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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/s2. (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

 

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