plank. The gaps grow, because the marble’s velocity increases by the same amount every second.

Objects do not always move with uniform velocity. A bus speeds up and slows down; a marble rolling down an even slope, pulled by gravity, gains velocity every second. The rate of change of velocity, the change in velocity in each unit of time, is called acceleration:

Acceleration (a)=final velocity (v)−initial velocity (u)time taken (t)=v−ut \text{Acceleration } (a) = \frac{\text{final velocity } (v) - \text{initial velocity } (u)}{\text{time taken } (t)} = \frac{v-u}{t}

Its SI unit is m/s ÷ s = m/s² (meters per second, per second). An acceleration of 3 m/s² means the velocity grows by 3 m/s every second. Like velocity, acceleration is a vector.

When the velocity decreases, as when a bus brakes, v−u v - u is negative and so is the acceleration. A decrease of velocity per unit time is called retardation (or deceleration, or negative acceleration).

Acceleration due to gravity

An object falling freely is pulled by the Earth’s gravity, and gains velocity at a steady rate. This acceleration is the acceleration due to gravity, written g. Near the Earth’s surface its average value is 9.8 m/s²: a falling stone goes 9.8 m/s faster every second.

0 m2 m4 m6 m8 m10 mdistance fallent = 0.0 sfallen 0.0 mv = 0.0 m/st = 0.5 sfallen 1.2 mv = 4.9 m/st = 1.0 sfallen 4.9 mv = 9.8 m/st = 1.5 sfallen 11.0 mv = 14.7 m/s
Figure 7.3 A falling ball drawn every quarter of a second. The gaps grow because its velocity increases by 9.8 m/s every second.

Here are the measurements for an object falling freely from rest, every second:

Time, t (s) Distance fallen, y (m) Velocity, v (m/s) Acceleration, a (m/s²)
0 0 0 9.8
1 4.9 9.8 9.8
2 19.6 19.6 9.8
3 44.1 29.4 9.8

A car’s velocity changes from 10 m/s to 25 m/s in 5 s. What is its acceleration?

Pick an answer to see why.

A cyclist’s acceleration is −2 m/s². What is happening?

Pick an answer to see why.

Equations of linear motion

Suppose an object moves in a straight line with uniform acceleration a. In time t its velocity changes from the initial velocity u to the final velocity v, and it covers a displacement s. These five quantities, u, v, a, t and s, are linked by three equations. Here is where each one comes from.

1. Relation among u, v, a and t

By the definition of acceleration,
a=v−ut a = \frac{v-u}{t}
Multiply both sides by t:
at=v−u at = v - u
v=u+at...(i) v = u + at \qquad \text{...(i)}

2. Relation among u, v, a and s

For uniform acceleration, the average velocity is u+v2 \tfrac{u+v}{2} , so
displacement=average velocity×time \text{displacement} = \text{average velocity} \times \text{time}
s=(u+v2)t...(ii) s = \left( \frac{u+v}{2} \right) t \qquad \text{...(ii)}
From (i), t=v−ua t = \dfrac{v-u}{a} . Put this in (ii):
s=(v+u2)(v−ua)=v2−u22a s = \left( \frac{v+u}{2} \right) \left( \frac{v-u}{a} \right) = \frac{v^2 - u^2}{2a}
(using (v+u)(v−u)=v2−u2 (v+u)(v-u) = v^2 - u^2 ). Multiply both sides by 2a:
2as=v2−u2 2as = v^2 - u^2
v2=u2+2as...(iii) v^2 = u^2 + 2as \qquad \text{...(iii)}

3. Relation among u, a, t and s

From (ii), s=(u+v2)t s = \left( \dfrac{u+v}{2} \right) t . Put v=u+at v = u + at from (i):
s=(u+u+at2)t=(2u+at2)t=(u+12at)t s = \left( \frac{u + u + at}{2} \right) t = \left( \frac{2u + at}{2} \right) t = \left( u + \tfrac{1}{2}at \right) t
s=ut+12at2...(iv) s = ut + \tfrac{1}{2}at^2 \qquad \text{...(iv)}

A stone is dropped from a bridge and hits the water after 3 s. Which equation finds the bridge’s height most directly?

Pick an answer to see why.

Graphs of linear motion

A graph shows a whole journey in one picture. We use two kinds: displacement–time graphs and velocity–time graphs. Time always goes along the horizontal x-axis.

Displacement–time graph

Rohan and Binaya are in a school race. At the start, Rohan is ahead of Binaya; a little later, Binaya overtakes him. Figure 7.4 shows the race as a graph, with displacement (s) up the y-axis and time (t) along the x-axis.

0123051015time, t (s)displacement, s (m)ABCD
Figure 7.4 Displacement–time graph of the race. One line is Rohan’s and one is Binaya’s.

A displacement–time graph (s–t graph) shows how the displacement of an object changes with time: displacement on the y-axis, time on the x-axis.

  • The slope of the line gives the velocity of the object.
  • A steeper line means faster motion; a gentler line, slower motion.
  • Where two lines cross, the two objects are at the same place at the same time.

The slope of a line is its “rise over run”:
Slope=riserun=change in displacement (Δs)change in time (Δt) \text{Slope} = \frac{\text{rise}}{\text{run}} = \frac{\text{change in displacement } (\Delta s)}{\text{change in time } (\Delta t)}
(Δ, the Greek capital letter delta, means “change in”.)

012340481216time, t (s)displacement, s (m)Δt = 2 s (run)Δs = 8 m (rise)
Figure 7.5 Finding the slope of a displacement–time line: pick two points, and divide the rise (Δs) by the run (Δt).

In Figure 7.5, between 1 s and 3 s the displacement rises from 4 m to 12 m:
Slope=ΔsΔt=8 m2 s=4 m/s \text{Slope} = \frac{\Delta s}{\Delta t} = \frac{8\ \text{m}}{2\ \text{s}} = 4\ \text{m/s}
So the body is moving with a uniform velocity of 4 m/s. The slope of an s–t graph is the velocity.

02040608010012014002004006008001,000time, t (s)displacement, s (m)Δt = 20 sΔs = +400 mΔt = 40 sΔs = −400 mABCDGEHJmoving forwardat restcoming back
Figure 7.6 Displacement–time graph of the motorcycle in Example 7.3: moving forward (AB), at rest (BC), coming back (CD).
01234567891002468101214time, t (s)displacement, s (m)Δt = 2 sΔs = 2 mABCDEJHI
Figure 7.7 Displacement–time graph of the man chased by a dog.

So, on a displacement–time graph:

  • a line parallel to the x-axis (flat) means the displacement does not change: the object is at rest;
  • a straight sloping line means uniform velocity;
  • a curved line means a variable velocity: curving upwards, speeding up; curving over, slowing down.
stflat: at rest(zero velocity)ststraight, up:uniform velocityststraight, down:uniform velocity backstcurving up:speeding upstcurving over:slowing down
Figure 7.8 The shapes of displacement–time graphs.

On a displacement–time graph, a straight line slopes downwards. What does it show?

Pick an answer to see why.

Velocity–time graph

01234501020304050time, t (s)velocity, v (m/s)motorcyclefalling objectplane
Figure 7.9 The three velocity–time graphs of Activity 7.5.

When the velocity of a moving object is plotted against time, the graph is a velocity–time graph (v–t graph): time on the x-axis, velocity on the y-axis. It shows at a glance whether a body is at rest, moving steadily, speeding up or braking.

  • The slope of a velocity–time line is the acceleration: slope=ΔvΔt=a \text{slope} = \dfrac{\Delta v}{\Delta t} = a .
  • A flat line (parallel to the x-axis): the velocity doesn’t change, so the object moves with uniform velocity and no acceleration.
  • A straight sloping line: uniform acceleration (sloping up) or uniform retardation (sloping down).
  • A curved line: variable (non-uniform) acceleration.
  • Comparing the slopes of two lines tells you which velocity is changing faster.
vtflat: uniform velocity(no acceleration)vtstraight, up:uniform accelerationvtstraight, down:uniform retardationvtcurved:variable acceleration
Figure 7.10 The shapes of velocity–time graphs.
01234567891011121301234567time, t (s)velocity, v (m/s)ABCDEFGH
Figure 7.11 A velocity–time graph with segments of every kind.

Make your own motion graphs

Choose a starting velocity and an acceleration (or one of the four buttons), then press Go. The car drives for 10 seconds while both of its graphs are drawn. Can you make an s–t graph that is a straight line? A v–t line that slopes down?

0 m100 m200 m300 m400 m01002003004000246810displacement, s (m)time, t (s)010203040500246810velocity, v (m/s)time, t (s)
Time
0.0 s
Displacement
0.0 m
Velocity
10.0 m/s
Acceleration now
+2 m/s²

Uniform acceleration: the velocity rises by the same amount every second (a straight, sloping v–t line), so the s–t line curves upward.

A velocity–time graph is a straight line sloping upwards from the origin. What is the object doing?

Pick an answer to see why.

On a velocity–time graph, what does a line parallel to the time axis at v = 10 m/s mean?

Pick an answer to see why.

Inertia

Talk about these in class: what is going on, and why?

  1. On a swing, it takes a little while to get the swing moving, and a while for it to come to rest again.
  2. A ceiling fan doesn’t reach full speed the moment you switch it on, and doesn’t stop the moment you switch it off.

When the forces on an object are balanced (they cancel out), its state of motion doesn’t change: its velocity stays the same in size and direction. To start it, stop it, or change its speed or direction, an unbalanced force must act on it. Objects “resist” any change in their state: a body at rest tends to stay at rest, and a moving body tends to keep moving in the same direction with the same velocity.

This property of a body, its inability to change by itself its state of rest or of uniform motion in a straight line, is called inertia.

Relation between mass and inertia

Every object has inertia, and the inertia of an object depends on its mass: the greater the mass, the greater the inertia. Mass is, in fact, the measure of inertia. That is why it is much harder to start, stop or turn a loaded truck than a bicycle.

Which has the most inertia?

Pick an answer to see why.

Types of inertia

1. Inertia of rest

1. Before2. A quick flickflickcoincardglass mugcard flies offthe coin stays put,then falls in
Figure 7.12 Inertia of rest. Flick the card quickly and the coin, which was at rest, stays where it was, then falls into the mug.

The tendency of a body at rest to stay at rest is called inertia of rest. Examples:

  • i.Dust comes out of a carpet when it is hung up and beaten with a stick: the carpet moves suddenly, but the dust stays at rest and separates from it.
  • ii.When the branches of a mango tree are shaken, the mangoes fall: the branches move, but the mangoes tend to stay at rest, and their stalks break.
  • iii.When the bottom coin of a pile of carrom men is struck hard with the striker, only that coin flies out; the rest of the pile stays where it is and drops straight down.
  • iv.When a bus starts suddenly, a standing passenger falls backwards: the feet move with the bus, but the upper body tends to stay at rest.
The bus starts suddenlyThe bus brakes suddenlybus moves offfeet move with the bus;the upper body stays at rest: falls backbrakes on: bus slowsfeet stop with the bus;the upper body keeps moving: falls forward
Figure 7.13 Inertia in a bus. Starting suddenly, a passenger falls back (inertia of rest); braking suddenly, a passenger falls forward (inertia of motion).

2. Inertia of motion

A cyclist’s front wheel hits a stone. The bicycle is stopped suddenly, but the rider’s body tends to go on moving forward at the same speed, so the rider is thrown over the handlebars. The tendency of a moving body to keep moving with the same velocity in the same direction is called inertia of motion. Examples:

  • i.A bicycle keeps rolling for some time after the rider stops pedaling.
  • ii.When a moving bus brakes suddenly, the passengers are thrown forward.
  • iii.When a bus turns sharply on a bend, the passengers lean towards the outside of the bend: their bodies tend to keep going straight.
  • iv.Jumping off a moving bus is dangerous: your feet stop when they touch the ground, but your body keeps moving forward, so you fall.
  • v.A long-jumper runs up before jumping: the speed she has gained carries her farther through the air.

Newton’s laws of motion

Sir Isaac Newton (1643–1727), an English physicist and mathematician, set out three laws of motion in 1687, in his book Principia. The first law tells us what a force is. The second law relates the force on a body to its mass and its acceleration, so it lets us measure force. The third law tells us a property of forces: they always come in pairs.

The title page of a 17th-century book in Latin, Philosophiae Naturalis Principia Mathematica, by Isaac Newton, printed in London in 1687
Figure 7.14 The title page of Newton’s Principia (1687), the book in which he set out his three laws of motion and his law of gravitation. Photo: Isaac Newton, 1687; scan from English Wikisource, Public domain

Newton’s first law of motion

Long before Newton, Galileo Galilei had found the law of inertia: a ball rolling on a smooth level surface would roll on for ever if nothing slowed it. Newton made it the first of his laws.

Newton’s first law of motion: an object at rest remains at rest, and an object in motion continues to move in a straight line with uniform velocity, unless an unbalanced (resultant) force acts on it.

So if the resultant force on a body is zero, it carries on as it is. To bring a body from rest to motion, or from motion to rest, or to change its speed or direction, an unbalanced force is needed. This gives us the meaning of force: a force is what changes, or tries to change, the state of rest or of uniform motion of a body. Because the first law describes the property of matter to stay at rest or in uniform motion, it is also called the law of inertia.

A puck slides across perfectly smooth ice with nothing at all acting on it sideways. What happens to it?

Pick an answer to see why.

Newton’s second law of motion

Think about this: a cyclist starts from rest. How will her acceleration change if a friend sits on the carrier? And on a level road, how will it change if she pushes the pedals harder?

If the force on the pedals stays the same, the acceleration is smaller when the mass is greater (two people). If the mass stays the same, the acceleration is greater when the force is greater. Here are more examples:

Acceleration changes with the mass Acceleration changes with the force
We can walk faster if we put down our load. We run faster when our muscles push harder.
A loaded truck on a steep uphill road speeds up more easily once it is unloaded. When the engine pushes harder, a vehicle accelerates more. When the brakes press harder, it slows down faster.
An empty wheelbarrow is easier to push into motion than a loaded one. To make a ball go fast, we must kick it harder.

Activity 7.13 on the screen

Choose the masses on the hanger and the mass of the trolley, then press Release. Each run goes into the table. Double the pulling force: what happens to a? Double the trolley’s mass: what then?

start1 m2 × 50 gstopwatch: 0.00 s
Mass of the trolley
Pulling force (N)Trolley (kg)Time for 1 m (s)a = 2s/t² (m/s²)
 
 
 
 
 
 

From the activity:

  • with the mass kept the same, the acceleration increases when the pulling force increases: a∝F a \propto F (mass constant) ...(i)
  • with the force kept the same, the acceleration decreases when the mass increases: a∝1m a \propto \dfrac{1}{m} (force constant) ...(ii)

Newton’s second law of motion: the acceleration produced in a body by a force is directly proportional to the force and inversely proportional to the mass of the body.

Putting (i) and (ii) together:
a∝Fm⇒F∝ma a \propto \frac{F}{m} \quad \Rightarrow \quad F \propto ma
F=k ma...(iii) F = k\,ma \qquad \text{...(iii)}
where k is a constant of proportionality. The unit of force is chosen to make k = 1: one newton (1 N) is the force that gives a body of mass 1 kg an acceleration of 1 m/s². Putting F = 1 N, m = 1 kg and a = 1 m/s² in (iii) gives 1 = k × 1 × 1, so k = 1, and
F=ma F = ma

The same force acts on a 2 kg ball and on a 6 kg ball. How do their accelerations compare?

Pick an answer to see why.

Newton’s third law of motion

Can you walk forward without pushing the ground with your feet? Try it, on a polished floor in socks, or think of walking on ice.

BAthe hand pullsA pulls B: 5 NB pulls A: 5 Nreads 5 Nreads 5 N
Figure 7.15 Two spring balances pulling on each other. Both read the same: A pulls B with 5 N, and B pulls A back with 5 N.

When one object pushes or pulls a second object, the second always pushes or pulls back on the first, with a force equal in size and opposite in direction. One force of the pair is called the action and the other the reaction. When two balls collide, ball A pushes B and ball B pushes A, equally and oppositely, even when one ball is much bigger than the other. A volleyball player’s hands push the ball up, and the ball pushes her hands down.

Newton’s third law of motion: to every action there is an equal and opposite reaction. If body A exerts a force on body B, then B exerts an equal force on A in the opposite direction.

The action and the reaction are equal and opposite, yet they don’t cancel each other out, because they act on two different objects. Forces cancel only when they act on the same object. That is why things can move at all: when you push the ground backwards, the ground’s forward push acts on you, and you walk forward.

Rowing a boataction: oar pushes water backwardreaction: water pushes boat forwardSwimmingaction: hands push water backwardreaction: water pushes swimmer forward
Figure 7.16 Rowing and swimming: the oar or the hands push the water back (action); the water pushes the boat or swimmer forward (reaction).
A rocket taking offaction: rocket pushes gases downreaction: gases push rocket upFiring a gunaction: gun pushes bullet forwardreaction: bullet pushes gun back
Figure 7.17 A rocket and a gun: each pushes something out one way and is pushed the other way.
action: balloon pushes air outreaction: air pushes balloon forwardstrawtight threadballoon
Figure 7.18 A balloon rocket on a thread. The air is pushed out backwards; the balloon is pushed forwards.

A horse pulls a cart. By the third law, the cart pulls back on the horse with an equal force. So why do they move?

Pick an answer to see why.

Elasticity and plasticity

What things come to mind when you hear the word “elastic”? And “plastic”? Make two lists.

When you pull a rubber band gently, it gets longer; let go and it goes back to its original length. A pressed rubber ball or balloon changes shape and springs back when the force is removed.

Forces applied to a body from different directions act together like a single resultant force, and a resultant force can change the shape and size of a body. How much it changes depends on what the body is made of, and on the size and direction of the forces. A force that changes the shape or size of a body is called a deforming force.

Does a steel spring go back to its shape if it is stretched as far as possible? No: there is a limit. A body deformed up to a certain extent goes back to its original shape and size when the deforming force is removed. That extent is its elastic limit.

Elasticity is the property of matter by which a body regains its original shape and size when the deforming force is removed. When a body is deformed, a restoring force develops inside it, which pulls it back.

The elasticity of different materials is different. The more elastic a material is, the harder it is to change its shape or size, and the better it returns. Steel is very elastic, so it is used in buildings, bridges and machines.

A road bridge with a tall tower and steel cables fanning down to hold up its deck
Figure 7.19 The Karnali Bridge at Chisapani, where the East-West Highway crosses the Karnali River. Its deck hangs from steel cables: steel stretches only a tiny bit under the weight of traffic and springs back as soon as the load has passed. Photo: Ganesh Paudel, CC BY-SA 3.0

If the deforming force goes beyond the elastic limit, the body cannot regain its shape: it is permanently deformed. Plasticity is the property of matter by which a body keeps its new, deformed shape even after the deforming force is removed. Plasticity is the opposite of elasticity. Wet mud, clay and dough are very plastic: that’s why mud can be pressed into bricks and clay shaped into pots. Pushed further still, a deformed body breaks.

Stretch it, then let it go

Pick a material. Pull a little and let go; then pull harder, past its elastic limit, and let go again. Which one is elastic only for the smallest pull?

0 cm5 cm10 cm15 cmoriginal length
Stretched by
0.0 cm
Strongest pull so far
0.0 N
Longer for good
0.0 cm

Pull with the slider, then press Let go.

Which of these uses plasticity?

Pick an answer to see why.

Exercise

1. Choose the correct answer

a. In a velocity–time graph, what does the slope of the line give?

Pick an answer to see why.

sti)stii)stiii)vtiv)
Figure 7.20 The four graphs for question 1 b.

b. Which of the graphs i–iv above is a displacement–time graph of an object moving with uniform velocity?

Pick an answer to see why.

c. Stepping down from a moving bus is dangerous. Which of these explains it?

Pick an answer to see why.

d. Which statement about action and reaction is true?

Pick an answer to see why.

e. Which of the following is an application of elasticity?

Pick an answer to see why.

f. A loaded truck and a car are moving at the same velocity on a road. Brakes are applied to both with equal force. Which statement is true?

Pick an answer to see why.

2. State the differences between

i.

A displacement–time graph and a velocity–time graph

Answer
Displacement–time graph Velocity–time graph
Displacement is plotted on the y-axis against time. Velocity is plotted on the y-axis against time.
Its slope gives the velocity. Its slope gives the acceleration; the area under it gives the displacement.
A flat line means the body is at rest. A flat line means the body moves with uniform velocity.
A straight sloping line means uniform velocity. A straight sloping line means uniform acceleration (or retardation).
ii.

Inertia of rest and inertia of motion

Answer
Inertia of rest Inertia of motion
The tendency of a body at rest to stay at rest. The tendency of a moving body to keep moving with the same velocity.
It resists being set in motion. It resists being stopped or turned.
e.g. a passenger falls back when a bus starts suddenly; dust leaves a beaten carpet. e.g. a passenger falls forward when a bus brakes suddenly; a cyclist is thrown forward when the wheel hits a stone.
iii.

Elasticity and plasticity

Answer
Elasticity Plasticity
The property by which a body regains its original shape and size when the deforming force is removed. The property by which a body keeps its deformed shape after the deforming force is removed.
The change is temporary. The change is permanent.
A restoring force develops. No restoring force develops.
e.g. steel spring, rubber band, racket strings e.g. clay, wet mud, dough

3. Give reasons

i.

It takes different lengths of time to bring a motorcycle, a car, a bus and a train, all moving with the same uniform velocity, to rest.

Answer

They have different masses, so different inertia. A more massive vehicle has more inertia of motion, and for a given braking force its retardation is smaller (a = F/m), so it takes longer to stop. A train, with an enormous mass, takes the longest; a motorcycle the shortest.

ii.

When the branches of a tree are shaken, its leaves and fruits fall.

Answer

Because of inertia of rest. The shaking suddenly moves the branches, but the leaves and fruits tend to stay at rest. The stalks are pulled hard, snap, and the leaves and fruits fall.

iii.

A bag placed in the aisle of a bus slides forward when the bus suddenly slows down or stops.

Answer

Because of inertia of motion. The bag was moving forward with the bus. When the bus slows suddenly, the floor’s small friction can’t slow the bag as quickly, so the bag keeps moving forward and slides along the floor.

iv.

A gun must be held firmly against a strong support when it is fired.

Answer

By Newton’s third law, when the gun pushes the bullet forward (action), the bullet pushes the gun backward with an equal force (reaction): the gun recoils. Held loosely, it would jerk back hard and could hurt the shooter. Held firmly against the shoulder, the recoil is shared with the much larger mass of the body, so it moves back only a little (a = F/m).

v.

Two rubber balls of the same size are dropped on the floor, and one bounces higher than the other.

Answer

They have different elasticity (different rubber, or different air pressure inside). On hitting the floor, each ball is squashed and springs back. The more elastic ball returns more of the energy of its fall as it regains its shape, and wastes less as heat, so it bounces higher.

vi.

A rubber band should not be stretched beyond a certain limit.

Answer

A rubber band regains its shape only while it is stretched within its elastic limit. Stretched beyond it, it is permanently deformed (it stays longer and loose), and stretched further still, it snaps.

4. Answer the following questions

i.

Define average velocity and acceleration.

Answer

Average velocity is the total displacement of a body divided by the total time taken: average velocity = total displacement ÷ total time. It has a direction (that of the displacement). For uniform acceleration it also equals u+v2 \tfrac{u+v}{2} .
Acceleration is the rate of change of velocity: the change in velocity per unit time, a=v−ut a = \dfrac{v-u}{t} . Its SI unit is m/s².

ii.

Prove these equations of linear motion:

  • a.v=u+at v = u + at
  • b.v2=u2+2as v^2 = u^2 + 2as
  • c.s=ut+12at2 s = ut + \tfrac{1}{2}at^2
Answer

Let a body moving in a straight line with uniform acceleration a change its velocity from u to v in time t, covering a displacement s.

  • a.By definition, a=v−ut a = \dfrac{v-u}{t} . So at=v−u at = v - u , and v=u+at v = u + at .
  • b.For uniform acceleration, average velocity = u+v2 \dfrac{u+v}{2} , so s=(u+v2)t s = \left(\dfrac{u+v}{2}\right) t . From (a), t=v−ua t = \dfrac{v-u}{a} . Then

s=(v+u)2×(v−u)a=v2−u22a s = \dfrac{(v+u)}{2} \times \dfrac{(v-u)}{a} = \dfrac{v^2-u^2}{2a} , so 2as=v2−u2 2as = v^2 - u^2 , and v2=u2+2as v^2 = u^2 + 2as .
c. From s=(u+v2)t s = \left(\dfrac{u+v}{2}\right) t and v=u+at v = u + at :

s=(u+u+at2)t=(u+12at)t s = \left(\dfrac{u + u + at}{2}\right) t = \left(u + \tfrac{1}{2}at\right) t , so s=ut+12at2 s = ut + \tfrac{1}{2}at^2 .

iii.

Draw a displacement–time graph and a velocity–time graph for a body moving in a straight line with uniform velocity.

Answer

Displacement–time graph: a straight line sloping upward (starting from the origin if the body starts at the reference point). Its slope is the uniform velocity.

Velocity–time graph: a horizontal straight line, parallel to the time axis, at the height of the velocity. Its slope is zero: no acceleration.

They are the second picture in Figure 7.8 and the first in Figure 7.10.

iv.

Draw a displacement–time graph from this data. From the slope of the line, describe the motion. Find the average velocity of the object in the first 4 s from the slope.

Time (s) 0 2 4 6 8 10 12 14
Displacement (m) 0 4 8 8 12 8 4 0
Answer
02468101214024681012time, t (s)displacement, s (m)2 m/sat rest2 m/s−2 m/s (back)
Figure 7.21 The graph of the data.

0–4 s: a straight line rising; uniform velocity forward. 4–6 s: flat: at rest at 8 m. 6–8 s: straight, rising: uniform velocity forward again (4 m in 2 s = 2 m/s). 8–14 s: straight, falling: it comes back to the start with uniform velocity (−12 m in 6 s = −2 m/s).
Average velocity in the first 4 s = slope = 8−04−0 \dfrac{8 - 0}{4 - 0} = 2 m/s forward.

0123456789101112020406080100time, t (minutes)displacement, s (m)tortoisehareACDEFBfinish
Figure 7.22 The race of the hare and the tortoise.
v.

The graph shows the motion of the hare and the tortoise from the famous story. Study it and tell the story again, step by step, from the start to the finish.

Answer

The hare (A to C) sets off very fast: in 3 minutes it is 60 m along the 100 m track, far ahead of the tortoise. Sure of winning, it lies down for a nap: from C to D its line is flat, so it is at rest for 7 minutes. Meanwhile the tortoise’s straight line climbs steadily: it walks at the same slow, uniform speed all the way. At about 7 minutes the tortoise’s line crosses the hare’s: it passes the sleeping hare. At D (10 minutes) the hare wakes and runs faster and faster (D to E to F, a curve getting steeper), but at 12 minutes the tortoise reaches the finish, B, while the hare, at F, is still a few meters short. Slow and steady wins the race.

vi.

Define inertia. Give two examples each of inertia of rest and inertia of motion.

Answer

Inertia is the property of a body by which it cannot by itself change its state of rest or of uniform motion in a straight line.

Inertia of rest: (1) a passenger falls backward when a bus starts suddenly; (2) dust comes out of a carpet when it is beaten with a stick.

Inertia of motion: (1) a passenger falls forward when a moving bus brakes suddenly; (2) a long-jumper runs before jumping so as to jump farther.

vii.

State the relation between mass and inertia.

Answer

The inertia of a body depends on its mass: the greater the mass, the greater the inertia, and the smaller the mass, the smaller the inertia. Mass is a measure of inertia. For example, a loaded truck is much harder to start or stop than an empty one.

viii.

State Newton’s first law of motion.

Answer

A body at rest remains at rest, and a body in motion continues to move in a straight line with uniform velocity, unless an external unbalanced force acts on it.

ix.

Give two examples of a resultant force changing the state of motion of an object.

Answer
  • 1.A footballer kicks a ball at rest: the force from the foot sets it moving.
  • 2.The brakes of a bicycle and the friction of the road bring a moving bicycle to rest.
  • 3.A batsman’s hit changes the direction of a cricket ball.
x.

Using high speed and the first law of motion, explain why bus accidents are more likely on winding roads in hilly areas.

Answer

By Newton’s first law, a moving bus tends to keep going in a straight line. To go round a bend, it needs a sideways unbalanced force, which comes from the friction of the road on its tyres. The faster the bus goes, the larger this force must be (and the sharper the bend, the larger still). On a sharp mountain bend at high speed, the force needed can be more than the friction can give, especially on a wet, gravelly or muddy road. The bus then carries on almost straight, off the edge of the road. Its large mass (inertia) also means it takes a long distance to slow down, and passengers are thrown sideways. So buses must go slowly on winding roads.

xi.

State Newton’s second law of motion and prove that F = ma.

Answer

Second law: the acceleration produced in a body is directly proportional to the force acting on it and inversely proportional to its mass.

Proof: a∝F a \propto F (mass constant) and a∝1m a \propto \dfrac{1}{m} (force constant). Combining them, a∝Fm a \propto \dfrac{F}{m} , so F∝ma F \propto ma and F=k ma F = k\,ma , where k is a constant.

One newton is defined as the force that gives a 1 kg mass an acceleration of 1 m/s². Putting F = 1 N, m = 1 kg, a = 1 m/s²: 1 = k × 1 × 1, so k = 1. Therefore F = ma.

xii.

State Newton’s third law of motion.

Answer

To every action there is an equal and opposite reaction. If a body A exerts a force on a body B, then B exerts a force on A that is equal in magnitude and opposite in direction. The two forces act on different bodies.

xiii.

Give two examples from daily life for each of the three laws of motion.

Answer

First law: (1) passengers fall forward when a bus brakes suddenly; (2) a carpet is cleaned by beating it.

Second law: (1) an empty cart is easier to accelerate than a loaded one with the same push; (2) a cricket ball hit harder goes faster.

Third law: (1) we walk by pushing the ground backward, and the ground pushes us forward; (2) a gun recoils when it fires a bullet.

xiv.

Find two uses of each of Newton’s first, second and third laws, and explain them in a table.

Answer
Activity Newton’s law Explanation
Wearing a seat belt first In a sudden stop the body keeps moving (inertia of motion); the belt supplies the force to stop it safely.
Shaking a thermometer, or hitting a ketchup bottle, to bring the liquid down first The glass stops suddenly but the liquid keeps moving down by its inertia of motion.
A fielder pulls back his hands while catching a fast ball second Taking longer to stop the ball makes its retardation, and so the force on the hands (F = ma), smaller.
Lighter racing bicycles and cars second Less mass gives more acceleration for the same force.
Launching a rocket third The rocket pushes gases out downward; the gases push it up.
Rowing a boat third The oars push the water back; the water pushes the boat forward.
Hitting a ball with a racketaction: racket pushes ballreaction: ball pushes racket backStepping off a boataction: feet push boat backwardreaction: boat pushes man forward
Figure 7.23 The two activities of question xv.
xv.

Name the action and the reaction in each activity: (a) a tennis racket hitting a ball; (b) a man stepping from a boat on to the bank.

Answer
  • a.Action: the racket pushes the ball forward. Reaction: the ball pushes the racket backward (you feel it in your hand).
  • b.Action: the man’s feet push the boat backward. Reaction: the boat pushes the man forward, towards the bank. Because the boat is free to move on water, it moves backward away from the bank, and if the man isn’t careful he falls into the water.
xvi.

Define elasticity and plasticity, with examples.

Answer

Elasticity is the property of matter by which a body regains its original shape and size when the deforming force is removed. Examples: a rubber band, a steel spring, the strings of a racket, a football.
Plasticity is the property of matter by which a body keeps its deformed shape after the deforming force is removed. Examples: clay, wet mud (bricks), dough, a toothpaste tube.

5. Numerical problems

a.

A plane starts from rest and runs south along a runway, taking off after 30 s. Its acceleration during the run is 1.5 m/s². Find the distance it runs along the runway, and its velocity at take-off.

Answer

u = 0, a = 1.5 m/s², t = 30 s.
s=ut+12at2=0+12×1.5×302=0.75×900 s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 1.5 \times 30^2 = 0.75 \times 900 = 675 m.
v=u+at=0+1.5×30 v = u + at = 0 + 1.5 \times 30 = 45 m/s towards the south.

b.

Rohan dropped a stone from a bridge into the river below, and Suraj saw that the stone took 2 s to reach the water. Taking the acceleration of the falling stone as 9.8 m/s², find the height of the bridge above the water.

Answer

The stone is dropped, so u = 0; t = 2 s; g = 9.8 m/s².
h=ut+12gt2=0+12×9.8×22 h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2} \times 9.8 \times 2^2 = 19.6 m.

c.

Suraj has a mass of 50 kg and his bicycle 12 kg. Cycling down a hill, they have an acceleration of 2 m/s². Find the resultant force acting on them.

Answer

The total mass moving is m = 50 + 12 = 62 kg.
F = ma = 62 × 2 = 124 N (down the hill).

d.

A car of mass 1500 kg is moving at 72 km/h. The driver brakes, and its speed falls to 10 km/h over a distance of 50 m. Find the resultant force on the car.

Answer

u = 72 km/h = 72 × 1000 ÷ 3600 = 20 m/s; v = 10 km/h = 10 × 1000 ÷ 3600 = 2.78 m/s; s = 50 m.
v2=u2+2as⇒2.782=202+2×a×50⇒7.72=400+100a⇒a=−3.92 m/s2 v^2 = u^2 + 2as \Rightarrow 2.78^2 = 20^2 + 2 \times a \times 50 \Rightarrow 7.72 = 400 + 100a \Rightarrow a = -3.92\ \text{m/s}^2
F = ma = 1500 × (−3.92) = −5880 N: a braking force of about 5900 N, opposite to the motion.
(If you round the retardation to 3.9 m/s², you get 1500 × 3.9 = 5850 N. Either is right with the working shown.)

02468101214161820051015202530time, t (s)velocity, v (m/s)ABCDEFG
Figure 7.24 The velocity–time graph of the bus for question 5 e.
e.

The velocity–time graph of a bus is shown.

  • i.Name the segments of uniform velocity, uniform acceleration, variable acceleration and retardation.
  • ii.Among the segments of uniform acceleration, in which does the bus’s velocity change fastest?
  • iii.Find the acceleration from C to D.
  • iv.Find the displacement of the bus from C to D.
Hint

For (iv), the displacement is the area under the velocity–time line between C and D.

Answer
  • i.Uniform velocity: BC and DE (flat). Uniform acceleration: AB, CD and EF (straight, sloping up). Retardation: FG; its line is curved, so it is a variable retardation (the velocity falls more and more quickly). The only segment where the acceleration is variable is this curved one, FG.
  • ii.The steepest of AB (10 m/s in 4 s = 2.5 m/s²), CD (15 m/s in 4 s = 3.75 m/s²) and EF (5 m/s in 4 s = 1.25 m/s²) is CD.
  • iii.a=v−ut=25−1010−6=154 a = \dfrac{v-u}{t} = \dfrac{25-10}{10-6} = \dfrac{15}{4} = 3.75 m/s².
  • iv.The area under CD is a trapezium with parallel sides 10 m/s and 25 m/s, 4 s wide:

displacement = 10+252×4 \dfrac{10 + 25}{2} \times 4 = 70 m. (Check: s=ut+12at2=10×4+12×3.75×16=40+30=70 s = ut + \tfrac{1}{2}at^2 = 10 \times 4 + \tfrac{1}{2} \times 3.75 \times 16 = 40 + 30 = 70 m.)

02468101214161820051015202530time, t (s)velocity, v (m/s)70 mCD
Figure 7.25 The displacement from C to D is the shaded area under the line.