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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



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Lesson 9, Topic 2
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Newton’s Laws of Motion

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Sir Isaac Newton studied the behaviour of bodies in motion and was able to come up with three laws governing motion and they are as follows:

Newton’s First Law of Motion:

A body will continue to be in a state of rest or of uniform motion unless it is acted upon by an external force.

Newton’s Second Law of Motion:

The time rate of change of momentum of a body is directly proportional to the applied force and this takes place in the direction of the force.

Therefore, \( \scriptsize F \propto \normalsize \frac{change \: in \: momentum}{time} \)

\( \scriptsize F \propto \normalsize \frac{mv \: – \: mu}{t} \)

\( \scriptsize F \propto \normalsize \frac{m(v \: – \: u)}{t} \)

Recall that a = \(\frac{final \: velocity\: – \: initial \: velocity}{time} \)

a = \(\frac{v\: – \:u}{t} \), from equation of motion

\( \scriptsize F \propto ma \)

Introducing a constant, k, = 1

F = kma, k = 1

F = ma

The unit of force is Newton, and it is a vector quantity.

Newton is a force which gives us a mass of 1kg on acceleration of 1ms-2

From second law

F = \( \frac{mv\; – \;mu}{t} \)

Ft = mv – mu.

Recall that Ft = Impulse of a force

Impulse of a force = the change in momentum of the body.

Example 1:

A body of mass 5 kg that is moving at a velocity of 36m/s is suddenly hit by a force of 8 N for 0.05 seconds. Find the new velocity of the object.

Solution:

Values given:

Mass = 5 kg, velocity = 36 m/s, force = 8N, time = 0.05 second

Note:

We are asked to calculate the new velocity and we have been given the initial velocity of 36m/s.

Initial velocity = u = 36m/s

Fina velocity v = ? m/s

F = \( \frac{mv\: – \:mu}{t} \\ = \frac{m(v\: – \: u)}{t}\)

Since we are looking for v let’s make v – u, then v the subject of the formula

v – u = \( \frac{Ft}{m} \)

v = \( \frac{Ft}{m} \scriptsize \: + \: u\)

Substitute in the values given in the question

v = \( \frac{8 \; \times \; 0.05}{5} \scriptsize \: + \: 36 \\ = \frac{0.4}{5} \scriptsize \: + \: 36\)

v = 36.08 m/s

Example 2:

A ball of mass 0.8 kg dropped from a height of 14 m onto a wooden floor bounces back to a height of 4.5 m.
a. Calculate the change of momentum. 
b. If the ball is in contact with the floor for 0.3 seconds, what is the force exerted on the ball?

Solution:

Values given:

Mass of ball = 0.8kg, height = 14m, bounce back height = 4.5m,
time = 0.3 seconds,

To calculate the change in momentum, you must first calculate:

i. The final velocity v of the object as it hits the ground

ii. The bouncing velocity u

Therefore, final velocity of the object just before it hits the ground:

v² = u² + 2gs

Before the ball was dropped, its initial velocity was zero ( u = 0 m/s )

s = 14 m , g = 10 m/s²

Substitution the values into the equation         

v² = 0² + 2 x 10 x 14

v² = 280    

v = √280

v = 16.7 m/s

You have to calculate the velocity with which the ball bounced back

At the bounce back height of 4.5m, final velocity v = 0 m/s.

Then, Using the formula:

v² = u² + 2gv

Substitution:          

0² = u² – 2 x 10 x 4.5

Make u² the subject:    

u² = 2 x 10 x 4.5

u² = 90    

u = √90  

u = 9.5 m/s

Change in momentum = mv – mu

Note that Since the ball bounced back, its bouncing momentum is opposite to the momentum on hitting the ground. Therefore

Change in momentum = mv – ( – mu )

Change in momentum = mv + Mv

Substitute the values in to the equation

Change in momentum = 5 x 16.7 – ( -5 x 9.5)

Change in momentum = 83.5 + 47.5 = 131 kgm/s

To calculate force, we use the formula,

Force = \( \frac{change\:in\:momentum}{t} \)

Force = \( \frac{131}{0.3} \)

Force = 436.7 Newtons

Third Law of Newton:

Actions and reactions are equal and opposite. When a wooden block is resting on a table, the weight of the block is acting vertically down on the table while the table is also exerting equal and opposite forces to balance the downward force so that the block can remain at rest.

Newtons 3rd law of motion

This opposite force that allows the object to balance is called reactional force (W = -R).

  • Newton’s third law states, ‘to every action, there is an equal and opposite reaction.’
  • In other words, if item A exerts a force on object B, object B must respond with a force of similar magnitude and in the reverse direction.
  • This can be represented mathematically by the formula of Newton’s third law of motion as

Fa = – Fb

Fa ( force exerted by an object ‘a‘ on object ‘b‘)
Fb( force exerted by the object ‘b‘ on object ‘a‘)

Other examples of third law explanation are:

The recoil of a gun, jet propulsion and walking on the ground.

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