Chapter 1 (2-mark type).
A1. What is a force? Give its unit.
A force is a push or pull that can change the state of motion or shape of an object. Its SI unit is the newton (N).
A2. State Newton's First Law of Motion.
A body remains at rest or continues in uniform motion in a straight line unless acted upon by an external unbalanced force.
A3. What is an inertial frame of reference?
An inertial frame of reference is one in which a body not acted on by any net force moves with constant velocity (i.e., Newton’s laws hold true).
A4. State Newton's Second Law of Motion.
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.
A5. How do you calculate the net force acting on an object using the superposition principle?
The net force is found by taking the vector sum of all individual forces acting on the object.
A6. Give an example of Newton’s Second Law in action.
A heavier object requires a larger force than a lighter one to produce the same acceleration.
A7. What is the difference between mass and weight?
Mass is the amount of matter in a body (constant), while weight is the gravitational force acting on the body (variable).
A8. How does the acceleration due to gravity (g) vary with location on Earth?
The value of g is maximum at the poles and minimum at the equator because of Earth’s shape and rotation.
A9. How is mass measured in a laboratory setting?
Mass is measured using a beam balance or electronic balance, by comparing it with standard known masses.
A10. State Newton’s Third Law of Motion.
For every action, there is an equal and opposite reaction.
A11. How is weight measured, and what factors affect its value?
Weight is measured using a spring balance. It depends on the mass of the body and the local value of gravitational acceleration (g).
A12. What is the mathematical relationship between mass, force, and acceleration?
The relationship is F = m a, where F is force, m is mass, and a is acceleration.
A13. Give any two examples of applications of Newton’s third law in everyday life.
-
A gun recoils when a bullet is fired.
-
A swimmer pushes water backward and moves forward.
A14. How does Newton’s Third Law explain the motion of a rocket?
The rocket expels gases backward with great force (action), and the gases push the rocket forward with an equal and opposite force (reaction).
A15. What is the acceleration of an object in free fall under Earth's gravitational field? Provide its value.
An object in free fall accelerates downward due to gravity with acceleration g ≈ 9.8 m/s².
A16. Define contact force. Give an example.
A contact force is a force that acts only when two bodies physically touch each other.
Example: Frictional force or normal force.
A17. Give any two properties of forces.
-
Force has magnitude and direction (it is a vector).
-
Force can change the shape, speed, or direction of a body.
A18. What is a long-range force? Give an example.
A long-range force is a force that acts without physical contact between bodies.
Example: Gravitational force or electrostatic force.
A19. What does the principle of superposition of forces state?
The total force on a body is the vector sum of all the individual forces acting on it independently.
A20. Is force a scalar or vector? Justify your answer.
Force is a vector quantity because it has both magnitude and direction and follows vector addition.
Below are clear, concise, and exam-ready answers for A21–A34, written in standard definitions suitable for scoring full marks.
A21. How do forces combine when acting on the same object?
Forces acting on the same object combine by taking their vector sum, considering both magnitude and direction.
A22. If two forces F₁ and F₂ act in opposite directions how do you determine the resultant force?
The resultant force is the difference of the two forces, taken in the direction of the larger force.
A23. How do you calculate the net force if two forces act at right angles to each other?
The net force is found using the Pythagorean theorem:
[
F_\text{net}=\sqrt{F_1^2+F_2^2}
]
A24. What happens to an object in motion according to Newton’s First Law if no external force is applied?
It continues to move with constant speed in a straight line.
A25. How does Newton's First Law relate to balanced forces?
Newton’s First Law states that when forces on a body are balanced (net force = 0), the body remains at rest or moves with constant velocity.
A26. What does it mean when an object is in a state of equilibrium?
An object is in equilibrium when the net force acting on it is zero and there is no change in its state of motion.
A27. What is an inertial frame of reference?
An inertial frame of reference is a frame of reference in which a body not acted on by a net force moves with constant velocity.
A28. In an inertial frame, how does a body behave in the absence of external forces?
It either remains at rest or moves with constant velocity in a straight line.
A29. Name four common types of forces.
-
Gravitational force
-
Frictional force
-
Tension force
-
Normal force
A30. What happens to acceleration if the mass of an object is doubled and the force remains constant?
Acceleration becomes half of its original value.
A31. How is weight calculated using the acceleration due to gravity (g)?
Weight is calculated by the formula:
[
W = mg
]
where m is mass and g is gravitational acceleration.
A32. How does the value of gravitational acceleration (g) vary with altitude?
The value of g decreases as altitude increases.
A33. What factors cause the variation of acceleration due to gravity (g) on Earth?
-
Altitude (g decreases with height)
-
Latitude (g is more at poles, less at equator)
-
Earth’s rotation
A34. What is an "action–reaction" force pair?
It is a pair of forces in which two bodies exert equal and opposite forces on each other simultaneously, as stated by Newton’s Third Law.
Below are clear, concise, exam-ready long-answer explanations for A1–A13, written in standard Class 11 / intermediate-level physics language to help you score full marks.
A1. Motion of a raindrop and why it attains constant velocity
A raindrop initially accelerates downward under gravity with acceleration g. As its speed increases, the air resistance (viscous drag) acting upward also increases. After some time, the upward force of air resistance becomes equal to the downward gravitational force.
Thus,
[
mg = F_\text{drag}
]
and the net force becomes zero.
According to Newton’s First Law, when net force is zero, acceleration becomes zero.
[
a = 0
]
Hence, the raindrop continues to fall with a constant velocity, called the terminal velocity.
A2. Apparent weight and why astronauts feel weightlessness
Apparent weight is the normal reaction exerted by a surface (like a floor or weighing scale) on a body. It is what we “feel” as weight.
An astronaut in a space station feels weightlessness because both the astronaut and the space station are in continuous free fall around Earth. Since there is no normal reaction acting on the astronaut, their apparent weight becomes zero, even though gravitational force is still acting.
A3. Distinguish between apparent weight and actual weight; effect when tossing an object upward on a scale
-
Actual weight (W = mg): The true gravitational force acting on the body.
-
Apparent weight: The normal reaction felt by the body; depends on acceleration.
When you toss a heavy object upward while standing on a scale:
You push the object upward, and by Newton’s Third Law it pushes you downward.
Thus, the scale reads more weight temporarily.
A4. Terminal speed; graphs of velocity and acceleration vs time
Terminal speed is the constant speed attained by a body falling through a fluid when the net force becomes zero.
Graphical Explanation
-
Velocity–time graph: Velocity increases initially and then gradually approaches a constant value (terminal velocity).
-
Acceleration–time graph: Acceleration starts at g, decreases gradually due to increasing drag, and becomes zero at terminal velocity.
(Draw simple standard exam graphs: velocity curve rising and flattening; acceleration curve falling from g to 0.)
A5. Difference between kinetic and static friction; why friction increases on very smooth surfaces
-
Static friction: Acts when surfaces are at rest relative to each other; adjusts itself up to a maximum value.
-
Kinetic friction: Acts when surfaces slide; usually smaller than maximum static friction.
Why friction increases on extremely smooth surfaces?
Very smooth surfaces have more microscopic contact and allow electromagnetic attractive forces (like van der Waals forces) to dominate. These adhesive forces increase friction even though the surface is smooth.
A6. Graphical variation of frictional force with applied force; define limiting friction
As applied force increases:
-
Friction increases linearly to match the applied force (static friction region).
-
Just before motion begins, friction reaches a maximum value called limiting friction.
-
After motion starts, friction drops to a lower constant value (kinetic friction).
Thus, limiting friction is the maximum static friction that must be overcome to start motion.
A7. Why speed does not change in uniform circular motion even though net force exists
In uniform circular motion, the net force (centripetal force) is always directed towards the centre. This force is perpendicular to the velocity of the particle.
A force perpendicular to velocity changes only direction, not magnitude.
Hence, the speed remains constant while the direction continuously changes.
A8. Fundamental forces in nature and their order of strength
The four fundamental forces are:
-
Gravitational force (weakest)
-
Weak nuclear force
-
Electromagnetic force
-
Strong nuclear force (strongest)
Order (ascending strength):
[
\text{Gravitational} < \text{Weak} < \text{Electromagnetic} < \text{Strong}
]
A9. Centripetal force on a bucket in horizontal circular motion; why work done is zero
The tension in the string provides the centripetal force needed to keep the bucket moving in a circle.
Why work done by the string is zero?
Work done is
[
W = \vec{F} \cdot \vec{d}
]
The tension force acts toward the centre, while the bucket’s displacement is tangential.
Since the force is perpendicular to displacement,
[
W = 0.
]
A10. Terminal speed; table-tennis ball vs lead ball
Terminal speed is the constant speed a falling object reaches when drag force equals weight.
For two balls of the same radius:
-
Same air resistance (same cross-section)
-
But weight of the lead ball is greater
Thus, the lead ball requires a larger drag force to balance its weight, thus it attains a higher terminal speed than the table-tennis ball.
A11. Why a falling cat hits the safety net no faster from 50 stories than from 20 stories
As the cat falls, air resistance increases with speed. After some height, the cat reaches terminal velocity.
Once terminal velocity is reached, it falls with constant speed.
Since both 20 and 50 stories are much greater than the height needed to reach terminal velocity, the cat hits the net at approximately the same speed in both cases.
A12. Terminal speed; why heavy objects fall faster than light objects in air
Heavy objects have higher weight mg but experience the same drag force at the same speed.
Thus, they require a higher speed for drag to balance weight. Hence, they attain a higher terminal speed and fall faster in air (but not in vacuum).
**A13. (i) Fundamental forces of nature
(ii) Theory of everything**
(i) Fundamental Forces
The four fundamental forces are:
-
Gravitational
-
Electromagnetic
-
Weak nuclear
-
Strong nuclear
All physical interactions arise from these forces.
(ii) Theory of Everything (TOE)
A Theory of Everything is a theoretical framework that aims to unify all four fundamental forces into a single comprehensive theory.
Modern attempts include string theory and quantum gravity models.
If you want, I can also provide diagrams/graphs for A4 and A6, or combine everything (A1–A34) into one clean revision booklet.
Chapter 1 (6-mark type).
Below are well-explained, simple, exam-ready 6-mark answers for B1–B12.
Each answer includes the concept, steps, calculations, and final conclusion—exactly what is needed for full marks.
B1. Acceleration of a 13 kg object acted upon by 5 N right and 12 N upward
Forces act perpendicular to each other, so the net force is found using the Pythagorean theorem:
[
F_\text{net}=\sqrt{5^2 + 12^2}
]
[
F_\text{net}=\sqrt{25+144}= \sqrt{169}=13\text{ N}
]
Using Newton’s Second Law:
[
a=\frac{F_\text{net}}{m}=\frac{13}{13}=1\text{ m/s}^2
]
Direction
Angle with the horizontal:
[
\theta=\tan^{-1}\left(\frac{12}{5}\right)\approx 67.4^\circ
]
Final Answer
Acceleration = 1 m/s² at 67.4° above the horizontal (towards up-right direction).
B2. Velocity of two balls dropped from 50 m (no air resistance)
Using
[
v^2 = u^2 + 2gh
]
Here:
(u = 0), (g = 10,\text{m/s}^2), (h = 50\text{ m})
[
v=\sqrt{2gh}=\sqrt{2 \times 10 \times 50}=\sqrt{1000}=31.6\text{ m/s}
]
Both balls have the same final velocity because mass does not matter in free fall.
Final Answer
Each ball hits the ground with 31.6 m/s downward.
B3. Net force on car accelerating from rest to 72 km/h in 10 s
Step 1: Convert speed
[
72\text{ km/h} = 20\text{ m/s}
]
Step 2: Compute acceleration
[
a=\frac{v-u}{t}=\frac{20-0}{10}=2\text{ m/s}^2
]
Step 3: Apply Newton’s Second Law
[
F = ma=1200 \times 2=2400\text{ N}
]
Final Answer
Net force = 2400 N in the forward direction.
B4. Effect of varying gravitational acceleration; find weight on two planets
Weight depends on gravity:
[
W = mg
]
Planet 1: g = 12 m/s²
[
W_1 = 5 \times 12 = 60\text{ N}
]
Planet 2: g = 5 m/s²
[
W_2 = 5 \times 5 = 25\text{ N}
]
Explanation
If gravitational acceleration increases, weight increases.
If gravitational acceleration decreases, weight decreases.
Final Answer
Weight is 60 N on planet 1 and 25 N on planet 2.
B5. Forces acting on a book in static equilibrium on a shelf
Forces on the book
-
Weight (mg) acting downward.
-
Normal reaction from the shelf acting upward.
Reaction forces by Newton’s Third Law
-
The book pulls the Earth downward with force mg (reaction to weight).
-
The book pushes the shelf downward with an equal force N (reaction to normal force).
Action–reaction pairs
-
Earth pulls book downward ↔ Book pulls Earth upward
-
Shelf pushes book upward ↔ Book pushes shelf downward
Since forces balance,
[
N = mg
]
and the book remains at rest.
B6. Forces while pushing a shopping cart
When pushing the cart forward:
Newton’s Third Law
The force you apply on the cart is equal in magnitude and opposite in direction to the force the cart applies on your hands.
They form an action–reaction pair.
When the cart moves at constant speed
Net force = 0
Acceleration = 0
This means the applied force equals friction but action–reaction between hands and cart remains unchanged.
Final Answer
Your push on the cart and the cart’s push on your hands are always equal and opposite. Even when moving at constant speed, these forces remain equal.
B7. Is an object moving in a circular path at constant speed accelerating?
Student 1: Says no acceleration because speed is constant.
Student 2: Says acceleration exists.
Correct Student: Student 2
Reason
Acceleration is change in velocity, not just speed.
In circular motion, the direction of velocity continuously changes.
Thus, there is centripetal acceleration directed toward the centre, even though the speed is constant.
Final Answer
The object is accelerating because its direction changes continuously.
B8. Force of friction on a cyclist
Given:
Mass = 75 kg
Initial velocity = 5 m/s
Final velocity = 0
Time = 4.5 s
Step 1: Acceleration
[
a = \frac{v-u}{t} = \frac{0 - 5}{4.5} = -1.11\text{ m/s}^2
]
Step 2: Force of friction
[
F = ma = 75 \times (-1.11) = -83.25\text{ N}
]
Negative sign → force opposite motion.
Final Answer
Friction force ≈ 83.3 N opposite the direction of motion.
B9. Acceleration of a Rolls-Royce Phantom
Given:
Weight = (2.49 \times 10^4) N
Force = ( -1.83 \times 10^4) N
g = 9.8 m/s²
Step 1: Find mass
[
m=\frac{W}{g}=\frac{2.49 \times 10^4}{9.8}=2541\text{ kg}
]
Step 2: Acceleration
[
a = \frac{F}{m} = \frac{-1.83 \times 10^4}{2541} = -7.2\text{ m/s}^2
]
Final Answer
Acceleration = –7.2 m/s², i.e., slowing down.
B10. Resultant force of two dogs pulling at 60°
Given:
(F_1 = 270\text{ N}), (F_2 = 300\text{ N})
Step 1: Magnitude using vector addition
[
F_R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos60^\circ}
]
[
= \sqrt{270^2 + 300^2 + 2(270)(300)(0.5)}
]
[
= \sqrt{72900 + 90000 + 81000}
]
[
= \sqrt{243000} = 493\text{ N}
]
Step 2: Angle with Rover’s rope
[
\tan \theta = \frac{F_2 \sin 60^\circ}{F_1 + F_2 \cos 60^\circ}
]
[
\tan \theta = \frac{300(0.866)}{270 + 300(0.5)}
= \frac{259.8}{420}
= 0.618
]
[
\theta = 31.7^\circ
]
Final Answer
Resultant force = 493 N
Angle with Rover’s rope = 31.7°
B11. Elevator passenger: normal force 620 N, weight 650 N
Step 1: Reaction forces
-
Reaction to normal force: passenger pushes floor downward with 620 N.
-
Reaction to weight: passenger pulls Earth upward with 650 N.
Step 2: Net force
[
F_\text{net} = N - W = 620 - 650 = -30\text{ N}
]
Negative → downward net force.
Step 3: Acceleration
[
a=\frac{F}{m}=\frac{-30}{650/9.8}
]
[
m = \frac{650}{9.8} = 66.3\text{ kg}
]
[
a = \frac{-30}{66.3} = -0.45\text{ m/s}^2
]
Final Answer
Passenger accelerates downward at 0.45 m/s².
B12. Astronaut's pack weight
Earth weight = 17.5 N
Moon weight = 3.24 N
(a) Acceleration due to gravity on the moon
Weight = mg
[
g_m = \frac{W_m}{m}
]
First find mass using Earth value:
[
m = \frac{W_e}{g} = \frac{17.5}{9.8} = 1.79\text{ kg}
]
Now,
[
g_m = \frac{3.24}{1.79} = 1.81\text{ m/s}^2
]
(b) Mass on moon
Mass is constant everywhere: 1.79 kg
Final Answers
(a) (g_m = 1.81\text{ m/s}^2)
(b) Mass of pack = 1.79 kg
Below are clear, well-explained, full-length 6-mark answers for B1–B4, B6, B9, B10, written in simple language but with complete reasoning, steps, and formulas—exactly what is needed to score full marks.
⭐ B1. Elevator weighing machine problem
A student weighs 550 N on Earth.
Thus, the true weight = mg = 550 N
So,
[
m=\frac{550}{9.8}=56.1\text{ kg}
]
A weighing machine measures the normal reaction (apparent weight).
(i) Machine reads 450 N
Normal reaction:
[
N = 450\text{ N}
]
Using Newton's second law (taking upward as positive):
[
N - mg = ma
]
[
450 - 550 = 56.1a
]
[
-100 = 56.1a
]
[
a = -1.78\text{ m/s}^2
]
Negative sign = downward acceleration.
✔ Answer (i):
Acceleration = 1.78 m/s² downward
(ii) Machine reads 670 N
[
N = 670\text{ N}
]
[
N - mg = ma
]
[
670 - 550 = 56.1a
]
[
120 = 56.1a
]
[
a = 2.14 \text{ m/s}^2
]
✔ Answer (ii):
Acceleration = 2.14 m/s² upward
(iii) Machine reads ZERO — should the student worry?
If the weighing machine reads zero, then:
[
N = 0
]
[
0 - 550 = 56.1a
]
[
a = -9.8 \text{ m/s}^2
]
This means the elevator (and the student) are in free fall.
This is what happens when the cable breaks.
✔ Answer (iii):
Yes, the student should worry — a zero reading means the elevator is falling freely, and the student is weightless, which is dangerous.
⭐ B2. Terminal speed and fluid resistance
Definition
Terminal speed is the constant speed attained by a falling body when the net force becomes zero because fluid resistance = weight.
Fluid resistance for small bodies at low speed
For tiny objects moving slowly (e.g., pollen, raindrop initially):
Fluid resistance is proportional to speed:
[
F_{\text{drag}} = kv
]
At terminal speed:
[
mg = kv_t
]
[
v_t = \frac{mg}{k}
]
Fluid resistance for large bodies at high speed
For large objects moving fast (e.g., skydivers):
Fluid resistance is proportional to square of speed:
[
F_{\text{drag}} = bv^2
]
At terminal speed:
[
mg = bv_t^2
]
[
v_t = \sqrt{\frac{mg}{b}}
]
✔ Final Answer (B2):
Terminal speed exists when weight = drag force.
For small, slow objects:
[
v_t = \frac{mg}{k}
]
For large, fast objects:
[
v_t = \sqrt{\frac{mg}{b}}
]
⭐ B3. Speed approaches terminal speed as time → ∞
Consider a body falling through a viscous fluid where drag is proportional to velocity:
[
m\frac{dv}{dt} = mg - kv
]
Rearrange:
[
\frac{dv}{mg - kv} = \frac{dt}{m}
]
Integrate:
[
v(t) = v_t \left(1 - e^{-\frac{kt}{m}}\right)
]
Where
[
v_t = \frac{mg}{k}
]
As
[
t \to \infty,\quad e^{-kt/m} \to 0
]
Thus:
[
v(t) \to v_t
]
✔ Final Answer:
Mathematically, velocity approaches terminal speed exponentially, reaching it fully only as time → ∞.
⭐ B4. Rocket-propelled object on frictionless ice
Given:
Mass = 45 kg
Force = (F(t) = 16.8t)
Time = 5 s
Surface = frictionless
Step 1: Find acceleration
[
a(t) = \frac{F(t)}{m} = \frac{16.8t}{45} = 0.373t
]
Step 2: Velocity by integration
[
v(t) = \int a(t),dt = \int 0.373t,dt
]
[
v(t) = 0.373 \frac{t^2}{2} = 0.1865t^2
]
Step 3: Displacement
[
x(t)=\int v(t),dt = \int 0.1865 t^2, dt
]
[
x(t) = 0.1865\frac{t^3}{3} = 0.06217 t^3
]
At (t = 5 s):
[
x = 0.06217(125)=7.77\text{ m}
]
✔ Final Answer:
The object travels 7.77 m in the first 5 seconds.
⭐ B6. Worker pushing a 16.8 kg box at constant speed
Given:
Mass = 16.8 kg
Coefficient of kinetic friction = 0.20
Speed = constant → acceleration = 0
(a) Force needed to keep constant speed
Friction force:
[
f_k = \mu_k mg = 0.20 \times 16.8 \times 9.8 = 32.9\text{ N}
]
Since speed is constant:
[
F_{\text{applied}} = f_k = 32.9\text{ N}
]
(b) Distance traveled after force is removed
When force is removed, friction decelerates the box.
Deceleration:
[
a = \frac{f_k}{m}= \frac{32.9}{16.8} = 1.96\text{ m/s}^2
]
Initial velocity = 3.5 m/s
Final velocity = 0
Using:
[
v^2 = u^2 - 2ad
]
[
0 = (3.5)^2 - 2(1.96)d
]
[
d = \frac{12.25}{3.92} = 3.12\text{ m}
]
✔ Final Answers
(a) Required pushing force = 32.9 N
(b) Distance before stopping = 3.12 m
⭐ B9. Coefficients of static and kinetic friction
Given:
Weight of crate = 500 N
Mass = (500/9.8 = 51.02\text{ kg})
Static friction (just before motion)
Minimum force to start motion = 230 N
[
f_s = \mu_s N = \mu_s (500)
]
[
\mu_s = \frac{230}{500} = 0.46
]
Kinetic friction (during motion)
Force to keep constant speed = 200 N
[
f_k = \mu_k (500)
]
[
\mu_k = \frac{200}{500} = 0.40
]
✔ Final Answers:
[
\mu_s = 0.46,\qquad \mu_k = 0.40
]
⭐ B10. Object in vertical circle vs horizontal circle
Horizontal Circular Motion
-
Speed is constant.
-
Only the centripetal force acts (provided by tension).
-
Tension is constant.
Vertical Circular Motion
-
Object moves in a vertical plane.
-
Weight acts downward at all points.
-
Tension + weight together determine centripetal force.
-
Speed changes because gravity speeds up the object on the way down and slows it on the way up.
-
Tension is maximum at the bottom and minimum at the top of the circle.
Comparison
| Feature | Horizontal Circle | Vertical Circle |
|---|---|---|
| Speed | Constant | Changes continuously |
| Gravity effect | Irrelevant | Significant |
| Tension | Constant | Maximum at bottom, minimum at top |
| Centripetal force | Provided only by tension | Provided by tension + gravity component |
| Motion | Uniform | Non-uniform |
✔ Final Answer:
Horizontal circular motion is uniform, but vertical circular motion is non-uniform due to gravity. The tension varies in the vertical circle, making its dynamics more complex than the horizontal case.
Chapter 1 (14-mark type).
Below are clear, very simple, well-explained 14-mark (C-type) answers for C1–C11.
Each answer is written in easy language, with proper structure, diagrams explained in words, examples, equations, and enough content to score full marks.
⭐ C1. Describe Newton's three laws of motion with examples. (7 marks)
Newton’s First Law (Law of Inertia)
A body continues in its state of rest or uniform motion in a straight line unless an external unbalanced force acts on it.
Explanation:
Objects resist changes in motion. They keep doing what they are doing unless a force acts.
Example:
A book on a table remains at rest until someone pushes it.
Newton’s Second Law (Law of Acceleration)
The rate of change of momentum of a body is directly proportional to the force applied and occurs in the direction of the force.
For constant mass:
[
F = ma
]
Explanation:
More force means more acceleration. Heavier objects need more force.
Example:
Pushing an empty trolley accelerates faster than pushing a loaded trolley.
Newton’s Third Law (Action–Reaction Law)
For every action, there is an equal and opposite reaction.
Explanation:
Forces always occur in pairs acting on different bodies.
Example:
When you jump from the ground, you push the ground downward and the ground pushes you upward.
⭐ Final Summary:
-
1st Law → inertia
-
2nd Law → relation of force and acceleration
-
3rd Law → equal and opposite forces
⭐ C2. Newton’s laws and walking on a rough surface (with diagram). (7 marks)
1. Newton’s Third Law in walking
When a person walks, they push the ground backward with their foot.
The ground pushes the person forward with an equal and opposite force.
This forward reaction force helps us move.
2. Role of Friction
Friction on a rough surface acts forward, helping us move.
Without friction, the foot would slip backward.
3. Newton’s First Law
If friction disappears (e.g., ice), the person cannot start walking because there is no unbalanced forward force.
4. Newton’s Second Law
The stronger the push backward, the larger the forward reaction, so the greater the acceleration.
Diagram (describe in words for exam):
-
A foot shown behind pushing backward.
-
Backward arrow labelled "action force".
-
Forward arrow from ground labelled "reaction → friction".
-
Body moving forward.
⭐ Final Summary:
Walking is possible only because of friction and Newton’s Third Law (action–reaction pair between foot and ground).
⭐ C3. Inertial vs Non-inertial frames of reference (7 marks)
Inertial Frame
A frame that is either at rest or moving with constant velocity.
Newton’s laws of motion hold true here.
Example:
A train moving at constant speed.
Objects do not accelerate unless a force acts.
Non-Inertial Frame
A frame that is accelerating.
Newton’s laws do not hold unless we introduce a fictitious (pseudo) force.
Example:
Inside a car that suddenly starts, you feel pushed backward due to a pseudo force.
Why inertial frames are important in Newtonian mechanics?
-
Newton’s laws are valid only in inertial frames.
-
Non-inertial frames require special corrections (pseudo forces).
-
All basic definitions—force, momentum, and acceleration—are simple in inertial frames.
⭐ Final Summary:
Inertial → no acceleration; Newton’s laws work.
Non-inertial → accelerating frame; require pseudo forces.
⭐ C4. Difference between mass and weight; how they are measured. (7 marks)
Mass
-
Amount of matter.
-
Scalar.
-
Constant everywhere.
-
Unit: kg
-
Measured using: Beam balance (comparison method).
Weight
-
Gravitational force acting on a body.
-
Vector.
-
Changes from place to place.
-
Unit: N
-
Measured using: Spring balance.
[
W = mg
]
Example
On Earth, a 5 kg object weighs:
[
W = 5 \times 9.8 = 49 N
]
On the Moon (g = 1.6 m/s²):
[
W = 5 \times 1.6 = 8 N
]
Mass remains 5 kg, but weight changes.
⭐ C5. Newton’s Third Law + Horse–wagon explanation (7 marks)
(i) Newton’s Third Law
For every action, there is an equal and opposite reaction.
Examples
-
A rocket moves upward by pushing gases downward.
-
Swimming: The swimmer pushes water backward and water pushes them forward.
(ii) Why does the wagon move if forces are equal?
Action–reaction forces act on different bodies.
-
Horse pulls wagon → wagon moves forward.
-
Wagon pulls horse backward → force on horse.
The horse pushes the ground backward.
The ground pushes the horse forward (reaction).
This forward force moves both the horse and the wagon.
So motion depends on ground friction, not on the action–reaction pair between horse and wagon.
⭐ C6. Dynamics of uniform circular motion & centripetal vs centrifugal force (7 marks)
Uniform circular motion
A body moves with constant speed but changing direction.
There is an inward acceleration:
[
a = \frac{v^2}{r}
]
To produce this, a centripetal force acts towards the center:
[
F_c = \frac{mv^2}{r}
]
Centripetal Force
-
Real force
-
Direction: towards center
-
Needed to keep object moving in circle
-
Examples: tension, gravity, friction depending on situation
Centrifugal Force
-
Not a real force
-
Felt only in a rotating (non-inertial) frame
-
Acts outward
-
It is a pseudo force due to inertia.
Difference
| Centripetal | Centrifugal |
|---|---|
| Real force | Pseudo force |
| Inward | Outward |
| Needed for circular motion | Felt due to inertia |
| Seen in ground frame | Seen in rotating frame |
⭐ C7. Banking of roads ensures safety (7 marks)
Banking means raising the outer edge of a curved road.
Why needed?
On a flat road, friction alone provides centripetal force:
[
F = \frac{mv^2}{r}
]
At high speed, friction may not be enough → skidding.
How banking works
When the road is tilted:
-
The normal force has a horizontal component.
-
This component provides centripetal force.
-
Vehicle gets required force even without friction.
This:
-
Prevents skidding
-
Reduces wear and tear of tyres
-
Allows safer turning at higher speeds
Diagram explanation
Draw a tilted road; weight downward, normal force perpendicular, resolve normal force into vertical and horizontal components.
⭐ C8. Terminal speed and drag forces (7 marks)
(Same as earlier but expanded)
Definition
Terminal speed is the constant speed a falling body reaches when the drag force equals its weight:
[
mg = F_\text{drag}
]
(i) For small, low-speed objects
Drag is proportional to velocity:
[
F_d = kv
]
Equating:
[
mg = kv_t
]
[
v_t = \frac{mg}{k}
]
(ii) For large, high-speed objects
Drag is proportional to velocity squared:
[
F_d = bv^2
]
Equating:
[
mg = bv_t^2
]
[
v_t = \sqrt{\frac{mg}{b}}
]
⭐ C9. Speed approaches terminal value as time → ∞ (7 marks)
Equation of motion:
[
m\frac{dv}{dt} = mg - kv
]
Solution:
[
v(t) = v_t\left(1 - e^{-\frac{kt}{m}}\right)
]
As ( t \to \infty ):
[
e^{-kt/m} \to 0
]
[
v(t) \to v_t
]
Thus, velocity increases gradually but never exceeds terminal speed.
⭐ C10. Types of friction + friction graph (7 marks)
Types of Friction
-
Static friction (prevents motion)
-
Limiting friction (maximum static friction)
-
Kinetic friction (during motion)
-
Rolling friction (lowest friction)
Graph Explanation
-
Applied force increases → static friction increases equally.
-
At limiting friction, motion begins.
-
After motion starts, friction suddenly drops to kinetic friction (lower value).
Graph is a rising straight line till peak, then drops to a constant plateau.
⭐ C11. Angle of friction & angle of repose + relation (7 marks)
Angle of friction (θₑ)
The angle between the normal reaction and the resultant reaction force.
Angle of repose (θᵣ)
The angle of an inclined plane at which a body just begins to slide.
Relation
At the verge of sliding:
[
f_s = \mu_s N
]
On an incline:
[
\tan \theta = \frac{f_s}{N} = \mu_s
]
Thus,
[
\tan(\text{angle of friction}) = \mu_s
]
Also,
[
\theta_e = \theta_r
]