MOTION IN A STRAIGHT LINE

MOTION IN A STRAIGHT LINE

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

 

  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 =?

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

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.

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

   

 OADC  = t x u

           = ut

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

•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

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 =?

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

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

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

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

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

        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

 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

This is the equation of straight line with 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) =?

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)

thus

F=ma

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;

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

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

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

2..      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