# Lesson Notes By Weeks and Term - Senior Secondary 1

Elastic properties of solids

Term: 2nd Term

Week: 11

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject:      Physics

Topic:-       Elastic properties of solids

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

1. Define elasticity
2. State Hooke’s law
3. Solve exercises on the elasticity of solids

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

 PRESENTATION TEACHER’S ACTIVITY STUDENT’S ACTIVITY STEP 1 INTRODUCTION The teacher reviews the previous lesson on the states of matter Students pay attention STEP 2 EXPLANATION He defines and explains elasticity and states Hooke’s law. Students pay attention and participates STEP 3 DEMONSTRATION He solves exercises on the elasticity of solids Students pay attention and participate STEP 4 NOTE TAKING The teacher writes a summarized note on the board The students copy the note in their books

NOTE

ELASTIC PROPERTIES OF SOLIDS
Elasticity is the ability of a substance to regain its original shape and size after being distorted by an external force.

HOOKES LAW

Hooke’s law states that Provided that the elastic limit of an elastic material is not exceeded, the extension of the material is directly proportional to the applied force or load.

F α e

F= Ke

Where

F= Force

K= Stiffness or force constant or elastic constant

e = Extension

Example:

Calculate the elastic constant of an elastic cord stretched by a loan of 8.0N by 250cm.

Solution:

F=8N

e = 250cm = 2.5m

F = ke

K =    F

e

= 8

2.5

= 3.2 Nm-2

Example: A Spring extends to 1.86cm when a mass of 20g was hung from it, if Hooke’s law is obeyed. What is the extension when additional mass of 10g was hung, given that original length of the string is 1.0 cm.

Solution:

L2 = 1.86cm

L1 = 1.0 cm

M1 = 20g = 0.2N

M2 = 20 + 10 = 30g = 0.3N

e = L2 – L1

1.86 – 1.00 = 0.86cm

=0.0086m

For the first force,

F1 = 0.2N

e1 = 0.0086m

F1 = K e1

K =   F

e

K=  0.2

0.0086

K = 232.56Nm-2

F2 = k e2

e2 = F =      0.3

k         232.56

= 0.0129m

EVALUATION:   1. Define elasticity

1. State Hooke’s law
2. How much force is needed to pull a spring with a spring constant of 20 N/m a distance of 25 cm?

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively