Tuesday, October 23, 2007

Study guide H C Verma JEE Physics Ch. 5 NEWTON'S LAWS OF MOTION

Newton’s laws of motion;
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Main Sections

5.1 Newton's First law
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia

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Subsections under each main section

5.1 First law of motion
Inertial frames other than earth
5.2 Newton's second law
5.3 Working with Newton's laws
5.4 Newton's third law of motion
5.5 Pseudo forces
5.6 The Horse and the cart
5.7 Inertia

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Study Plan

Day 1

5.1 First law of motion


Example 5.1

W.O.E. 1 and 2

http://iit-jee-physics.blogspot.com/2008/08/newtons-laws-of-motion-study-plan.html

Day 2

5.2 Newton's second law
5.3 Working with Newton's laws

W.O.E. 3 and 4

Points to be noted
http://iit-jee-physics.blogspot.com/2008/09/newtons-laws-of-motion-study-plan.html

Day 3

5.4 Newton's third law of motion

W.O.E. 5,6

Day 4

5.5 Pseudo forces

W.O.E. 7,8

Day 5

5.6 The Horse and the cart
5.7 Inertia

W.O.E. 9,10,11

Day 6

Chapter 5
Objective I

Day 7
Chapter 5
Objective II

Day 8
Chapter 5
Exercises 1 to 5

Day 9
Chapter 5
Exercises 6 to 10

Day 10
Chapter 5
Exercises 11 to 15

Day 11
Chapter 5 Revision
Exercises 16 to 18

Day 12
Chapter 5 Revision
Exercises 19 to 21

Day 13
Chapter 5 Revision
Exercises 22 to 24

Day 14
Chapter 5 Revision
Exercises 25 to 27

Day 15
Chapter 5 Revision
Exercises 28 to 30

Day 16
Chapter 5 Revision
Exercises 31 to 33

Day 17
Chapter 5 Revision
Exercises 34 to 36

Day 18
Chapter 5 Revision
Exercises 37 to 39

Day 19
Chapter 5 Revision
Exercises 40 to 42

Day 20
Chapter 5 Revision
Attempt Questions for short Answer 1 to 17



Audiovisual lecture
Newton's first law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson03/Container.html

Newton's second law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson04/Container.html

Newton's thrid law
http://www.curriki.org/nroc/Introductory_Physics_1/lesson05/Container.html

Newton's laws - applications
http://www.curriki.org/nroc/Introductory_Physics_1/lesson06/Container.html

Websites

plasma4.sr.unh.edu/ng/phys407/phys407-6-23-05.pdf
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JEE Question 2007 Paper II

Statement - 1

A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table.

Because

Statement - 2

For every action there is an equal an opposite reaction.

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for statement – 1
(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is Not a correct explanation for Statement – 1.
(C) Statement – 1 is True, Statement – 2 is False
(D) Statement – 1 is False, Statement – 2 is True

Correct choice: (B)
S1 is correct.
S2 is also correct but it does not explain S1. S1 is possible because of inertia.
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JEE Question 2007 Paper I

Two particles of mass m each are tied at the ends of a light string of
length 2a. The whole system is kept on a frictionless horizontal surface
with the string held tight so that each mass is at a distance ‘a’ from the
centre P (as shown in the figure. *figure not given in this post). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result,the particles move towards each other on the surface. The magnitude of
acceleration, when the separation between them becomes 2x, is

(A) (F/2m)(a/SQRT(a^2-x^2))
(B) (F/2m)(x/SQRT(a^2-x^2))
(C) (F/2m)(x/a)
(D) (F/2m)(SQRT(a^2-x^2)/x)

Solution: B

Imagine the string being pulled upward from midpoint, but only horizontal movement is there.Assume that the string makes with the horizontal surface is θ (theta) , and the tension in the string is T.

F = 2Tsinθ (Tension in each side of the string)
Force in the horintal direction on each particle = F' = ma = T Cosθ
So a = T Cosθ/m
T = F/2Sinθ
a = (F/2m)cotθ

In the triange formed between half of the string and horizontal surface,and the imaginary vertical line through which force is being applied, diagonal is half of the string hence a, and the horizontal side is x. The third side is SQRT(a^2-x^2)
Hence cotθ = x/SQRT(a^2-x^2)

a = (F/2m)(x/SQRT(a^2-x^2))
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