Class 9 Physics Chapter 3 Numerical Problems

Class 9 Physics Chapter 3 numerical problems are according to the new syllabus of the Punjab Board. Chapter 3 is related to Dynamics which is the branch of Physics that deals with the study of motion with discussing its effects.

Class 9 Physics Chapter 3 Numerical Problems

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

F = ma

ma = T – mg

T = 2m₁m₂g / (m₁ + m₂)

a = (m₁ – m₂)g / (m₁ + m₂)

T = m₁m₂g / (m₁ + m₂)

a =  m₁g / (m₁ + m₂)

F = (P_f – P_i) / t

F_S = µ R = µmg                          (where R = mg)

F_c = mv²/r

Numerical Problems Chapter 3 – Dynamics

3.1 A 10 kg block is placed on a smooth horizontal surface. A horizontal force of 5 N is applied to the block. Find:
(a) the acceleration produced in the block.
(b) the velocity of block after 5 seconds

Solution:   

Mass (m) = 10 kg
Force (F) = 5 N

(a)
Using formula,
F = ma

a = F/m
a = 5/10
a = 0.5 m s⁻²

(b)
Using formula,
v_f = v_i + at

where v_i = 0
v_f = 0 + (0.5 × 5)
v_f = 2.5 m s⁻¹

3.2 The mass of a person is 80 kg. What will be his weight on the Earth? What will be his weight on the Moon? The value of acceleration due to gravity of Moon is 1.6 m s⁻².

Solution:

Mass = 80 kg
Acceleration due to gravity on the Earth: g = 9.8 m s⁻²
Acceleration due to gravity on the Moon: g = 1.6 m s⁻²

Weight on Earth
w = mg
w = 80 × 9.8
w = 800 N

Weight on Moon
w = mg
w = 80 × 1.6
w = 128 N

3.3 What force is required to increase the velocity of 800 kg car from 10 m s⁻¹ to 30 m s⁻¹ in 10 seconds?

Solution:

Mass = 800 kg
Initial velocity (v_i) = 10 m s⁻¹
Final velocity (v_f) = 30 m s⁻¹
Time (t) = 10 s

where a = (v_f-v_i)/t
a = (30-10)/10 = 2 m s⁻²

Using formula,
F = ma

F = 800 × 2 = 1600 N

3.4 A 5 g bullet is fired by a gun. The bullet moves with a velocity of 300 m s⁻¹. If the mass ofthe gun is 10 kg, find the recoil speed of the gun.

Solution:

Bullet mass (m₁) = 5 g = 0.005 kg
Bullet velocity (v₁) = 300 m s⁻¹
Gun mass (m₂) = 10 kg

Using the law of conservation of momentum:
m₁v₁ + m₂v₂ = 0
0.005 × 300 + 10v₂ = 0
v₂ = -(0.005 × 300)/10
v₂ = -0.15 m s⁻¹

3.5 An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m s⁻¹. Determine:
(a) the speed of astronaut as he recoils away from the wrench.
(b) the distance covered by the astronaut in 30 minutes.

Solution:

Astronaut mass (M) = 70 kg
Wrench mass (m) = 300 g = 0.3 kg
Wrench velocity (v₂)= 3.5 m s⁻¹

a)
Using the law of conservation of momentum:

Mv₁ = mv₂
70v₁ = 0.3 × 3.5
v₁ = 0.3 × 3.5/70
v₁ = -1.5 × 10⁻² m s⁻¹

b)
Distance in 30 minutes:
t = 30 × 60 = 1800 s

Using formula,
d = vt
d = (-1.5 × 10⁻²) × 1800
d = 27 m

3.6 A 6.5 × 10³ kg bogie of a goods train is moving with a velocity of 0.8 m s⁻¹. Another bogie of mass 9.2×10³ kg coming from behind with a velocity of 1.2 m s⁻¹ collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.

Solution:

m₁ = 6.5 × 10⁴ kg
v₁ = 0.8 m s⁻¹
m₂ = 9.2 × 10⁴ kg
v₂ = 1.2 m s⁻¹

Using momentum conservation:
m₁v₁ + m₂v₂ = (m₁ + m₂)v

(6.5 × 10⁴ × 0.8) + (9.2 × 10⁴ × 1.2) = (6.5 × 10⁴ + 9.2 × 10⁴)v

v = 1.03 m s⁻¹

3.7 A cyclist weighing 55 kg rides a bicycle of mass 5 kg. He starts from rest and applies a force of 90 N for 8 seconds. Then he continues at a constant speed for another 8 seconds. Calculate the total distance travelled by the cyclist.

Solution: :

Total mass (m) = 55 + 5 = 60 kg
Force (F) = 90 N
Time (t) = 8 s

Using formula,
F = ma

a = F/m
a = 90/60
a = 1.5 m s⁻²

First 8 seconds: d₁ = v_it + ½at²
S₁ = 0 + ½ × 1.5 × 8²
S₁ = 48 m

Final velocity = v = at
v = 1.5 × 8
v = 12 m s⁻¹

Next 8 seconds at constant speed:
S₂ = vt
S₂ = 12 × 8
S₂ = 96 m

Total distance S = S₁ + S₂
S = 48 + 96
S = 144 m

3.8 A ball of mass 0.4 kg is dropped on the floor from a height of 1.8 m. The ball rebounds straight upward to a height of 0.8 m. What is the magnitude and direction of the impulse applied to the ball by the floor?

Solution:

Mass (m) = 0.4 kg
Initial height (h₁) = 1.8 m
Rebound height (h₂) = 0.8 m
g = 9.8 m s⁻²

Step 1: Find velocities

Initial velocity before impact:
v₁ = √(2gh₁)
v₁ = √(2 × 9.8 × 1.8)
v₁ = -5.94 m s⁻¹ (downward)

Velocity after rebound:
v₂ = √(2gh₂)
v₂ = √(2 × 9.8 × 0.8)
v₂ = 3.96 m s⁻¹ (upward)

Step 2: Calculate impulse using change in momentum

Impulse = m(v₂ – v₁)

Impulse = 0.4(3.96 – (-5.94))

Impulse = 0.4 × 9.9

Impulse = 4 Ns (upward)

3.9 Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities 20 m s⁻¹ and 5 m s⁻¹ respectively. After collision, the velocity of 0.2 kg ball becomes 6 m s⁻¹. What will be the velocity of 0.4 kg ball?

Solution:

m₁ = 0.2 kg
v₁ = 20 m s⁻¹
m₂ = 0.4 kg
v₂ = 5 m s⁻¹ (opposite direction)

After collision: v₁’ = 6 m s⁻¹
Find: v₂’

Using conservation of momentum:
m₁v₁ + m₂v₂ = m₁v₁’ + m₂v₂’

Step 1: Write momentum equation

0.2(20) + 0.4(-5) = 0.2(6) + 0.4v₂’
4 – 2 = 1.2 + 0.4v₂’
2 = 1.2 + 0.4v₂’

Step 2: Solve for v₂’

0.4v₂’ = 0.8
v₂’ = 2 m s⁻¹

Class 9 Physics Full Book Numerical Problems with Solution