Concise Physics Class 9 ICSE Solutions for Chapter 3 - Laws of Motion

Table of contents covered in Chapter 3 Laws of Motion:

• 3.1 - Effect of force

• 3.2 - Contact forces

• 3.3 - Non-Contact forces

• 3.4 - Newton’s first law of motion

• 3.5 - Mass and inertia

• 3.6 - Inertia types and its illustrative examples

• 3.7 - Linear momentum

• 3.8 - Change in momentum

• 3.9 - Rate of change of momentum

• 3.10 - Newton’s second law of motion

• 3.11 - C.G.S and S.I. units of force

• 3.12 - Newton’s 2nd law of motion in terms of rate of change of momentum

• 3.13 - Newton’s 3rd law of motion

• 3.14 - The universal law of Gravitation

• 3.15 - Force due to Gravity

• 3.16 - Acceleration due to gravity

• 3.17 - Freefall

• 3.18 - Mass and weight

• 3.19 - Gravitational units of force

List of Exercise in Chapter 3 Laws of Motion

• Exercise - 3(A) - 10 Questions - Page.no - 59

• Exercise - 3(B) - 24 Questions - Page.no - 63

• Exercise - 3(C) - 36 Questions - Page.no - 69

• Exercise - 3(D) - 18 Questions - Page.no - 73

• Exercise - 3(E) - 47 Questions - Page.no - 79

Some Sample Questions and Answers

1: A ball is thrown vertically upwards. It goes to the height of 40 m and returns to the ground. Taking g = 10 ms2.Find:

1. The initial velocity of the ball

2.  The final velocity of the ball

3. The total journey of the ball

Our faculties concisely explained this question.

Given: h = 20 m, g = 10 ms2, v = 0

To find - Initial velocity (u), 

Final velocity (v), and 

Time (t)

Solution:

(a) Initial velocity

 Using the third equation of motion:

V2 = u2 + 2ah

Here,  a = g because the ball falls freely under Gravity.

V2 = u2 + 2gh

0 = u2 +2 x 10 x 20

U = - \[\sqrt{400}\] = -20 m/s

The negative sign indicates that motion is against Gravity.

(b) Final velocity

When the ball is coming down from the highest point. Then its final velocity is:

u = 0, h = 20 m, g = 10ms2 

As V2 = u2 + 2ah

V2- u2 = 2ah

V2 - 0 = 2 x 10 x 20 

\[\Rightarrow v = \sqrt{2}\times 10\times 20 = 20 m/s\]

(c) Total journey (Total time)

2u = g x t = t = 2u/ g

T = \[\frac{2\times 20}{10}\] = 4s

Q2: How is the gravitational force between two bodies affected if the separation between them is doubled?

Our faculties explained this question in the following way:

Solution: If the distance between two mass m and m doubles, then force reduces to 1/4th of its original value, i.e.,

If  \[F \propto \frac{1}{r^{2}}\] ….(1)

Then, \[F \propto \frac{1}{2r^{2}}\]

\[\Rightarrow F \propto \frac{1}{r^{2}}\]…(2)

Now, (2) ÷ (1)

\[\frac{F{}'}{F} = \frac{r^{2}}{4r^{2}} = \frac{1}{4}\]

= F' = F/4

We see that the new value F’ is ¼ of F.

You saw how these solutions were explained concisely. For more such solutions, download Selina concise Class 9 Physics ICSE solutions of chapter 3 Laws of motion solved by our faculties.

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