SUBJECT Physical Sciences WEEK 4 TOPIC Newton’s Second Law : Time 60 mins Lesson

LESSON SUMMARY FOR: DATE STARTED: DATE COMPLETED:

LESSON OBJECTIVES

At the end of the lesson learners should be able to:

• Discuss the relationship between net force and change in momentum • State Newton’s Law in a special case of constant mass

• Add all the forces in x – direction and y – direction for uniform velocity and static situations TEACHING and LEARNING ACTIVITIES

1. TEACHING METHOD/S USED IN THIS LESSON:

Induction method, Demonstration method, Question and Answer method

2. LESSON DEVELOPMENT 2.1 Introduction

a) PRE-KNOWLEDGE learners need understanding of the following:

b) BASELINE ASSESSMENT (educator to design a worksheet/ transparency or write questions on the board [preferably a worksheet to save time] to gauge the learners memory of their relevant prior knowledge)

QUESTIONS for the BASELINE ASSESSMENT [5 min]

(i) How will the momentum change if the force on the object is increased? (ii) What is the vector sum of the weight and the normal force of an object at rest? (iii) Under which condition(s) will the motion and the momentum of an object change? (iv) Compare the force acting on an object with the momentum of an object

Corrections

i) The momentum of an object will also increase

ii) They add up to zero since Normal force and the weight of an object are equal in magnitude and are different in their directions iii) If there is net (resultant) force acting on the object./ Unbalanced forces acting on the object

iv) There is a direct relationship between Momentum and Force. The rate at which momentum changes is equal to the net force applied to the object. A net force acting for a certain time produces change in momentum in the direction of net force

2.2 Main Body (Lesson presentation) [30 min]

From the definition of Newton’s second law: The net force on an object (or system of objects) equals to the rate at which the object's momentum changes,

t

v

m

F

∆

∆

=

There is a direct relationship between Momentum and net Force. The rate at which momentum changes is equal to the net force applied to the object. A net force acting for a certain time produces change of momentum in the direction of net force. Change in momentum comes as a result of change in velocity, and therefore acceleration is produced. Another definition of Newton’s second Law : Constant (net) resultant force produces acceleration in the direction of force. Acceleration is directly proportional to the net force , and acceleration is inversely proportional to the mass of an object.

For a constant mass

When an object accelerates, net Force applied is directly proportional to the acceleration it produces. If the force is increased, the rate of change of velocity increases. The acceleration of an object is inversely proportional to the mass of an object. If mass is increased, acceleration of an object will decrease . For the constant mass of an object, the equation

ma

F

=

becomes relevant and appropriate. And must be applied separately in the x- and y -direction s. In systems with more than one object of interest, a free body diagram for each object must be drawn and Newton second law must be applied to each object separately.

For objects that are stationary and objects with zero acceleration (uniform velocity), all forces in the x-direction together with all the forces in y-direction must add up to zero. How is principle of superposition used with Newton’s 2nd Law?

F1 + F2 + F3 + ---- = F(total) if m is constant and then a1 +a2 + a3 +--- = a(total Example

Solve the following force problems

1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m∙s-2_? Fnet = ma

= (60.0 )(1.15 ) = 69.0 N

2 . A net force of 225 N accelerates a bike and rider at 2.20 m∙s-2_{. What is the combined mass of the bike and rider?} Fnet = ma ,

225 = (2.20 ) m = 102 , 27kg

2.2 Learners Activities [15 min]

2.2.1 A box weighing 70 N rests on a table. A rope attached to that box runs up over a pulley and a second box is attached to the other end. Determine the Normal force on the box on that table if the hanging box weighs (a) 30 N, (b) 60 N, and (c) 90 N.

2.2.2 Two blocks on a horizontal surface are in contact with each other as shown. The surface has friction. A force F is applied to block 1 2.2.3 calculate the mass of a body when a force of 225 N produces an acceleration of 2.5 m∙s-2_?

2.2.4 What is the acceleration of a 4 kg box if an unbalanced force of 12 N acts on it?

2.2.5 When an object of unknown mass has an unbalanced force of 200N acting on it , it accelerates at 2 m∙s-2_{. What is its mass?} Corrections

2.2.1 For the 70 N block: Since F = 0, th e n T + FN - 70 N = 0 FN = 70 N - T

For the hanging block: Since F = 0, th e n T = Fg Therefore: FN = 70 N - Fg

a) Fg= 30 N

FN = 70 N - Fg = 70 N - 30 N = 40 N b) Fg= 60 N

c) Fg= 90 N

FN = 70 N - Fg = 70 N - 90 N = -20 N =0 The block has lifted off the table, therefore the normal force is equal to zero 2.2.2

2.2.3 Force (F) = 225 N Acceleration (a) = 2.5 m/s2 Force (F) = ma

Mass of the body = 90 kg 2.2.4 F = ma 12 = (4)a a = 12/4 = 3 m∙s-2 2.2.5 F = ma 200 = m(2) m = 200/2 = 100 kg 2.3 Conclusion

Reflection/Notes:

Name of Teacher: HOD:

Sign: Sign:

Date: Date:

Activity to Re-enforce lesson (Educator may summarise the main aspects of the lesson) [5 min.]

HOMEWORK QUESTIONS/ ACTIVITY (educator must give learners a few questions to answer at home by either writing them on the chalkboard or giving an exercise from the prescribed textbook) [30 min]