SS2 FURTHER MATHS SECOND TERM DYNAMICS

FURTHER MATHEMATICS, SS2, SECOND TERM ,

WEEK SIX

SS2 FURTHER MATHS SECOND TERM

DYNAMICS

CONTENT

(a) Newton’s Laws of Motion

(b) Motion along Inclined Plane.

(c) Motion of Connected Particles

SUB TOPIC: NEWTON’S LAW OF MOTION

PARTICLE: This is a body or thing that has a negligible dimension. It is usually denoted by a point or dot.

MASS OF A BODY: This is the amount of matter contained in a body. It is measured in Kilogram me (Kg).

LINEAR MOMENTUM: The linear momentum of a particle is the product of the mass of the particle and its velocity. It is also known as momentum.

If mass= m, velocity= v and momentum= p.

Then, p=mv

NEWTON’S LAWS

FIRST LAW: Everybody or object remains at rest or of uniform motion unless compelled by an external force to act otherwise.

SECOND LAW: The rate of change of momentum of a body is proportional to the applied force and is in the direction of the force. (F =ma)

If a force F acts on a body of a mass m (kg) it produces acceleration (a) in the mass given by the relation. F=ma

THIRD LAW: To every action, there is always an opposite and equal reaction.

Examples:

1. A force P acts on a body of mass 5kg on a smooth horizontal floor. If it produces an acceleration of 4.5ms^-2, find the magnitude of P.

Solution: a=4.5ms^-2 = 4 ½ ms^-2 = 9/2 ms^-2.

M= 5kg.

P= ma = 5 X 9/2 = 22.5N

1. A body of mass 100kg is placed in a lift. find the reaction between the floor of the lift and the body when the lift moves upward
2. At constant velocity
3. With an acceleration of 3.5ms^-2 (g=10ms^-2)

Solution

Let R₁ be the reaction between the floor of the lift and the load when the lift moves upward at constant velocity

The net force on the load = R₁ – mg

But,

R₁ – mg = ma

R₁– mg = 0 (a = 0).

1. R₁ = mg = 100 x 9.8 = 980 N
2. Let R₂ = reaction between the floor and the load at constant acceleration.

R₂ – 100 x 9.8 = 100 x 2

R₂ – 980 = 200

R₂ = 200 + 980

R₂ = 1180N

CLASS ACTIVITY:

1. A body of mass 20kg is placed in a lift. Find the reaction between the floor of the lift and the body when the lift moves downward with retardation of 2. 5m/s^2. (Take g= 10m/s²).
2. A body of mass 15kg is placed on a smooth plane which is inclined at 60⁰ to the horizontal. Find (a) the acceleration of the body as it moves down the plane
3. The velocity that the body attains after 5seconds if (i) it start from rest; (ii) it moves with an initial velocity of 4 ms^-1 .

SUB TOPIC: MOTION ALONG AN INCLINED PLANE.

DIAGRAM

F – mg sin α = ma, if F > mg sinα (a = acceleration upward plane)

Mg sinα – F = ma, if F <mgsinα (a = acceleration downward plane)

Example 1. An object whose weight is 20kg is placed on a smooth plane inclined at 60⁰ to the horinzontal.

Find: (a) the acceleration of the object as it moves down the plane ; (b) the velocity attained after 6 seconds if : (i) it starts from rest ; (ii) it moves with an initial velocity of 10ms^-1. (Take g= 10ms^-1)

Solution

Net force acting on the body down the plane is mgsin 60⁰

The force acting on the body upward is zero.

Hence, mg sin 60⁰= 20 x a.[mediator_tech]

10 x 10 sin 60⁰ = 20 x a

100√3/2 = 20 a

50 √3 = 20 a

a=25√3 ms^-2

OR a = 21. 7ms^-1

b(i) when the body start from rest u=0

using equation of motion

v = u + at

v = 0 + 10 x 6 = 60 ms^-1

b(ii) if the body move with an initial velocity of 10 ms^-1 , then;

v = u + at

v = 10 + 6 x 10 = 70 ms^-1

CLASS ACTIVITY:

1. A body of mass m is placed on the surface of a smooth plane which is inclined at an angle to the horizontal. A force F whose line of action is parallel to the surface of the inclined plane acts on the body to first prevent it from slipping down the plane. If R is the reaction between the surface of the inclined plane and the body, show that F = R tan θ

SUB TOPIC: MOTION OF CONNECTED PARTICLES

Example: Two vehicles of mass m₁ and m₂ are connected by an inextensible chord whose mass can be neglected. The vehicle of m₁ has broken down and the vehicle of mass m₂ is towing the vehicle of mass m₁, with a reactive force of T Newton. The two vehicles are moving with acceleration a ms^-2 . If T is the tension in the chord and any frictional force is to be neglected, show that

(i) a = F/ m₁+ m₂ (ii) T = m₁f/m₁+m₂

Solution

Vehicle with mass m₁; T = m₁ a ……………………… (1)

Vehicle with mass m₂ ; F – T = m₂a …………………..(2)

Adding (1) and (2)

F – T + T = m₁a + m₁a

F = a ( m₁+ m₂)

a = f/m₁ + m₂

(ii) substituting the value of “a” into equation …………………….. (i)

Therefore T= m₁ x f/m₁ + m₂ = m₁f/m₁ + m₂

CLASS ACTIVITY:

1. Two particle of masses 10kg and 8kg are connected by a light in extensible string which is passed over a light frictionless pulley. Find the tension in the string and the acceleration with which the particles move when released.
2. Two bodies of masses 2m and 3m are connected by a light in elastic string which is passed over a smooth fixed pulley. The body of mass 2m lies on a smooth horizontal table while the body of mass 3m. hangs freely. If the two bodies move with an acceleration a and T is the tension in the string, find expression for
3. a in terms of g
4. T in terms of g and m where g is the acceleration due to gravity.

PRACTICE EXERCISE:

Objectives

1. Any body whose dimensions can be neglected is called a ……….
2. ……… is the amount of matter contained in a body
3. ……… is the product of the mass of the particle and its velocity.
4. Action and reaction are ………… and ………..

Essay

1. A force P acts on a body of mass 5kg on a smooth horizontal floor. If it produces an acceleration of 4.5m/s², find the magnitude of P.
2. A body of mass 100kg is placed in a lift. Find the reaction between the floor of the lift and the body when the lift moves upward: (i) at constant velocity (ii) with an acceleration of 3.5m/s², (Take g = 10m/s²).
3. A body of mass 50kg is placed in a lift. If the lift moves upward with a retardation of 2m/s²,find the reaction between the floor of the lift and the body. (Take g = 9.8m/s²).
4. Two bodies of masses 2m and 3m are connected by o light inelastic string which is passed over a smooth fixed pulley. The body of mass 2m lies on a smooth horizontal table while the body of mass 3m hangs freely. If the two bodies moves with an acceleration a and T is the tension in the string, find the expression for: (i) a in terms of g (ii)T in terms of g and m where g is the acceleration due to gravity.

ASSIGNMENT

1. State the three law of motions
2. Derive the formula F = ma
3. What will be an acceleration of mass of 50kg when acted upon by a force of 30N?
4. A ball fall from rest down a smooth plane inclined at an angle to the horizontal. How long in seconds, will it take the ball to cover a distance of dm along the plane?
5. A body of mass 2kg moving with the velocity 5m/s due East collides with another body with velocity 4m/s. if the heavier body is brought to rest by the collision, find the velocity of the lighter body after collision.

KEY WORDS

• PARTICLE
• MASS
• MOMENTUM
• VELOCITY
• INCLINED PLANE
• ACCELERATION
• DECCELERATION
• RETARD
• MOTION
• COLLIDES
• GRAVITY
• NEWTON
• FORCE
• MAGNITUDE
• TENSION
• INEXTENSIBLE