NEWTON'S LAWS OF MOTION [SS2 1ST TERM]

 

NEWTON'S LAW OF MOTION

Sir Isaac Newton ( a 17th century Scientist) put forward a variety of laws that explain why objects move or don't move as they do. These 3 laws have become known as Newton's 3 laws of motion. These 3 physical laws that together laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to said forces.

NEWTON'S FIRST LAW OF MOTION

The law states that every object continues in its state of rest or uniform motion in a straight line unless acted upon by an external force. It may be seen as a statement of inertia, tendency of  objects to remain in their state of motion unless a force act to change the motion. Any change in motion requires an acceleration.

There are two parts to this statement: one that predicts the behaviour of stationary objects and the other that predicts the behaviour of moving objects. The behaviour of all objects can be described by saying that objects tend to keep on doing what they are doing unless acted upon by an unbalanced force.

Application of Newton's 1st law

1.        Behaviour of tea in a tea cup filled to the rim while starting a car from rest or while bringing a car to rest from a state of motion.

2.        Headrest are placed in a car to prevent whiplash injuries during rear-end collision.

3.        Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator.

4.        Inertia in a automobile when it is breaking to stop.

NEWTON'S SECOND LAW OF MOTION

It states that the rate of change of momentum is proportional to the applied force and take place in the direction of that force.

This law provides the explanation for the behaviour of objects upon which the force do not balance. The law states that unbalanced forces cause objects to accelerate with an acceleration that is directly proportional to the net force  and inversely proportional to the mass.

Mathematically,

F α change in momentum  ⁄ time taken for the change

IMPULSE (I): is defined as the product of the average force acting on a particle and the time during which it acts. It is usually associated with collision. I = F x t.

MOMENTUM (p): of a body is defined as the product of its mass, m, and its velocity, v. its unit is Kgms-1. it is a vector quantity. P = m x v.

Example

1.        What is the impulse of a body lifted by a force of 30N within time of 2 seconds.

Solution

I = F x t = 30N x 2s = 60Ns

2.        A body of mass 0.5Kg initially at rest and is subjected to a force of 2N for 1 second. Calculate (a) the change in momentum of the body during the time. (b) a change in kinetic energy of the body during the time.

Solution

a). Momentum  = Impulse , momentum = F x t = 2N x 1s = 2Ns

b). Momentum = m x V, V = momentum / mass = 2Ns / 0.5Kg = 4ms-1.

Kinetic energy = 1/2 mv2  = 1/2 x 0.5Kg x (4ms-1)2 = 4J.

NEWTON'S THIRD LAW OF MOTION

It states that action and reaction are equal and opposite. It can also be stated as to every action there is an equal and opposite reaction.

The laws implies that all forces exist in pairs. If one object A exert a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal and opposite: FA = - FB. The law means that all forces have interactions between different bodies, and thus that there is no such things as unidirectional force or a force that acts on only one body.

For example, in swimming , a person interacts with the water, pushing the water backward, while simultaneously pushing the person forward- both the person and the water push against each other. These forces depend on friction.

CONSERVATION OF LINEAR MOMENTUM

The principle of the conservation of linear momentum states that in any system of colliding objects the total momentum is always conserved provided that there is no net external force acting on the system.

Thus, the conservation of momentum can be stated as a system in which the momentum before collision is equal after collision.

COLLISION

There are two principal types of collisions. The elastic and inelastic collision.

Elastic Collision: in this type of collision, both energy and momentum are conserved i.e they are the same before and after collision. Example , a tennis ball that bounce back to its original height after hitting a surface.

M₁U₁ + M₂U₂ = M₁V₁ + M₂V₂ ;  1/2M₁U₁² + 1/2M₂U₂² = 1/2M₁V₁² + 1/2M₂V₂²

Inelastic Collision: The kinetic energy decreases after collision but the momentum is still conserved. Inelastic collision, the colliding bodies stick together and move as a unit after collision. This means that the velocities of the two bodies after collision are V1 = V2 = V

From conservation of linear momentum we have

M₁U₁ + M₂U₂ = (M₁ + M₂)V ;  1/2M₁U₁² + 1/2M₂U₂² = 1/2(M₁ + M₂ )V²

For a complete inelastic collision, the kinetic elastic before collision is greater than the kinetic energy after collision.

Example

1.  A ball of mass 0.5 kg moving at 10 ms-1 collides with another ball of equal mass at rest. If the two balls  move off together after the impact, calculate their common velocity.

Solution

M1 = 0.5kg , V1 = 10 m/s , M2 = 0.5Kg,  V2 =0, V=?

M₁U₁ + M₂U₂ = (M₁ + M₂)V ;   ( 0.5 x 10 ) + (0.5 x 0) = (0.5+0.5)V ;  5 + 0 = V; V = 5m/s.

APPLICATION OF NEWTON'S AND CONSERVATION OF MOMENTUM LAWS

1.        Recoil of a gun  2. Jet and Rocket propulsion

Lift / Elevator

Two forces acts on a man standing in a lift or elevator:

i.  The man's true weight W (mg) acting vertically downward.

ii. The reaction R of the floor of the elevator on the man acting upward.

(a) when a lift is moves upward with uniform acceleration R = m(g + a).

(b) when a lift is moves downward with uniform acceleration. R = m(a -g).

(c ) when a lift is at rest R = mg

(d) when a lift is moving up or down with constant velocity R = mg

(e) when a lift is falling freely, the resultant force becomes zero (weightlessness). R = 0.

 Example

A rocket of mass 1000 Kg containing a propellant gas of 3000 Kg is to launched vertically. If the fuel is consumed at a steady rate of 60 Kgs-1, calculate the least velocity of the exhaust gases if the rocket and content will just  lift off the launching pad immediately after firing.

Solution

M_R = 1000 Kg , M_P = 3000 Kg; mass of gas per second = 60kgs^-1.

W_max  = W_R  +  W_P  = (1000 + 3000) x 10  =  40, 000 Kgms^-2.     Least velocity =   =  = 666.67 m/s

Assignment

1.  A ball of mass 6.0 Kg moving with a velocity of 10.0 m/s collides with a 2.0 Kg ball moving in the opposite direction with a velocity of 5.0 m/s. after the collision the two balls coalesce and move in the same direction. Calculate the velocity of the composite body.

2. An object of mass 1 Kg falls a distance of 5 m onto a horizontal surface and rebounds to a vertical height of 2 m. calculate the change in momentum.