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 Draw the representative lines of the following vectors:
(a) A velocity of 400 m s⁻¹, making an angle of 60^° with the x-axis.
(b) A force of 50 N making an angle of 120^° with the x-axis.

Solution:      

a) Velocity vector (400 m s⁻¹ at 60° with x-axis)

• This would be drawn as an arrow with:
  • Length proportional to 400 units
  • Angle of 60° above the positive x-axis

b) Force vector (50 N at 120° with x-axis)

Angle of 120° above the positive x-axis

This would be drawn as an arrow with:

Length proportional to 50 units

2.2 A car is moving with an average speed of 72 km h⁻¹. How much time will it take to cover a distance of 360 km?

Solution:         

Time to cover 360 km at 72 km h⁻¹
Distance = 360 km
Speed = 72 km h⁻¹

Using formula,
Time = Distance/Speed
Time = 360/72 = 5 h

2.3 A truck starts from rest. It reaches a velocity of 90 km h⁻¹ in 50 seconds. Find its average acceleration.

Solution:

Acceleration from rest to 90 km h⁻¹ in 50 seconds

First convert 90 km h⁻¹ to m s⁻¹
90 km h⁻¹ = 90 × 1000 / 3600 m s⁻¹
90 km h⁻¹ = 25 m s⁻¹

Initial velocity (v_i) = 0 m s⁻¹
Final velocity (v_f) = 25 m s⁻¹
Time (t) = 50 s

Using formula,
a = (v_f-v_i)/t

a = (25-0)/50
a = 0.5 m s⁻²

2.4 A car passes a green traffic signal while moving with a velocity of 5 m s⁻¹. It then accelerates to 1.5 m s⁻². What is the velocity of car after 5 seconds?

Solution:

Final velocity (v_f) after acceleration = ?
Initial velocity (v_i) = 5 m s⁻¹
Acceleration (a) = 1.5 m s⁻²
Time (t) = 5 s

Using formula,
v_f = v_i + at

v_f = 5 + (1.5 × 5)
v_f = 5 + 7.5
v_f = 12.5 m s⁻¹

2.5 A motorcycle initially travelling at 18 km h⁻¹ accelerates at a constant rate of 2 m s⁻². How far will the motorcycle go in 10 seconds?

Solution:

Distance covered (S) with acceleration = ?
Initial velocity (v_i) = 18 km h⁻¹ = 5 m s⁻¹
Acceleration (a) = 2 m s⁻²
Time (t) = 10 s

Using formula,
S = v_it + ½at²

S = (5 × 10) + (½ × 2 × 10²)
S = 50 + 100
S = 150 m

2.6 A wagon is moving on the road with a velocity of 54 km h_⁻¹. Brakes are applied suddenly. The wagon covers a distance of 25 m before stopping. Determine the acceleration of the wagon.

Solution:

Deceleration (a) to stop = ?

Initial velocity (v_i) = 54 km h⁻¹ = 15 m s⁻¹
Final velocity (v_f) = 0 m s⁻¹
Distance (s) = 25 m

Using v_f² = v_i² + 2as

0 = 15² + 2a(25)
-225 = 50a
a = -4.5 m s⁻²

2.7 A stone is dropped from a height of 45 m. How long will it take to reach the ground? What will be its velocity just before hitting the ground?

Solution:

For free fall problem
Using g = 9.8 m s⁻²
Height (h) = 45 m

Using formula,
h = ½gt²

45 = ½ 9.8 t²
45 = 4.9t²
t = 3 s
Final velocity v = gt
v = 9.8 × 3
v = 30 m s⁻¹

2.8 A car travels 10 km with an average velocity of 20 m s⁻¹. Then it travels in the same direction through a diversion at an average velocity of 4 m s⁻¹ for the next 0.8 km. Determine the average velocity of the car for the total journey.

Solution:

Average velocity calculation

First section: 10 km at 20 m s⁻¹
S₁ = 10 km = 10000 m
v₁ = 20 m s⁻¹
Time for first section
t₁ = S₁ / v₁
t₁ = 10000/20 = 500 s

Second section: 0.8 km at 4 m s⁻¹
S₂ = 0.8 km = 800 m
v₂ = 4 m s⁻¹
Time for second section
t₂ = S₂ / v₂
t₂ = 800/4 = 200 s

Total distance S = S₁ + S₂
S = 10000 m + 800 m
S = 10800 m

Total time t = t₁ + t₂
t = 700 s

Using formula,
Average velocity v = S/t

v = 10800/700
v = 15.4 m s⁻¹

2.9 A ball is dropped from the top of a tower. The ball reaches the ground in 5 seconds. Find the height of the tower and the velocity of the ball with which it strikes the ground.

Solution:

Time taken t = 5 s

Using formula,
h = ½gt²

h = ½ × 9.8 × 5²
h = 125 m

Using formula,
velocity v = gt
v = 9.8 × 5
v = 50 m s⁻¹

2.10 A cricket ball is hit so that it travels straight up in the air. An observer notes that it took 3 seconds to reach the highest point. What was the initial velocity of the ball? If the ball was hit 1 m above the ground, how high did it rise from the ground?

Solution:

Time to reach highest point t = 1.5 s (half of 3 s)
Using v_f = v_i – gt for highest point

0 = v_i – 9.8 × 1.5
v_i = 30 m s⁻¹

Maximum height h = v_i²/(2g)
h = 30²/(2 × 9.8)
h = 46 m

Class 9 Physics Full Book Numerical Problems with Solution