š Newton’s Laws of Motion
š Chapter 19: Dynamics | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Newton’s Laws, Equations of Motion, Rectilinear Motion
ā
Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
š Dynamics deals with the study of motion of objects and the forces that cause this motion. This chapter explores velocity, acceleration, equations of motion, and applies calculus principles (differentiation and integration) to analyze particle motion under various forces. The notes have been updated according to the latest syllabus of 2080.
š Exercises
š Newton’s Laws of Motion
First Law (Law of Inertia): Every body continues in its state of rest or uniform motion in a straight line unless compelled by an external force to change that state.
Second Law (Law of Acceleration): The rate of change of momentum is proportional to the applied force and takes place in the direction of the force.
\(F = ma\)
\(F = ma\)
Third Law (Action-Reaction): For every action, there is an equal and opposite reaction.
š Equations of Motion (Constant Acceleration)
\[
v = u + at
\]
\[
s = ut + \frac{1}{2}at^2
\]
\[
v^2 = u^2 + 2as
\]
\[
s = \frac{u + v}{2} \cdot t
\]
where \(u\) = initial velocity, \(v\) = final velocity, \(a\) = acceleration, \(t\) = time, \(s\) = displacement.
š Motion Under Gravity
For vertical motion under gravity (taking upward as positive):
\[
v = u – gt
\]
\[
h = ut – \frac{1}{2}gt^2
\]
\[
v^2 = u^2 – 2gh
\]
where \(g \approx 9.8 \text{ m/s}^2\) or \(\approx 32 \text{ ft/s}^2\)
š Solved Problems (15 Important Questions)
Q1 A car starts from rest and accelerates uniformly at 2 m/sĀ² for 5 seconds. Find its final velocity and distance covered.
\(u = 0, a = 2 \text{ m/s}^2, t = 5 \text{ s}\)
\(v = u + at = 0 + 2 \times 5 = 10 \text{ m/s}\)
\(s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 25 = 25 \text{ m}\)
\(v = u + at = 0 + 2 \times 5 = 10 \text{ m/s}\)
\(s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 25 = 25 \text{ m}\)
ā
v = 10 m/s, s = 25 m
Q2 A body moving with velocity 20 m/s is brought to rest after 4 seconds. Find acceleration and distance.
\(u = 20, v = 0, t = 4\)
\(a = \frac{v-u}{t} = \frac{0-20}{4} = -5 \text{ m/s}^2\) (retardation)
\(s = \frac{u+v}{2} \times t = \frac{20+0}{2} \times 4 = 40 \text{ m}\)
\(a = \frac{v-u}{t} = \frac{0-20}{4} = -5 \text{ m/s}^2\) (retardation)
\(s = \frac{u+v}{2} \times t = \frac{20+0}{2} \times 4 = 40 \text{ m}\)
ā
a = -5 m/sĀ², s = 40 m
Q3 A ball is thrown upward with velocity 20 m/s. Find maximum height and time to reach there. (g = 10 m/sĀ²)
\(u = 20, v = 0, a = -g = -10\)
\(v^2 = u^2 – 2gh\) ā \(0 = 400 – 20h\) ā \(h = 20 \text{ m}\)
\(v = u – gt\) ā \(0 = 20 – 10t\) ā \(t = 2 \text{ s}\)
\(v^2 = u^2 – 2gh\) ā \(0 = 400 – 20h\) ā \(h = 20 \text{ m}\)
\(v = u – gt\) ā \(0 = 20 – 10t\) ā \(t = 2 \text{ s}\)
ā
h = 20 m, t = 2 s
Q4 A stone is dropped from a height of 80 m. Find time to reach ground and velocity on impact. (g = 10 m/sĀ²)
\(u = 0, s = 80, a = g = 10\)
\(s = \frac{1}{2}gt^2\) ā \(80 = \frac{1}{2} \times 10 \times t^2 = 5t^2\) ā \(t^2 = 16\) ā \(t = 4 \text{ s}\)
\(v = u + gt = 0 + 10 \times 4 = 40 \text{ m/s}\)
\(s = \frac{1}{2}gt^2\) ā \(80 = \frac{1}{2} \times 10 \times t^2 = 5t^2\) ā \(t^2 = 16\) ā \(t = 4 \text{ s}\)
\(v = u + gt = 0 + 10 \times 4 = 40 \text{ m/s}\)
ā
t = 4 s, v = 40 m/s
Q5 A force of 10 N acts on a mass of 2 kg. Find acceleration.
\(F = ma\) ā \(a = \frac{F}{m} = \frac{10}{2} = 5 \text{ m/s}^2\)
ā
a = 5 m/sĀ²
Q6 A car accelerates from rest to 30 m/s in 6 seconds. Find force if mass is 1000 kg.
\(a = \frac{v-u}{t} = \frac{30-0}{6} = 5 \text{ m/s}^2\)
\(F = ma = 1000 \times 5 = 5000 \text{ N}\)
\(F = ma = 1000 \times 5 = 5000 \text{ N}\)
ā
F = 5000 N
Q7 Two masses 3 kg and 2 kg are connected by a string over a smooth pulley. Find acceleration and tension. (g = 10 m/sĀ²)
\(a = \frac{(m_1 – m_2)g}{m_1 + m_2} = \frac{(3-2) \times 10}{3+2} = \frac{10}{5} = 2 \text{ m/s}^2\)
\(T = m_1(g – a) = 3(10 – 2) = 24 \text{ N}\) or \(T = m_2(g + a) = 2(10 + 2) = 24 \text{ N}\)
\(T = m_1(g – a) = 3(10 – 2) = 24 \text{ N}\) or \(T = m_2(g + a) = 2(10 + 2) = 24 \text{ N}\)
ā
a = 2 m/sĀ², T = 24 N
Q8 A body of mass 5 kg is pulled with a force of 30 N on a horizontal frictionless surface. Find acceleration.
\(a = \frac{F}{m} = \frac{30}{5} = 6 \text{ m/s}^2\)
ā
a = 6 m/sĀ²
Q9 A ball is thrown upward with velocity 30 m/s. For how long will it remain in air? (g = 10 m/sĀ²)
Time of ascent: \(t = \frac{u}{g} = \frac{30}{10} = 3 \text{ s}\)
Total time in air = \(2t = 6 \text{ s}\)
Total time in air = \(2t = 6 \text{ s}\)
ā
Total time = 6 s
Q10 A particle moves with uniform acceleration. Its velocities at time t=2s and t=5s are 10 m/s and 25 m/s. Find acceleration and initial velocity.
\(v_2 = u + a(2) = u + 2a = 10\)
\(v_5 = u + a(5) = u + 5a = 25\)
Subtracting: \(3a = 15\) ā \(a = 5 \text{ m/s}^2\)
\(u = 10 – 10 = 0 \text{ m/s}\)
\(v_5 = u + a(5) = u + 5a = 25\)
Subtracting: \(3a = 15\) ā \(a = 5 \text{ m/s}^2\)
\(u = 10 – 10 = 0 \text{ m/s}\)
ā
a = 5 m/sĀ², u = 0 m/s
Q11 A body sliding on an inclined plane has acceleration 4 m/sĀ². If it starts from rest, find velocity after 3 seconds and distance covered.
\(v = u + at = 0 + 4 \times 3 = 12 \text{ m/s}\)
\(s = \frac{1}{2}at^2 = 0.5 \times 4 \times 9 = 18 \text{ m}\)
\(s = \frac{1}{2}at^2 = 0.5 \times 4 \times 9 = 18 \text{ m}\)
ā
v = 12 m/s, s = 18 m
Q12 A 10 kg mass is pulled by a 50 N force on a rough surface with friction coefficient 0.2. Find acceleration. (g = 10 m/sĀ²)
Friction = \(\mu mg = 0.2 \times 10 \times 10 = 20 \text{ N}\)
Net force = \(50 – 20 = 30 \text{ N}\)
\(a = \frac{30}{10} = 3 \text{ m/s}^2\)
Net force = \(50 – 20 = 30 \text{ N}\)
\(a = \frac{30}{10} = 3 \text{ m/s}^2\)
ā
a = 3 m/sĀ²
Q13 A car moving at 20 m/s applies brakes and stops in 5 seconds. Find retardation and distance.
\(a = \frac{v-u}{t} = \frac{0-20}{5} = -4 \text{ m/s}^2\) (retardation 4 m/sĀ²)
\(s = \frac{u+v}{2} \times t = \frac{20+0}{2} \times 5 = 50 \text{ m}\)
\(s = \frac{u+v}{2} \times t = \frac{20+0}{2} \times 5 = 50 \text{ m}\)
ā
Retardation = 4 m/sĀ², s = 50 m
Q14 A stone is thrown vertically upward with velocity 40 m/s. Find height after 2 seconds. (g = 10 m/sĀ²)
\(h = ut – \frac{1}{2}gt^2 = 40 \times 2 – \frac{1}{2} \times 10 \times 4 = 80 – 20 = 60 \text{ m}\)
ā
h = 60 m
Q15 In the Atwood machine with masses 4 kg and 6 kg, find acceleration and tension. (g = 10 m/sĀ²)
\(a = \frac{(6-4)g}{6+4} = \frac{2 \times 10}{10} = 2 \text{ m/s}^2\)
\(T = \frac{2m_1 m_2}{m_1 + m_2} g = \frac{2 \times 4 \times 6}{10} \times 10 = \frac{48}{10} \times 10 = 48 \text{ N}\)
\(T = \frac{2m_1 m_2}{m_1 + m_2} g = \frac{2 \times 4 \times 6}{10} \times 10 = \frac{48}{10} \times 10 = 48 \text{ N}\)
ā
a = 2 m/sĀ², T = 48 N
āļø Practice Questions
P1 A train starts from rest and accelerates at 1.5 m/sĀ² for 10 seconds. Find final velocity and distance.
ā
Answer: v = 15 m/s, s = 75 m
P2 A ball thrown upward with 25 m/s. Find maximum height. (g = 10 m/sĀ²)
ā
Answer: h = 31.25 m
P3 A force of 25 N gives a mass of 5 kg an acceleration of 4 m/sĀ². Find frictional force.
ā
Answer: Friction = 5 N
P4 A car moving at 15 m/s accelerates at 2 m/sĀ² for 4 seconds. Find distance covered.
ā
Answer: s = 76 m
P5 Two masses 2 kg and 3 kg are connected by string over a pulley. Find acceleration. (g = 10 m/sĀ²)
ā
Answer: a = 2 m/sĀ², T = 24 N
š Multiple Choice Questions (HSEB Pattern)
MCQ 1 Newton’s first law of motion is also known as:
A) Law of Acceleration B) Law of Inertia C) Law of Action-Reaction D) Law of Gravity
ā
Answer: B) Law of Inertia
MCQ 2 The SI unit of force is:
A) Joule B) Watt C) Newton D) Pascal
ā
Answer: C) Newton
MCQ 3 If a body moving with constant velocity, the net force acting on it is:
A) Maximum B) Zero C) Minimum D) Constant non-zero
ā
Answer: B) Zero
MCQ 4 The equations of motion are valid only for:
A) Variable acceleration B) Constant acceleration C) Zero acceleration D) Negative acceleration
ā
Answer: B) Constant acceleration
MCQ 5 The value of acceleration due to gravity (g) on Earth’s surface is approximately:
A) 9.8 m/sĀ² B) 98 m/sĀ² C) 9.8 km/sĀ² D) 980 m/sĀ²
ā
Answer: A) 9.8 m/sĀ²
MCQ 6 The formula \(F = ma\) represents Newton’s:
A) First law B) Second law C) Third law D) Law of gravitation
ā
Answer: B) Second law
MCQ 7 When a body is thrown vertically upward, at the highest point:
A) Velocity is maximum B) Acceleration is zero C) Velocity is zero D) Both velocity and acceleration are zero
ā
Answer: C) Velocity is zero
MCQ 8 In an Atwood machine, the acceleration is given by:
A) \(\frac{m_1 + m_2}{m_1 – m_2}g\) B) \(\frac{m_1 – m_2}{m_1 + m_2}g\) C) \(\frac{m_1}{m_2}g\) D) \(\frac{m_2}{m_1}g\)
ā
Answer: B) \(\frac{m_1 – m_2}{m_1 + m_2}g\)
ā ļø Please do not repost this PDF on any website or social platform without permission.
This PDF provides the solutions of every question from the exercises of class 12 dynamics chapter.
This PDF provides the solutions of every question from the exercises of class 12 dynamics chapter.
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š Key Dynamics Summary:
ā¢ Newton’s Second Law: \(F = ma\)
ā¢ Equations of Motion: \(v = u + at\), \(s = ut + \frac{1}{2}at^2\), \(v^2 = u^2 + 2as\)
ā¢ Motion under Gravity: \(v = u – gt\), \(h = ut – \frac{1}{2}gt^2\), \(v^2 = u^2 – 2gh\)
ā¢ Atwood Machine: \(a = \frac{m_1 – m_2}{m_1 + m_2}g\), \(T = \frac{2m_1m_2}{m_1 + m_2}g\)
ā¢ Weight: \(W = mg\)
ā¢ Newton’s Second Law: \(F = ma\)
ā¢ Equations of Motion: \(v = u + at\), \(s = ut + \frac{1}{2}at^2\), \(v^2 = u^2 + 2as\)
ā¢ Motion under Gravity: \(v = u – gt\), \(h = ut – \frac{1}{2}gt^2\), \(v^2 = u^2 – 2gh\)
ā¢ Atwood Machine: \(a = \frac{m_1 – m_2}{m_1 + m_2}g\), \(T = \frac{2m_1m_2}{m_1 + m_2}g\)
ā¢ Weight: \(W = mg\)