# Class 9 Physics Chapter 2 Numerical Problems

Class 9 Physics Chapter 2 numerical problems are according to the new syllabus of the Punjab Board. Chapter 2 is related to Kinematics which is the branch of Physics that deals with the study of motion without discussing the effects.

## Class 9 Physics Chapter 2 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

S = v x t

vf = vi + at

S = vit + 1/2 at2

2aS = vf2 – vi2

Vav = (vf + vi)/2

### Numerical Problems Chapter 2 – Kinematics

#### 2.1 A train moves with a uniform velocity of 36 kmh-1 for 10 s. Find the distance travelled by it.                                                                                                                                                     (100 m)

Solution:

Velocity = v = 36 kmh-1 = 36 x 1000/ 60 x 60 = 36000/ 3600 = 10 ms-1

Time t = 10 s

Distance = S =?

S = v x t

S = 10 x 10

S = 100 m

#### 2.2 A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.?   (20 ms-1)

Solution:

Initial velocity V= 0 ms-1

Distance S = 1 km = 1000 m

Time = 100 s

Final velocity Vf =?

S = Vi x t +1/2 x a x t2

1000 = 0 x 100 + 1/2 x a x (100)2

1000 = 1/2 x 10000a

1000 = 5000a

a = 1000/5000 = 0.2 ms-2

Now using 1st equation of motion

Vf = Vi + at

Vf = 0 + 0.2 x 100

Vf = 20 ms-1

#### 2.3 A car has a velocity of 10 ms-1. It accelerates at 0.2 ms-2 for half minute. Find the distance travelled during this time and the final velocity of the car. (390 m, 16 ms-1)

Solution:

Initial velocity = Vi = 10 ms-1

Acceleration a = 0.2 ms-2

Time t = 0.5 min = 0.5 x 60 = 30 s

Distance S =?

Final velocity Vf =?

S = Vx t + 1/2 x a x t2

S= 10 x 30 + 1/2 x 0.2 x (30)2

S = 300 + 1/2 x 0.2 x 90

S = 300 + 1/2 x 2/10 x 90

S = 300 + 90

S = 390 m

Using 1st equation of motion

Vf = Vi + at

V= 10 + 0.2 x 30

Vf = 10 + 6

Vf = 16 ms-1

#### 2.4 A tennis ball is hit vertically upward with a velocity of 30 ms-1, it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to the ground? (45 m, 6 s)

Solution:

Initial velocity = Vi = 30 ms-1

Acceleration due to gravity g = -10 ms-2

Time to reach maximum height = t = 3 s

Final velocity v= 0 ms-1

Maximum height attained by the ball S =?

S = Vx t + 1/2 x g x t2

S = 30 x 3 + 1/2 x (-10) x (3)2

S = 90 – 5 x 9

S = 90 – 45

S = 45 m

Total time = time to reach maximum height + time to return to the ground

t = 3 s + 3 s

t = 6 s

#### (-4 ms-2, 400 m)

Solution:

Initial velocity = Vi = 40 ms-1

Time = t = 5 s

Final velocity = Vf = 0 ms-1

Time = 10 s

deceleration a =?

total distance S =?

Vf = Vi + at

Or

at = Vf – Vi

a = (Vf – Vi)/t

a = 0 – 40/10

a = -4 ms-2

Total distance travelled = S = S1 + S2

By using this relation

S1 = v x t

S1 = 40 x 5

S1 = 200 m ………………………………. (i)

Now by using 3rd equation of motion

2aS2 = Vf2 – Vi2

S2 = (Vf2 – Vi2)/2a

S2 = [(0)2 – (40)2]/2 x (-4)

S2 = -1600/-8

S2 = 200 m ……………………………………… (ii)

From (i) and (ii) we get;

S = S1 + S2

Or

S = 200 m + 200 m

S = 400 m

#### 2.6 A train starts from rest with an acceleration of 0.5 ms-2. Find its speed in kmh-1, when it has moved through 100 m.   (36 kmh-1)

Solution:

Initial velocity Vi = 0 ms-1

Acceleration a = 0.5 ms-2

Distance S = 100 m

Final velocity Vf =?

2aS = Vf2 – Vi2

2 x 0.5 x 100 = Vf2 – 0

Or

100 = Vf2

Or

Vf= 100 ms-1……………………..(I)

Speed in kmh-1:

From (I) we get;

v= 10 x 3600/1000

vf = 36 kmh-1

#### 2.7 A train starting from rest, accelerates uniformly and attains a velocity of 48 kmh-1 in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance travelled by train. (6000m)

Solution:

Case – I:

Initial velocity = Vi = 0ms-1

Time = t = 2 minutes = 2x 60 = 120 s

Final velocity = Vf = 48 kmh-1 = 48 x 1000/3600 = 13.333 ms-1

S1 = Vav x t

S1 = (Vf + Vi)/2 x t

S1 = 13.333 + 0/2 x 120

S1 = 6.6665 x 120

S1 = 799.99 m = 800 m

Case – II:

Uniform velocity = Vf = 13.333 ms-1

Time = t = 5 minutes = 5 x 60 = 300 s

S2 = v x t

S2 = 13.333 x 300

S2 = 3999.9 = 4000 m

Case – III:

Initial velocity = V= 13.333 ms-1

Final velocity = Vi = 0 ms-1

Time = t = 3 minutes = 3 x 60 = 180 s

S3 = Vav x t

S3 = (V+ Vi)/2 x 180

S3 = 13.333 + 0/2 x 180

S3 = 6.6665 x 180

S3 = 1199.97 = 1200 m

Total distance = S = S1 + S+ S3

S = 800 + 4000 + 1200

S = 6000 m

#### (ii)      initial velocity of the ball                                                   (45m, 30 ms-1)

Solution:

Acceleration due to gravity = g = -10 ms-1 (for upward motion)

Time to reach maximum height (one-sided time) = t = 6/2 = 3 s

Velocity at maximum height = Vf = 0 ms-1

The maximum height reached by the ball S = h =?

The maximum initial velocity of the ball = V=?

Since,

V= Vi + gxt

V= V– gxt

Vi = 0 – (-10) x 3

Vi = 30 ms-1

Now using 3rd equation of motion

2aS = Vf2 – Vi2

S =  (Vf2 – Vi2)/2a

S = [(0)2 – (30)2/2] x (-10)

S = -90/-20

S = 45 m

#### 2.9 When brakes are applied, the speed of a train decreases from 96 kmh-1 to 48 kmh-1 in 800 m. How much further will the train move before coming to rest? (Assuming the retardation to be constant).                                                                                                (266.66 m)

Solution:

Initial velocity = Vi = 96 kmh-1 = 96 x 1000/3600 = 96000/3600 = 26.66ms-1

Final velocity = Vf = 48 kmh-1 = 48 x 1000/3600 = 48000/3600 = 13.33ms-1

Distance = S = 800 m

Further Distance = S=?

First of all, we will find the value of acceleration a

2aS = Vf2 – Vi2

2 x a x 800 = (13.33)2 – (26.66)2

1600a = (13.33)2 – (26.66)2

1600a = 177.8 – 710.8

1600a = -533

a = -533/1600 = -0.333ms-2

Now, we will find the value of further distance S1:

V= 0              ,           S=?

2aS = Vf2 – Vi2

2 (-0.333) x S1 = (0)2 – (13.33)

-0.666 S1 = -(13.33)2

S1 = -177.7/-0.666

S1 = 266.66m

#### 2.10 In the above problem, find the time taken by the train to stop after the application of brakes.                                                                                                                              (80 s)

Solution:

Initial velocity = vi = 96 kmh-1 = 96 x 1000/3600 = 96000/3600 = 26.66ms-1

Final velocity = vf = 0 ms-1

a = -0.333 ms-2

time = t =?

Or   vf = vi + at

Or   at = vf – vi

t = (vf – vi)/a

t = (0 – 13.33) / (-0.333)

t = 2 x 40

t = 80 s

Class 9 Physics Full Book Numerical Problems with Solution

#### By Ahsa.Pk

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