All 8 numericals of 9th Class Physics Chapter 3, Dynamics, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.
Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.
Numerical 3.1
A force of 5 N acts on a 10 kg block resting on a smooth (frictionless) surface. Find (a) its acceleration (b) its velocity after 5 s.
Given Data
F = 5 N
m = 10 kg
vᵢ = 0
t = 5 s
To Find
(a) Acceleration a
(b) Velocity vf after 5 s
Formulas
a = F ÷ m
vf = vᵢ + at
Calculation
(a) a = F ÷ m = 5 ÷ 10 = 0.5 m s⁻².
(b) vf = vᵢ + at = 0 + 0.5 × 5 = 2.5 m s⁻¹.
Result
(a) 0.5 m s⁻² (b) 2.5 m s⁻¹
Numerical 3.2
A person has a mass of 80 kg. Find his weight on the Earth (g = 10 m s⁻²) and on the Moon (g = 1.6 m s⁻²).
Given Data
m = 80 kg
g (Earth) = 10 m s⁻²
g (Moon) = 1.6 m s⁻²
To Find
Weight on the Earth
Weight on the Moon
Formula
W = mg
Calculation
On the Earth: W = mg = 80 × 10 = 800 N.
On the Moon: W = mg = 80 × 1.6 = 128 N.
Result
800 N on the Earth, 128 N on the Moon
Note: His mass is 80 kg in both places; only his weight changes.
Numerical 3.3
What force is needed to increase the speed of an 800 kg car from 10 m s⁻¹ to 30 m s⁻¹ in 10 s?
Given Data
m = 800 kg
vᵢ = 10 m s⁻¹
vf = 30 m s⁻¹
t = 10 s
To Find
Force F
Formulas
a = (vf − vᵢ) ÷ t
F = ma
Calculation
a = (vf − vᵢ) ÷ t = (30 − 10) ÷ 10 = 2 m s⁻².
F = ma = 800 × 2.
Result
F = 1600 N
Numerical 3.4
A 5 g bullet is fired from a 10 kg gun with a speed of 300 m s⁻¹. Find the recoil speed of the gun.
Given Data
m₁ = 5 g = 0.005 kg
v₁ = 300 m s⁻¹
m₂ = 10 kg
total momentum before firing = 0
To Find
Recoil velocity of the gun v₂
Formula
m₁v₁ + m₂v₂ = 0
Calculation
By conservation of momentum: m₁v₁ + m₂v₂ = 0.
0.005 × 300 + 10 × v₂ = 0, so 1.5 + 10v₂ = 0.
v₂ = −1.5 ÷ 10 = −0.15 m s⁻¹.
Result
The gun recoils at 0.15 m s⁻¹, opposite to the bullet
Numerical 3.5
A 70 kg astronaut floating in space throws a 300 g wrench away at 3.5 m s⁻¹. Find (a) the astronaut's recoil speed (b) how far the astronaut moves in 30 minutes.
Given Data
m₁ (wrench) = 300 g = 0.3 kg
v₁ = 3.5 m s⁻¹
m₂ (astronaut) = 70 kg
t = 30 min = 1800 s
To Find
(a) Recoil speed of the astronaut
(b) Distance moved in 30 min
Formulas
m₁v₁ + m₂v₂ = 0
S = v × t
Calculation
(a) Momentum is conserved and was zero: 0.3 × 3.5 + 70 × v₂ = 0, so v₂ = −1.05 ÷ 70 = −0.015 m s⁻¹.
(b) In space nothing slows the astronaut, so the speed stays 0.015 m s⁻¹.
Distance = speed × time = 0.015 × 1800 = 27 m.
Result
(a) 0.015 m s⁻¹, opposite to the wrench (b) 27 m
Numerical 3.6
A railway bogie of mass 6.5 × 10³ kg moves at 0.8 m s⁻¹. A second bogie of mass 9.2 × 10³ kg moving at 1.2 m s⁻¹ in the same direction catches up and couples with it. Find their common velocity.
Given Data
m₁ = 6.5 × 10³ kg, u₁ = 0.8 m s⁻¹
m₂ = 9.2 × 10³ kg, u₂ = 1.2 m s⁻¹
coupled: one common velocity v
To Find
Common velocity v after coupling
Formula
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Calculation
Momentum before = 6500 × 0.8 + 9200 × 1.2 = 5200 + 11 040 = 16 240 kg m s⁻¹.
Momentum after = (6500 + 9200) × v = 15 700 v.
v = 16 240 ÷ 15 700 ≈ 1.03 m s⁻¹.
Result
v ≈ 1.03 m s⁻¹ in the same direction
Numerical 3.7
A 55 kg cyclist rides a 5 kg bicycle. Starting from rest, the cyclist applies a force of 90 N for 8 s and then rides at constant speed for another 8 s. Find the total distance travelled.
Given Data
total mass m = 55 + 5 = 60 kg
F = 90 N for t₁ = 8 s from rest
then constant speed for t₂ = 8 s
To Find
Total distance travelled
Formulas
a = F ÷ m
vf = vᵢ + at
S = vᵢt + ½at²
S = v × t
Calculation
Acceleration: a = F ÷ m = 90 ÷ 60 = 1.5 m s⁻².
Speed after 8 s: v = 0 + 1.5 × 8 = 12 m s⁻¹.
Distance while accelerating: S₁ = ½at² = ½ × 1.5 × 8² = 48 m.
Distance at constant speed: S₂ = 12 × 8 = 96 m.
Total = 48 + 96 = 144 m.
Result
Total distance = 144 m
Numerical 3.8
A 0.4 kg ball is dropped from a height of 1.8 m onto the floor and bounces back up to 0.8 m. Find the magnitude and direction of the impulse the floor gives it (g = 10 m s⁻²).
Given Data
m = 0.4 kg
dropped from 1.8 m
rebounds to 0.8 m
g = 10 m s⁻²
To Find
Impulse from the floor (size and direction)
Formulas
v² = 2gh
Impulse = Δp = m(vf − vᵢ)
Calculation
Speed on hitting the floor: v² = 2gh = 2 × 10 × 1.8 = 36, so v = 6 m s⁻¹ downward.
Speed leaving the floor: v² = 2 × 10 × 0.8 = 16, so v = 4 m s⁻¹ upward.
Take upward as positive: vᵢ = −6 m s⁻¹, vf = +4 m s⁻¹.
