TOPIC 6: MOTION IN A STRAIGHT LINE – PHYSICS NOTES FORM 2

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MOTION IN STRAIGHT LINE

MOTION

This is the changing of position of an object.

Distance and displacement

– If the body moves 5m east of the fixed point A. In the first case 5m represents a physical quantity, i.e. the magnitude.This is the distance; on the other hand 5m east represents a physical quantity which specifies magnitude and direction thus that is displacement.

•  Distance – Is the length between two points . SI unit of distance is meter (m).
•  Displacement – is the distance in a given direction. SI unit of displacement is metre(m)

SPEED AND VELOCITY

If a body covers a certain distance in a given time then the speed of the body may be found by dividing the distance traveled by the time taken.

Speed – Is the distance traveled per unit time.

Thus speed = 

The SI unit of speed is m/s (meter per second)

T = time

S = distance

V = speed

Velocity =

 SI unit = m/s

Velocity – Is the speed in a given direction.

Example 1

1. A body covers a distance of 480m in 6sec. Calculate its speed.

Solution

Data given:

Distance = 480m

Time taken = 6sec

Speed =?

Speed = 

Speed = 

Speed = 80m/s

ACCELERATION

Acceleration – Is the rate of change of velocity.

Thus;
Acceleration = 
Acceleration = 

a = 

Initial velocity – is the starting velocity.

Final velocity – is the finishing velocity.

The SI unit of acceleration is m/s² (meter per Second Square).

RETARDATION

Retardation – is the negative acceleration

Example 1

1. A car with a velocity of 90km/h under uniform retardation and brought to rest after 10s. Calculate its acceleration

Solution

Data given

Initial velocity = 90km/h

Final velocity = 0m/s

Time taken = 10s

Acceleration =?

To change the initial velocity to m/s

= 25m/s

Acceleration = 

= a = ^–2.5m/s²

VELOCITY TIME GRAPH

If a body moves with a uniform or constant velocity then for any given time, the velocity remains the same. The graph of velocity against time is a straight line parallel to the time axis.

Suppose a body is moving with uniform velocity is for time (t) then from the relation.

Velocity = 

Distance = Velocity x Time

 VELOCITY TIME GRAPH FOR UNIFORM MOTION

•Area under the graph AB is the area of the rectangle OABC which is also ut. Thus it follows that the area under the velocity time graph represents distance.

•For a body thrown vertically upwards, it retards by the force of gravity on its upward journey. If the body has the initial velocity U, its velocity becomes zero at the highest point of its motion. The body then accelerates downwards until it reaches the point of projection with velocity U.

•The graph of the velocity against time for a body thrown vertically upwards.

•The line AB represents upward motion and the line BC represents downward motion

•From point O to A, the body accelerates uniformly to velocity U, and it accelerates uniformly to a velocity V. The body then continues to move with this velocity up to point B.

•As the body moves from A to B its acceleration is zero. Finally decreases its velocity from V to zero, as it moves from B to C. In this case the velocity is retarded.

Acceleration = 

Acceleration = 

.^.. a =

From the figure above, the slope of the graph is given as     

EQUATIONS OF MOTION

Now it follows that;

a =     (cross multiplication)

at = v – u

V – U = at

v = u + at ……… (i)

–         • Distance traveled by a body is given by the area under the graph. The area covered by the triangle ABD and the rectangle  OADC.

–          Area of triangle ABD
=   x base x height
=   x t x (v – u)
=    (v – u) t

Area of a rectangle OADC

= base x height

OADC  = t x u

= ut

•Total area under the curve = area of a triangle ABD + area of rectangle OADC.

S =    (v – u) t + ut

S =    (vt – ut) + ut

S =    vt –    ut + ut

S =     vt +    ut

S =    t (v + u)
S = 

From the first equation;

V = u + at

S =  

S =   S = 

S = ut +   at²…………….. (ii)

•The first and second equations of the motion can be combined to give the third equation of motion as follows:

V = u + at

•By squaring both sides of the equation;

V² = (u + at) ²

V² = (u + at) (u + at)

V² = u² + 2uat + a²t²

V² = u² + 2a (ut +   at²)

S = ut +   at²

V² = u² + 2as ……………. (iii)

Therefore the three equations of motion are
1. a = v – u
t
2. s = ut + 1 at²
2
3. v² _{+ 2as}

EXAMPLES

1. A body moving with a velocity of 30m/s is accelerated uniformly to a velocity of 50m/s in 5s. calculate the acceleration and the distance traveled by the body.
Data given

Initial velocity (U) = 30m/s

Final velocity (V) = 50m/s

Time (t) = 5s

Acceleration =?

Distance traveled =?

a = a = 

a =  a = 4m/s²

.^.. Acceleration = 4m/s²

Distance traveled = ?

S = ut +    at²

S = 30m/s x 5s +   x 4m/s² x 5 x 5

= 30m x 5 +   x 4 x 25

= 150 + 50

= 200 m

.^.. Distance traveled = 200m

2. Starting from rest, a car accelerates uniformly at 2.5m/s² for 6s. The constant speed is maintained for one third of a minute. The brakes are then applied making the car to retard uniformly to rest in 4sec. Determine the maximum speed attained in km/h and the displacement covered In km.

Solution

Data given

Initial velocity (U) = 0m/s
Final velocity (V) = 2.5m/s
Time (t) = 6s
V = u + at

V = 0m/s + 2.5m/s² x 6s

V = 0m/s + 15

V = 15m/s

V = 

V = 9 X 6

V = 54km/hr

Distance (s) = ut +   at²

S = 0 x6 +   x 2.5 x 36

S = 45m

Convert into km

X = 45/1000

X = 0.045km

Distance covered from A to B

Distance = velocity x time

= 15m/s x 20s

= 300m

S₂ = 0.3 km

Distance covered from B to C

U = 15m/s

t = 4s

v = 0m/s

a =?
a = a = a = ^–3.75m/s²

v² = u² + 2as

0² = 15² + 2 x 3.75 x S

= 225 + 7.5S

^–225 = 7.5S

S = 30m

S₃ = 0.03km

Alternatively

Area under the graph = Distance

Area of trapezium OABC = (OC + AB) AD

= (30 + 20) x 15

=   x 50 x 15

= 25 x 15

= 375m

= 0.375 km

3. A train with a velocity of 40m/s is uniformly retarded and brought to rest after 5 seconds. Determine its deceleration and draw the graph

Solution
(u) = initial velocity
(v) = final velocity
Time (t) = 5sDeceleration =           = = ^–8m/s²
.^..Deceleration = ^–8m/s²

TOPIC 6: MOTION IN A STRAIGHT LINE – PHYSICS NOTES FORM 2

MOTION UNDER GRAVITY

All bodies on the earth will always fall down towards the earth’s surface when released from a point. What makes these bodies fall downwards is the acceleration due to gravity called acceleration of free falling body which is 9.8 or 10 N/kg.

Acceleration of free falling body is denoted by ‘g’. Light bodies like feathers, paper etc are observed to fall down more slowly than iron balls. This is because light bodies are very much affected by air resistance.

a) Free body falling from rest

Acceleration a = acceleration due to gravity g i.e. that is a = g

Initial velocity (u) = 0

After time (t) velocity (v) is given as:

V = u + at

But a = g

V = u + gt

But U = 0

V = 0 + gt

V = gt

As body is falling the distance traveled downward is the height (h) is given by

U = 0

S = h
S = ut +   at²

h = 0 +  gt²

h =   gt²

The displacement time graph will be a quadratic curve as shown in fig (a) above. Using the third equation of motion, which is v² = u² + 2as with u = 0, a = g, (for a falling body)

V² = u² + 2as

V² = 0 + 2gs

V² = 2gs

.   = 

V =   a)      An object thrown vertically upward

From the figure above a ball is thrown vertically upwards, it rises to a height ‘h’ on its journey upward; it has an acceleration negative due to gravity. That is a = ^–g (it is going against pull). If it is thrown with the initial velocity U, then velocity at time t is given by a = ^–g

V = u + at

V = u – gt

Distance S = ut +   at²

S = h, a = ^–g

h = ut –   gt²

Using the equation v² = u² + 2as

a = ^–g

Examples

1. A body is released from rest at a certain height above the ground. If the body strikes the ground with a velocity of 60m/s calculate the height from which the body was released and the time taken by the body to strike the ground.

Solution

Data given

Initial velocity (u) = 0m/s

Final velocity (v) = 60m/s

Gravity (g) = 10

Height =?

Time (t) =?

V² = u² + 2as

V² = u² + 2gs

V² – u² = 2gs

S = 

S = 

S = 180m

Tim:

a = 

t =    = 

t =   

Time = 6sec

2.      A stone is thrown vertically upwards from the ground with a velocity of 30m/s. Calculate

a.      The max height reached

b.      The time taken to reach the max height

c.       The time to reach the ground after the ball is thrown up

d.      The velocity reached half way to the max. height

a)      U = 30m/s

V = 0 (at turning point)

a = ^–g = ^–10 (upward)

V² = u² + 2as

V² = u² – 2gh

0² = 30² – 2 x 10 x h

0 = 900 – 20h

20h = 900

h = 900/20

h = 45m

b)      V = u + at

V = u – gt

0 = 30 – 10t

10t = 30

t = 3sec

c)  For a stone to reach the ground, it makes the journey from the max point to the ground

U = 0m/s

h =45m

S = ut +   at²

h = ut +   gt²

h = 0 x t +   gt²

h =   gt²

2h = gt²

t² = 2h/g

t =    =   = 

t = 

t= 3s

.^.. Time = 3s

d)  Distance to the half way to the max. height is h₂ = 

h₂ =   = 22.5m

v² = u² – 2gh

v² = 30² – 2 x 10 x 22.5

v² = 900 – 450

v² = 450

v = 

v = 21.2m/s²

SIMPLE PENDULUM

Simple pendulum is defined as a small heavy body suspended by a light in extensible string suspended from a fixed support. A simple pendulum is made by attaching a long thread to a spherical ball called a pendulum bob.

If the bob is slightly displaced to position B and then released and swings to and fro going to C through O and back to B through O. when the pendulum completes one cycle or revolution, the time taken is called the period (T) of the oscillations. The length of the string from the point of attachment to the center of gravity of the pendulum is called the length of the pendulum (i.e. AO)

At position B or C, where the pendulum bob attains the max. Height from O (the lowest or resting point) the pendulum bob said to have reached the maximum displacement called the amplitude and the angle made by the string and vertical axis is called angular amplitude.

It has been observed from the experiments that changing the weight of the bob and keeping the same length of the pendulum, the period is always constant

The period of the single pendulum is given by T =   , where T = period of the single pendulum.

^{L = length of the extensible string or length of the pendulum, g = acceleration due to gravity.}

. 

T² = 2^2 2 

T2 = 4 2

T²g = 4 ²l

Make l as a subject

L = This is the equation of straight line with the slope.
Slope of the graph is given by 

From the relationship between l and T², the slope    = 

EXERCISE

1. A plane drops a sack of maize 25kg from a height of 20m to reach the famine stricken areas of Somalia. What is the time taken for the sack to reach the ground and at what velocity.

Solution

Data given

Maize = 25kg
Height = 20m
Final velocity (v) =?
Gravity (g) = 10m/s²
Initial velocity (u) = 0m/s

From the third equation V² = u² + 2gs

V = √2gs

V = √2 x 10 x 20

V = √400

V = 20m/s

.^.. Velocity = 20m/s

Time taken is obtained from first equation

V=u + at where a=g

V=u-gt where u=0

V=gt

20=10t

T=2 seconds

2.      A man fires a bullet at ground level, vertically towards the sky. The bullet is ejected with the initial velocity of 200m/s. After how long will the bullet return to the ground? Calculate the maximum height by the bullet.

Solution

Data given

Initial velocity (u) = 200m/s
Gravity (g) = 10m/s²
Final velocity (v) = 0m/s
Maximum height =?
Time (t) =?  
V² = u² + 2gs

V² – u² = 2gs

Make s as a subject

S =  Since s = h

h = 

h = 

h = ^–2000m

.^.. Height = 2000m

To find time (t)

a = 

Make t as a subject

t = 

But a = ^–g

t = 

t = 

t = 

t = 20s

  Time = 20s

THE LAWS OF MOTION AND LINEAR MOMENTUM

INERTIA

Inertia is defined as the ability of a body at rest to resist motion or a body in motion to continue moving in a straight line when abruptly stopped.

Newton’s first law of motion

It sometimes called “the law of Inertia”

It states that “Everybody continues its state of rest or uniform motion in a straight line if there is no external force acting on it”

Momentum

A body is said to be in motion if it changes its position with time and when it has velocity

A body with zero velocity therefore it is not in motion and hence it is at rest

The motion of a body can be measured by multiplying out its mass ‘m’ and its velocity ‘v’ the product M.V is known as the linear momentum of a body

Linear momentum = Mass X Velocity

= kg X m/s

= kgm/s

.^.. The S.I unit of momentum is kgm/s

Newton’s second law of motion

It states that the “rate of change of linear momentum of a body is directly proportional to the applied force and takes place in the direction of the force”

• Suppose force F acts on a body of mass ‘m’ for time t. This force causes the velocity of the body to change from initial velocity ‘u’ to find velocity ‘v’ in that interval t

•The change in momentum will then be Mv – Mu (kgm/s)

•The rate of change of momentum is     , by Newton’s second law of motion  .  Hence   .  But   = a (acceleration of a body). F Ma

•If a constant of proportionality K is introduced in the above relation, then F = kMa. This equation can be used to define unit of force. If m = 1kg and a = 1m/s², then the unit of force is chosen in such a way that when F = 1 the constant K = 1, hence F = Ma

•If a mass of 1kg is accelerating with 1m/s², then a force 1N is said to be acting on the body. Therefore a force F of 1N can be defined as the force which when acting on the body of mass 1kg produces an acceleration of 1m/s², That  is 1N = 1Kgm/s² then constant of proportionality K will be equal to one.

Thus;

MASS AND WEIGHT

In the earth gravitational field, the acceleration due to gravity is given by the symbol ‘g’.

The gravitational force therefore which act on a body of mass ‘m’ is equal to mg. This is what is known as the weight of the body. This force tends to pull the body towards center of the earth.

Newton’s third law of motion

• It states that: “In every action there is an equal and opposite reaction”

•If a body of a mass ‘m’ is placed on a table as shown above, the body presses on the surface of the table with a force mg, the table supports the body with an upward force.

The force which is the weight of the body is called the action of force while the upward force provided by the table is known as the reaction force. By Newton’s third law of motion the body does not move upwards or downwards.

Impulse of a force

•When a constant force F acts on a body of mass ‘m’, it produces an acceleration such that F = Ma. If the velocity of the body changes from U to V in time t, then;

a = 

From F = Ma

F = 

F = 

Ft = Mv – Mu

The quantity Ft is called the impulse of a force. The S.I unit of impulse of a force is the Newton second (Ns)

EXAMPLES

1. Find the linear momentum of a body of mass 5kg moving with the velocity of 2m/s.

Solution

Data given

Mass = 5kg

Velocity = 2m/s

Linear momentum = mass X velocity

= 5kg X 2m/s

= 10Kgm/s

2. A football was kicked into hands of a goal keeper at 4m/s. The goal keeper stopped the ball in 2 seconds. If the mass of a ball is 0.5kg, calculate the average force exerted on the goal keeper

Solution

Data given

U = 4m/s
T = 2s
Mass = 0.5kg
V = 0m/s

F = Ma

F =   F =    F =    F = ^–1N

The average force exerted by the goal keeper is 1N

3. What force is required to give a mass of 0.2kg an acceleration of 0.5m/s²?

Solution

Data given

Mass = 0.2kg
Acceleration (a) = 0.5m/s²
From F = acceleration X mass
F = 0.2kg X 0.5m/s²
F = 0.1N

4. What acceleration will be given to a body of mass 6kg by a force of 15N?

Solution

Data given

Mass = 6kg
Force = 15N
Acceleration (a) =?
From force = mass x acceleration
Acceleration = 

Acceleration = 

A = 2.5m/s²

5.      What mass will be given an acceleration of 5m/s² by a force 2N?

Solution
A = 5m/s²
F = 2N
M =?
Force =mass x acceleration
Mass = Mass = Mass = 0.4kg

CONSERVATION OF A LINEAR MOMENTUM

Consider the case of firing a gun, as the bullet leaves the gun (reaction), the one holding it feels a backward force (reaction from the butt of the gun)

According to Newton’s third law of motion, these two forces are equal and opposite. Since these two forces act at the same time, the impulses (i.e. change in momentum) produced must be equal in magnitude and opposite in direction. The sum of the two momentum is equal to zero.

This implies that momentum cannot be produced some where without producing an equal and opposite momentum somewhere else. This is the law of conservation of linear momentum which states that “When two or more bodies act upon each other, their total momentum remains constant provided no external forces are acting”

Consider the collision of two balls moving in a straight line

The balls have the masses M₁ and M₂ and they are approaching each other with velocity U₁ and U₂ fig (a)

The balls have the velocities V₁ and V₂ after collision fig (b)

Let F₁ and F₂ be the forces acting on M₁ and M₂ during collision.

By Newton’s third law of motion the forces are equal and opposite since the two forces act during the same time t, the impulses produced are therefore equal and opposite

.^.. F₁t = ^–F₂t

But F₁t = M₁V₁ – M₁U₁

F₂t = M₂V₂ – M₂U₂

From;

F₁t = ^–F₂t

M₁V₁ – M₁U₁ = -M₂V₂ + M₂U₂

M₁V₁ + M₂V₂ =M₁U₁ + M₂U₂

M₁U₁ + M₂U₂ = M₁V₁ + M₂V₂

This shows that the total momentum before collision is equals to the total momentum after collision.

Question

1. A body of mass 8kg moving with velocity of 20m/s collides with another body of mass 4kg moving with a velocity of 10m/s in the same direction. The velocity of 8kg body is reduced to 15m/s after collision. Calculate the final velocity of the 4kg body

Solution

Data given

M₁ = 8kg

U₁ = 20m/s

M₂ = 4kg

U₂ = 10m/s

V₁ = 15m/s

V₂ =?

Apply

M₁U₁ + M₂U₂ = M₁V₁ + M₂V₂

8 x20 + 4 x 10 = 8 x 15 + 4 x V₂

160 + 40 = 120 + 4V₂

200 = 120 + 4V₂

200 – 120 = 4V₂

80 = 4V₂

V₂ = 20m/s

.^.. The final velocity is 20m/s

EXERCISE

1. Define the term momentum.

A car of mass 500kg is moving in a straight line with a velocity of 90km/h. calculate the linear momentum of the car

Solution

Momentum – is the product of mass and velocity

Data given

Mass = 500kg

Velocity = 90km/h

Linear momentum = mass X velocity

= Kg X m/s

Velocity = 90km/h

Convert km into m

Velocity = 90m/h = 

Velocity = 25m/s

Momentum = 500kg X 25m/s

.^.. Momentum is 12500kgm/s

2. After striking its target a bullet of mass 50g is brought to rest in 2 seconds by a force of 300N. Calculate the velocity of a bullet before striking a target

Solution

Data given

Mass = 50g

Time = 2s

Force = 300N

Velocity =?

Convert g into kg

50g = 0.05kg

F = 
300 = 

V = 

V = 12000m/s

3.      A man of mass 80kg jumps off a trolley of mass 160kg. If the initial speed of the man is 8m/s, at what initial speed will the trolley move?

Solution

Data given

Velocity = 0m/s

Mass (M₁) = 80kg

Speed (U₁) = 8m/s

Mass (M₂) = 160kg

M₁U₁ + M₂U₂ = M₁V₁ + M₂V₂

80 X8 + 160 X U₂ = 80 x 0 + 160 x 0

640 + 160U₂ = 0

160U₂ = 640

U₂ = 4m/s

Initial speed = 4m/s

4. What is the linear momentum of a body of mass 6kg caused to move when a constant force of 15N is acting on it for 3s? What is the acceleration developed?

Solution

Data given

Mass = 6kg

Force = 15N

Time = 3s

A =?
Acceleration = 

= = 2.5m/s²
Linear momentum = mass X velocity

2.5 = 

U = 7.5

Linear momentum = 6kg x 7.5m/s

Linear momentum = 45kgm/s