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

v_f = v_i + at

S = v_it + 1/2 at²

2aS = v_f² – v_i²

V_av = (v_f + v_i)/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_i = 0 ms^-1

Distance S = 1 km = 1000 m

Time = 100 s

Final velocity V_f =?

S = V_i 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 1^st equation of motion

V_f = V_i + at

V_f = 0 + 0.2 x 100

V_f = 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 = V_{i =} 10 ms^-1

Acceleration a = 0.2 ms^-2

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

Distance S =?

Final velocity V_f =?

S = V_i x 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 1^st equation of motion

V_f = V_i + at

V_f = 10 + 0.2 x 30

V_f = 10 + 6

V_f = 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 = V_{i =} 30 ms^-1

Acceleration due to gravity g = -10 ms^-2

Time to reach maximum height = t = 3 s

Final velocity v_f = 0 ms^-1

Maximum height attained by the ball S =?

Time taken to return to ground t =?

S = V_i x 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

2.5 A car moves with a uniform velocity of 40 ms^-1 for 5 s. It comes to rest in the next 10 s with uniform deceleration. Find deceleration and total distance travelled by car.

(-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 =?

V_f = V_i + at

Or                     

at = V_{f –} V_i

a = (V_{f –} V_i)/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 3^rd equation of motion

2aS₂ = V_f² – V_i²

S₂ = (V_f² – V_i²)/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 V_i = 0 ms^-1

Acceleration a = 0.5 ms^-2

Distance S = 100 m

Final velocity V_f =?

2aS = Vf² – V_i²

2 x 0.5 x 100 = Vf² – 0

Or           

100 = Vf²

Or             

Vf² = 100 ms^-1……………………..(I)

Speed in kmh^-1:

From (I) we get;

v_f = 10 x 3600/1000

v_f = 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 = V_i = 0ms^-1

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

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

S₁ = V_av x t

S₁ = (V_f + V_i)/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 = V_f = 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_f = 13.333 ms^-1

Final velocity = V_i = 0 ms^-1

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

S₃ = V_av x t

S₃ = (V_f + V_i)/2 x 180

S₃ = 13.333 + 0/2 x 180

S₃ = 6.6665 x 180

S₃ = 1199.97 = 1200 m

Total distance = S = S₁ + S₂ + S₃

S = 800 + 4000 + 1200

S = 6000 m

2.8 A cricket ball is hit vertically upwards and returns to the ground 6 s later. Calculate

(i)       Maximum height reached by the ball

(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 = V_f = 0 ms^-1

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

The maximum initial velocity of the ball = V_i =?

Since,     

V_f = V_i + gxt

V_i = V_f – gxt

V_i = 0 – (-10) x 3

V_i = 30 ms^-1

Now using 3^rd equation of motion

2aS = V_f2 – V_i2

S =  (V_f2 – V_i2)/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 = V_{i =} 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 = S₁ =?

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

2aS = V_f² – V_i²

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_f = 0              ,           S₁ =?

2aS = V_f² – V_i²

2 (-0.333) x S₁ = (0)² – (13.33)² 

-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 = v_i = 96 kmh^-1 = 96 x 1000/3600 = 96000/3600 = 26.66ms^-1

Final velocity = v_f = 0 ms^-1

a = -0.333 ms^-2

time = t =?

Or   v_f = v_i + at

Or   at = v_f – v_i

t = (v_f – v_i)/a

t = (0 – 13.33) / (-0.333)

t = 2 x 40

t = 80 s

Class 9 Physics Full Book Numerical Problems with Solution