📘 Chapter 3: Dynamics – Numerical Problems
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers numerical problems from Chapter 3 Dynamics including force, acceleration, weight, momentum, recoil, impulse, and conservation of momentum. Each problem is presented with Given Data, To Find, and step-by-step Solution as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 3: Dynamics
📋 Exercise Short Qs
📝 Short Qs
📖 Comprehensive Qs
🔢 Numerical Problems
✅ MCQs
✏️ Constructed Response
Dynamics covers force, Newton's laws, momentum, friction, and circular motion.
📐 Numerical Problems & Solutions (PECTAA 2026)
3.1
A \(10kg\) block is placed on a smooth horizontal surface. A horizontal force of \(5N\) is applied to the block. Find: (a) the acceleration produced in the block. (b) the velocity of block after 5 seconds.
Given Data
Mass of block\(= m = 10\ kg\)
Force\(= F = 5\ N\)
Initial velocity\(= v_i = 0\ ms^{-1}\)
Time\(= t = 5\ s\)
To Find
Acceleration \(= a = ?\)
Final velocity \(= v_f = ?\)
Solution
According to second law of motion:
\[\begin{aligned}
F &= ma \\
5 &= (10)(a) \\
a &= \frac{5}{10} \\
\boldsymbol{a} &= \boldsymbol{0.5\ ms^{-2}}
\end{aligned}\]
Now by using first equation of motion:
\[\begin{aligned}
v_f &= v_i + at \\
v_f &= 0 + (0.5)(5) \\
\boldsymbol{v_f} &= \boldsymbol{2.5\ ms^{-1}}
\end{aligned}\]
✅ (a) \(a = 0.5\ ms^{-2}\), (b) \(v_f = 2.5\ ms^{-1}\)
3.2
The mass of a person is \(80kg\). 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 \(1.6ms^{-2}\).
Given Data
Mass of body\(= m = 80\ kg\)
Value of \(g\) on Earth\(= g_E = 10\ ms^{-2}\)
Value of \(g\) on Moon\(= g_M = 1.6\ ms^{-2}\)
To Find
Weight on Earth \(= w_E = ?\)
Weight on Moon \(= w_M = ?\)
Solution
By using formula of weight \(w = mg\):
\[\begin{aligned}
w_E &= mg_E \\
w_E &= (80)(10) \\
\boldsymbol{w_E} &= \boldsymbol{800\ N}
\end{aligned}\]
Now again by using formula of weight:
\[\begin{aligned}
w_M &= mg_M \\
w_M &= (80)(1.6) \\
\boldsymbol{w_M} &= \boldsymbol{128\ N}
\end{aligned}\]
✅ Weight on Earth = \(800\ N\), Weight on Moon = \(128\ N\)
3.3
What force is required to increase the velocity of \(800kg\) car from \(10ms^{-1}\) to \(30ms^{-1}\) in 10 seconds?
Given Data
Mass of car\(= m = 800\ kg\)
Initial velocity\(= v_i = 10\ ms^{-1}\)
Final velocity\(= v_f = 30\ ms^{-1}\)
Time\(= t = 10\ s\)
To Find
Force \(= F = ?\)
Solution
According to second law of motion:
\[\begin{aligned}
F &= ma = m\left(\frac{v_f - v_i}{t}\right) \\
F &= (800)\left(\frac{30 - 10}{10}\right) \\
F &= (800)\left(\frac{20}{10}\right) \\
F &= (800)(2) \\
\boldsymbol{F} &= \boldsymbol{1600\ N}
\end{aligned}\]
✅ \(F = 1600\ N\)
3.4
A \(5g\) bullet is fired by a gun. The bullet moves with a velocity of \(300ms^{-1}\). If the mass of the gun is \(10kg\), find the recoil speed of the gun.
Given Data
Mass of bullet\(= m = 5\ g = 0.005\ kg\)
Velocity of bullet\(= v = 300\ ms^{-1}\)
Mass of gun\(= M = 10\ kg\)
To Find
Recoil speed of the gun \(= V = ?\)
Solution
According to law of conservation of momentum:
\[\begin{aligned}
\text{Total momentum before firing} &= \text{Total momentum after firing} \\
0 &= MV + mv \\
0 &= (10)V + (0.005)(300) \\
0 &= 10V + 1.5 \\
-1.5 &= 10V \\
V &= \frac{-1.5}{10} \\
\boldsymbol{V} &= \boldsymbol{-0.15\ ms^{-1}}
\end{aligned}\]
Negative sign indicates the gun recoils (moves in backward direction opposite to the motion of bullet).
✅ Recoil speed = \(0.15\ ms^{-1}\) (opposite direction)
3.5
An astronaut weighs \(70 kg\). He throws a wrench of mass \(300 g\) at a speed of \(3.5 ms^{-1}\). Determine: (a) the speed of astronaut as he recoils away from the wrench. (b) the distance covered by the astronaut in 30 minutes.
Given Data
Mass of astronaut\(= M = 70\ kg\)
Mass of wrench\(= m = 300\ g = 0.3\ kg\)
Speed of wrench\(= v = 3.5\ ms^{-1}\)
Time\(= t = 30\ min = 1800\ s\)
To Find
Recoil speed of astronaut \(= V = ?\)
Distance covered in 30 minutes \(= S = ?\)
Solution
According to law of conservation of momentum:
\[\begin{aligned}
\text{Total momentum before throwing} &= \text{Total momentum after throwing} \\
0 &= MV + mv \\
0 &= (70)V + (0.3)(3.5) \\
0 &= 70V + 1.05 \\
-1.05 &= 70V \\
V &= \frac{-1.05}{70} \\
\boldsymbol{V} &= \boldsymbol{-0.015\ ms^{-1}} = \boldsymbol{-1.5 \times 10^{-2}\ ms^{-1}}
\end{aligned}\]
Negative sign indicates the astronaut recoils (moves in the opposite direction of the wrench).
Now by using formula of distance \(S = Vt\):
\[\begin{aligned}
S &= (0.015)(1800) \\
\boldsymbol{S} &= \boldsymbol{27\ m}
\end{aligned}\]
Note: We don't take negative speed for distance because distance is always positive. The negative sign shows direction, not how far something moves.
✅ (a) \(V = 0.015\ ms^{-1}\) (opposite direction), (b) \(S = 27\ m\)
3.6
A \(6.5 \times 10^3 kg\) bogie of a goods train is moving with a velocity of \(0.8 ms^{-1}\). Another bogie of mass \(9.2 \times 10^3 kg\) coming from behind with a velocity of \(1.2 ms^{-1}\) collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.
Given Data
Mass of first bogie\(= m_1 = 6.5 \times 10^3\ kg\)
Velocity of first bogie\(= v_1 = 0.8\ ms^{-1}\)
Mass of second bogie\(= m_2 = 9.2 \times 10^3\ kg\)
Velocity of second bogie\(= v_2 = 1.2\ ms^{-1}\)
To Find
Common velocity after coupling \(= V = ?\)
Solution
According to law of conservation of momentum:
\[\begin{aligned}
m_1v_1 + m_2v_2 &= (m_1 + m_2)V \\
V &= \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\
V &= \frac{(6.5 \times 10^3)(0.8) + (9.2 \times 10^3)(1.2)}{6.5 \times 10^3 + 9.2 \times 10^3} \\
V &= \frac{5200 + 11040}{15700} \\
V &= \frac{16240}{15700} \\
\boldsymbol{V} &= \boldsymbol{1.03\ ms^{-1}}
\end{aligned}\]
✅ Common velocity = \(1.03\ ms^{-1}\)
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.
Given Data
Mass of cyclist\(= m_1 = 55\ kg\)
Mass of bicycle\(= m_2 = 5\ kg\)
Total mass\(= m = 60\ kg\)
Force applied\(= F = 90\ N\)
Time of acceleration\(= t_1 = 8\ s\)
Time of constant speed\(= t_2 = 8\ s\)
Initial speed\(= v_i = 0\ ms^{-1}\)
To Find
Total distance travelled \(= S = ?\)
Solution
According to second law of motion:
\[\begin{aligned}
F &= ma \\
90 &= (60)(a) \\
a &= \frac{90}{60} \\
a &= 1.5\ ms^{-2}
\end{aligned}\]
Now by using first equation of motion:
\[\begin{aligned}
v_f &= v_i + at_1 \\
v_f &= 0 + (1.5)(8) \\
v_f &= 12\ ms^{-1}
\end{aligned}\]
Distance covered during acceleration by using second equation of motion:
\[\begin{aligned}
S_1 &= v_i t_1 + \frac{1}{2} a t_1^2 \\
S_1 &= (0)(8) + \frac{1}{2}(1.5)(8)^2 \\
S_1 &= 0 + \frac{1}{2}(1.5)(64) \\
S_1 &= 48\ m
\end{aligned}\]
Distance covered at constant speed by formula \(S = vt\):
\[\begin{aligned}
S_2 &= v_f t_2 \\
S_2 &= (12)(8) \\
S_2 &= 96\ m
\end{aligned}\]
Total distance travelled:
\[\begin{aligned}
S &= S_1 + S_2 \\
S &= 48 + 96 \\
\boldsymbol{S} &= \boldsymbol{144\ m}
\end{aligned}\]
✅ Total distance = \(144\ m\)
3.8
A ball of mass \(0.4kg\) is dropped on the floor from a height of \(1.8m\). The ball rebounds straight upward to a height of \(0.8m\). What is the magnitude and direction of the impulse applied to the ball by the floor?
Given Data
Mass of ball\(= m = 0.4\ kg\)
Drop height\(= h_1 = 1.8\ m\)
Rebound height\(= h_2 = 0.8\ m\)
Acceleration due to gravity\(= g = 10\ ms^{-2}\)
To Find
Impulse (magnitude and direction) = ?
Solution
Since the ball is dropped, \(v_i = 0\ ms^{-1}\):
\[\begin{aligned}
2gh_1 &= v_f^2 - v_i^2 \\
2(10)(1.8) &= v_f^2 - (0)^2 \\
36 &= v_f^2 \\
v_f &= \pm 6\ ms^{-1}
\end{aligned}\]
But since the ball is moving downward, we select the negative root: \(v_f = v_{before} = -6\ ms^{-1}\).
At maximum rebound height, \(v_f = 0\ ms^{-1}\):
\[\begin{aligned}
2gh_2 &= v_f^2 - v_i^2 \\
2(-10)(0.8) &= (0)^2 - v_i^2 \quad (\because \text{ball moving upward}) \\
-16 &= -v_i^2 \\
v_i &= \pm 4\ ms^{-1}
\end{aligned}\]
But since the ball is moving upward, we select the positive root: \(v_i = v_{after} = 4\ ms^{-1}\).
Now by using formula of impulse:
\[\begin{aligned}
\text{Impulse} &= \text{Change in momentum} \\
&= \Delta p \\
&= p_f - p_i \\
&= mv_{after} - mv_{before} \\
&= m(v_{after} - v_{before}) \\
&= (0.4)[4 - (-6)] \\
&= (0.4)[10] \\
\boldsymbol{\text{Impulse}} &= \boldsymbol{4\ Ns}
\end{aligned}\]
The positive result means the impulse is upward (floor pushes the ball up).
✅ Impulse = \(4\ Ns\) (upward)
📘 Solved Examples (PECTAA 2026)
Example 3.1
A \(10kg\) block moves on a frictionless horizontal surface with an acceleration of \(2ms^{-2}\). What is the force acting on the block?
Given Data
Mass of block\(= m = 10\ kg\)
Acceleration\(= a = 2\ ms^{-2}\)
To Find
Force \(= F = ?\)
Solution
By Newton's second law of motion:
\[\begin{aligned}
F &= ma \\
F &= (10)(2) \\
\boldsymbol{F} &= \boldsymbol{20\ N}
\end{aligned}\]
✅ \(F = 20\ N\)
Example 3.2
A force of \(7500N\) is applied to move a truck of mass \(3000kg\). Find the acceleration produced in the truck. How long will it take to accelerate the truck from \(36kmh^{-1}\) to \(72kmh^{-1}\) speed?
Given Data
Force applied\(= F = 7500\ N\)
Mass of truck\(= m = 3000\ kg\)
Initial speed\(= v_i = 36\ kmh^{-1} = 36 \times \frac{10}{36} = 10\ ms^{-1}\)
Final speed\(= v_f = 72\ kmh^{-1} = 72 \times \frac{10}{36} = 20\ ms^{-1}\)
To Find
Acceleration \(= a = ?\)
Time \(= t = ?\)
Solution
By Newton's second law of motion:
\[\begin{aligned}
F &= ma \\
7500 &= (3000)(a) \\
a &= \frac{7500}{3000} \\
a &= 2.5\ ms^{-2}
\end{aligned}\]
Now by using first equation of motion:
\[\begin{aligned}
v_f &= v_i + at \\
20 &= 10 + (2.5)(t) \\
10 &= 2.5t \\
t &= \frac{10}{2.5} \\
\boldsymbol{t} &= \boldsymbol{4\ s}
\end{aligned}\]
✅ (a) \(a = 2.5\ ms^{-2}\), (b) \(t = 4\ s\)
Example 3.3
A bullet of mass \(15g\) is fired by a gun. If velocity of bullet is \(150ms^{-1}\), what is its momentum?
Given Data
Mass of bullet\(= m = 15\ g = 0.015\ kg\)
Velocity of bullet\(= v = 150\ ms^{-1}\)
To Find
Momentum \(= P = ?\)
Solution
Using formula of momentum \(P = mv\):
\[\begin{aligned}
P &= mv \\
P &= (0.015)(150) \\
\boldsymbol{P} &= \boldsymbol{2.25\ kgms^{-1}}
\end{aligned}\]
✅ \(P = 2.25\ kgms^{-1}\)
Example 3.4
A cricket ball of mass \(160g\) is hit by a bat. The ball leaves the bat with a velocity of \(52ms^{-1}\). If the ball strikes the bat with a velocity of \(-28ms^{-1}\) (opposite direction) before hitting, find the average force exerted on the ball by the bat. The ball remains in contact with the bat for \(4 \times 10^{-3}s\).
Given Data
Mass of ball\(= m = 160\ g = 0.16\ kg\)
Initial velocity\(= v_i = -28\ ms^{-1}\)
Final velocity\(= v_f = 52\ ms^{-1}\)
Time\(= t = 4 \times 10^{-3}\ s\)
To Find
Average force \(= F = ?\)
Solution
According to second law of motion:
\[\begin{aligned}
F &= ma = m\left(\frac{v_f - v_i}{t}\right) \\
F &= (0.16)\left[\frac{52 - (-28)}{4 \times 10^{-3}}\right] \\
F &= (0.16)\left(\frac{80}{4 \times 10^{-3}}\right) \\
F &= (0.16)(20000) \\
\boldsymbol{F} &= \boldsymbol{3200\ N}
\end{aligned}\]
✅ \(F = 3200\ N\)
Example 3.5
A bullet of mass \(m_1\) is fired by a gun of mass \(m_2\). Find the velocity of the gun in terms of velocity of bullet \(v_1\) just after firing.
Given Data
Mass of bullet\(= m_1\)
Mass of gun\(= m_2\)
Velocity of bullet after firing\(= v_1\)
Velocity of gun\(= v_2\)
To Find
Velocity of gun after firing \(= v_2 = ?\)
Solution
Before firing, both the bullet and gun are at rest. Therefore, the total momentum before firing is zero.
According to law of conservation of momentum:
\[\begin{aligned}
\text{Total momentum before firing} &= \text{Total momentum after firing} \\
0 &= m_1v_1 + m_2v_2 \\
\boldsymbol{v_2} &= \boldsymbol{-\frac{m_1}{m_2}v_1}
\end{aligned}\]
The negative sign shows that the gun moves backward, opposite to the direction of the bullet.
✅ \(v_2 = -\frac{m_1}{m_2}v_1\) (recoil velocity)
Example 3.6
A ball of mass \(3kg\) moving with a velocity of \(5ms^{-1}\) collides with a stationary ball of mass \(2kg\) and then both of them move together. If the friction is negligible, find out the velocity with which both the balls will move after collision.
Given Data
Mass of first ball\(= m_1 = 3\ kg\)
Mass of second ball\(= m_2 = 2\ kg\)
Initial velocity of first ball\(= v_1 = 5\ ms^{-1}\)
Initial velocity of second ball\(= v_2 = 0\ ms^{-1}\)
To Find
Velocity of both balls after collision \(= v = ?\)
Solution
According to law of conservation of momentum:
\[\begin{aligned}
m_1v_1 + m_2v_2 &= (m_1 + m_2)v \\
(3)(5) + (2)(0) &= v(3 + 2) \\
15 + 0 &= 5v \\
v &= \frac{15}{5} \\
\boldsymbol{v} &= \boldsymbol{3\ ms^{-1}}
\end{aligned}\]
✅ \(v = 3\ ms^{-1}\)
📐 Important Formulas – Dynamics
Newton's Second Law: \(F = ma\)
Weight: \(w = mg\)
Relation Between Force and Momentum: \(F = \frac{\Delta p}{\Delta t}\)
Centripetal Force: \(F_c = \frac{mv^2}{r}\)
Frictional Force: \(F_s = \mu mg\)
Impulse: \(\text{Impulse} = F \times \Delta t = \Delta p\)
Conservation of Momentum: Total momentum before = Total momentum after
📐 Key Formulas – Dynamics
Newton's Second Law: \(F = ma\)
Weight: \(w = mg\)
Momentum: \(p = mv\)
Impulse: \(F \Delta t = \Delta p\)
Conservation of Momentum: \(m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\)
Centripetal Force: \(F_c = \frac{mv^2}{r}\)
Frictional Force: \(F_s = \mu mg\)
💡 Exam Tip:
For numerical problems, always write the given data, the formula being used, and show the steps of your solution clearly. Pay attention to unit conversions (g to kg, kmh⁻¹ to ms⁻¹). Practice the equations of motion and conservation of momentum thoroughly as they are the foundation for most dynamics numericals. These problems follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Chapter 1Physical Quantities and Measurements
Chapter 2Kinematics
Chapter 3✨ Dynamics
Chapter 4Turning Effect of Forces
Chapter 5Work and Energy
Chapter 6Mechanical Properties of Matter
Chapter 7Thermal Properties of Matter
Chapter 8Magnetism
Chapter 9Nature of Science
📖 Complete syllabus coverage for Class 9 Physics (PECTAA 2026) – Chapters 1 to 9
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus