📝 Important Questions · Class 9 Science
Chapter 6: How Forces Affect Motion
15 Short Answer + 10 Long Answer Questions | Theory & Application Both
2–3 Marks SAQ
5 Marks LAQ
Newton’s Laws
Numericals Included
CBSE Pattern
5 Marks LAQ
Newton’s Laws
Numericals Included
CBSE Pattern
📚 Contents
How to Use This Q&A Sheet
- Read each question carefully. Try to answer it on your own first before reading the answer.
- Short Answer Questions (SAQ) carry 2–3 marks each. Keep answers concise — 3 to 5 sentences.
- Long Answer Questions (LAQ) carry 5 marks each. Write all steps for numericals to get full marks.
- Questions marked [Theoretical] test your understanding of laws and concepts.
- Questions marked [Practical/Application] test your ability to apply concepts to real-life situations and numerical problems.
- All questions in this sheet are strictly from Chapter 6 content. No outside material is included.
Exam Strategy
In questions about Newton’s Laws, always state the law first and then explain or apply it. Examiners award marks for the law statement itself!
In questions about Newton’s Laws, always state the law first and then explain or apply it. Examiners award marks for the law statement itself!
Short Answer Questions (15 Questions)
⚡ Forces — Concepts & Friction (Q1–Q5)
Q1. [Theoretical] What is force? State its SI unit and mention two effects it can produce on an object. (2 Marks)
Ans: A force (बल) is a push or pull that can make an object move from rest, change the speed or direction of a moving object, or change the shape of an object. Force is a physical quantity that requires both magnitude and direction to be fully described. The SI unit of force is newton (N). Two effects of force are: (i) it can set a stationary object into motion, and (ii) it can change the shape of an object (e.g., squeezing a lemon).
Q2. [Theoretical] What are balanced forces? Give one example from daily life. (2 Marks)
Ans: Balanced forces are two or more forces acting on an object such that the net force on the object is zero — meaning the forces cancel each other out and the object remains at rest or moves with constant velocity. When the net force is zero, the object does not accelerate. Example: In a tug-of-war game, if both teams pull the rope with equal force in opposite directions, the rope does not move — the forces are balanced.
Exam Tip
Balanced forces → zero net force → zero acceleration → object stays at rest or moves with constant velocity (Newton’s First Law).
Balanced forces → zero net force → zero acceleration → object stays at rest or moves with constant velocity (Newton’s First Law).
Q3. [Practical/Application] Two forces of 8 N and 5 N act on a block in opposite directions. What is the net force on the block? In which direction will the block accelerate? (2 Marks)
Ans: When two forces act in opposite directions, the net force equals the difference of their magnitudes, acting in the direction of the larger force.
Given: F₁ = 8 N (say, right), F₂ = 5 N (left)
Net Force = F₁ − F₂ = 8 − 5 = 3 N
∴ Net Force = 3 N, acting towards the right (direction of the larger force)
Net Force = F₁ − F₂ = 8 − 5 = 3 N
∴ Net Force = 3 N, acting towards the right (direction of the larger force)
Q4. [Theoretical] What is the force of friction? In which direction does it act on a moving object? (2 Marks)
Ans: The force of friction (घर्षण बल) is a force that opposes the relative motion between two surfaces in contact. It acts on a moving object in the direction opposite to its direction of motion. For example, when you push a box on the floor, friction acts backward on the box. Friction is caused by the roughness between the surfaces in contact and depends on the nature of the surfaces.
Q5. [Practical/Application] A bicycle slows down and stops after the rider stops pedalling. Which force is responsible for this? Explain briefly. (2 Marks)
Ans: The force of friction is responsible for slowing down and stopping the bicycle. Once the rider stops pedalling, there is no driving force. However, friction acts between the tyres and the road in the direction opposite to the motion of the bicycle. This friction produces a net decelerating force on the bicycle, gradually reducing its velocity until it comes to rest.
Bonus Point
Air resistance (drag) also acts on the bicycle and contributes to slowing it down, but its magnitude is usually much smaller than road friction and is often neglected.
Air resistance (drag) also acts on the bicycle and contributes to slowing it down, but its magnitude is usually much smaller than road friction and is often neglected.
⚡ Newton’s First & Second Laws (Q6–Q10)
Q6. [Theoretical] State Newton’s First Law of Motion. What does the term ‘inertia’ mean? (3 Marks)
Ans: Newton’s First Law of Motion (जड़त्व का नियम) states: “An object at rest remains at rest and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.” This means that if the net force on an object is zero, its acceleration is zero — it will not start moving, stop, or change direction on its own. The term inertia was used by Isaac Newton to describe the tendency of objects to resist any change in their state of rest or uniform motion. A heavier object has greater inertia than a lighter one.
Remember Verbatim
“An object at rest remains at rest and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.” — Learn this exact statement for exams.
“An object at rest remains at rest and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.” — Learn this exact statement for exams.
Q7. [Practical/Application] A box is being pushed along the floor with a force equal to the force of friction. Will the box move, remain stationary, or accelerate? Give reason. (2 Marks)
Ans: The box will continue moving with constant velocity (if it was already moving) or remain stationary (if it was at rest). Since the applied force equals the force of friction in magnitude but they act in opposite directions, the two forces balance each other. The net force on the box is zero. By Newton’s First Law, zero net force means zero acceleration — so the state of the box does not change.
Q8. [Theoretical] State Newton’s Second Law of Motion and write its mathematical expression. Define one newton. (3 Marks)
Ans: Newton’s Second Law of Motion (त्वरण का नियम) states: “When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the magnitude of the net force and inversely proportional to the mass of the object.” Mathematically:
F = m × a
where F = net force (N), m = mass (kg), a = acceleration (m s⁻²). One newton (1 N) is defined as the force that produces an acceleration of 1 m s⁻² in an object of mass 1 kg.
Q9. [Practical/Application] A net force of 40 N acts on an object of mass 8 kg. Calculate the acceleration produced. (2 Marks)
Ans: Using Newton’s Second Law:
Given: Net Force (F) = 40 N, Mass (m) = 8 kg
Formula: a = F ÷ m
Substituting: a = 40 ÷ 8
∴ Acceleration (a) = 5 m s⁻²
Formula: a = F ÷ m
Substituting: a = 40 ÷ 8
∴ Acceleration (a) = 5 m s⁻²
Q10. [Practical/Application] How do airbags in vehicles help protect passengers during a collision? Which law of motion is involved? (3 Marks)
Ans: This is an application of Newton’s Second Law of Motion. During a collision, the vehicle comes to an abrupt halt and the passenger’s body tends to continue moving forward due to inertia. Without an airbag, the passenger’s head hits the hard steering wheel in a very short time, causing a very large decelerating force. The airbag inflates quickly and acts as a soft cushion — it increases the time duration over which the passenger’s velocity reduces to zero. Since F = ma and a = Δv / t, a longer stopping time means smaller acceleration, and therefore a smaller force on the passenger’s body, reducing the risk of serious injury.
⚡ Newton’s Third Law & Systems (Q11–Q15)
Q11. [Theoretical] State Newton’s Third Law of Motion. Does it apply to non-contact forces? (2 Marks)
Ans: Newton’s Third Law of Motion (क्रिया-प्रतिक्रिया का नियम) states: “Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first object.” These forces are called action-reaction pairs. Importantly, the two forces always act on two different objects. Yes, Newton’s Third Law applies to all types of forces — both contact forces (push, friction) and non-contact forces (gravitational, magnetic, electrostatic).
Q12. [Practical/Application] Explain why a rocket is able to launch and move upward. Which law of motion explains this? (3 Marks)
Ans: The launch of a rocket is explained by Newton’s Third Law of Motion. The rocket engine burns fuel and expels hot gases downward at very high velocity. By Newton’s Third Law, the expelled gases exert an equal and opposite force on the rocket in the upward direction. This upward force (called thrust) is greater than the weight of the rocket, so the net force on the rocket is upward, and the rocket accelerates upward and lifts off.
Indian Context
India’s Chandrayaan-3 mission used the same principle to slow down and land near the south pole of the Moon. The Vikram lander fired its engine in the direction of its motion to reduce speed before soft landing — the exhaust force acted opposite to motion, decelerating the lander.
India’s Chandrayaan-3 mission used the same principle to slow down and land near the south pole of the Moon. The Vikram lander fired its engine in the direction of its motion to reduce speed before soft landing — the exhaust force acted opposite to motion, decelerating the lander.
Q13. [Theoretical] The Earth and a falling apple exert equal gravitational forces on each other. Why does only the apple appear to move towards the Earth? (2 Marks)
Ans: By Newton’s Third Law, the Earth pulls the apple downward and the apple pulls the Earth upward with equal forces. Both forces are equal in magnitude. However, by Newton’s Second Law, acceleration = Force / mass. The mass of the Earth is enormously large compared to the apple. Therefore, the acceleration of the Earth due to the apple’s pull is extremely tiny — far too small to be noticed. The apple, having a very small mass, gains a large acceleration (g = 9.8 m s⁻²) and visibly moves toward the Earth.
Q14. [Practical/Application] Why does a fielder pull their hands backward while catching a fast cricket ball? (2 Marks)
Ans: This is an application of Newton’s Second Law of Motion. When a fielder pulls their hands backward while catching the ball, they increase the time duration over which the ball comes to rest. Since acceleration a = Δv / t, a larger time gives a smaller acceleration. By F = ma, a smaller acceleration means a smaller force needs to be applied by the fielder on the ball to stop it. This smaller force also acts on the fielder’s hands (Newton’s 3rd Law), thus reducing pain and preventing injury.
Q15. [Theoretical] What is meant by internal and external forces in a system of objects? Give a brief example. (3 Marks)
Ans: When two or more connected objects are treated as a single system, the forces that act between the objects within the system are called internal forces, while forces acting from outside the system are called external forces. Internal forces cancel each other within the system and do not change the overall motion of the system. Example: Two boxes connected by a string, pulled by a force F — the tension T in the string is an internal force (between the boxes), while F is the external force. The acceleration of the system is a = F / (m₁ + m₂).
🌟
Fun Fact — Smallest Forces Ever Measured In everyday life, the smallest force we can feel is about 1 millinewton (10⁻³ N) — like a light touch. Scientists, however, have measured forces as tiny as yoctonewtons (10⁻²⁴ N) in specialised experiments!
Long Answer Questions (10 Questions)
⚡ Forces & Newton’s Laws — Theory (Q1–Q4)
Q1. [Theoretical] State and explain Newton’s First Law of Motion with two real-life examples. What role did Galileo Galilei play in its development? (5 Marks)
Ans:
- Statement: “An object at rest remains at rest and an object in motion continues to move with a constant velocity, unless a net force acts upon the object.”
- Explanation: The law means that objects do not change their state of rest or uniform motion on their own — they need a net force to do so. If the net force is zero, acceleration is zero.
- Example 1 (Rest → Motion): A book lying on a table will not move by itself. You need to apply a force to push it. Without a force, it stays at rest.
- Example 2 (Motion → Rest): When you stop pedalling a bicycle, the force of friction brings it to rest. If there were no friction (no net force), the bicycle would continue moving forever at constant velocity.
- Galileo’s Contribution: In ancient times, it was wrongly believed that a continuous force was needed to keep an object moving. In the 17th century, Galileo Galilei used thought experiments to show that if all impediments (like friction) are removed, a body moving on a horizontal plane would continue to move indefinitely without any applied force. Newton built on this idea to formally state his First Law.
Key Point for 5 Marks
Always include: (1) the law statement, (2) explanation with the idea of net force = 0, (3) at least two examples with reasoning, and (4) Galileo’s contribution for full marks.
Always include: (1) the law statement, (2) explanation with the idea of net force = 0, (3) at least two examples with reasoning, and (4) Galileo’s contribution for full marks.
Q2. [Theoretical] State Newton’s Second Law of Motion. Derive the expression F = ma. Define one newton. (5 Marks)
Ans:
- Statement: “When a net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is proportional to the net force and inversely proportional to the mass of the object.”
- Step 1 — Force proportional to acceleration (for fixed mass): From experiments, if mass m is constant, then acceleration a ∝ F.
- Step 2 — Acceleration inversely proportional to mass (for fixed force): If force F is constant, then a ∝ 1/m.
- Step 3 — Combining both relations: a ∝ F/m → F = k × m × a, where k is a proportionality constant.
- Step 4 — Choosing SI units: The unit of force (newton) is defined so that k = 1. Therefore: F = m × a
- Definition of 1 Newton: One newton is the force that produces an acceleration of 1 m s⁻² in an object of mass 1 kg. That is: 1 N = 1 kg × 1 m s⁻² = 1 kg m s⁻²
F = m × a
Advanced Tip (Bonus Mark)
The more complete form of Newton’s Second Law is: rate of change of momentum = net force. Momentum p = mv. This form applies even when mass is not constant.
The more complete form of Newton’s Second Law is: rate of change of momentum = net force. Momentum p = mv. This form applies even when mass is not constant.
Q3. [Theoretical] State Newton’s Third Law of Motion. With the help of two examples, explain that the action-reaction forces act on different objects and do not cancel each other. (5 Marks)
Ans:
- Statement: “Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first object.” These forces are called action-reaction pairs.
- Key Note: The two forces are equal in magnitude and opposite in direction, but they act on two different objects. Therefore, they cannot cancel each other (cancellation only happens when two forces act on the same object).
- Example 1 — Rowing a canoe: The canoeist pushes water backward with the paddle (action). The water pushes the paddle and canoe forward with an equal force (reaction). Action acts on water; reaction acts on the canoe. Both are equal in magnitude but do not cancel — the canoe moves forward.
- Example 2 — Walking: When you walk, your foot pushes the ground backward (action). The ground exerts an equal frictional force on your foot in the forward direction (reaction). This reaction pushes you forward. Without friction, your foot would slip and you could not walk.
- Conclusion: Newton’s Third Law applies to all forces — contact (friction, normal) and non-contact (gravity, magnetic, electrostatic).
Common Mistake
Many students think action-reaction forces cancel each other because they are equal and opposite. Remember: they act on two different objects, so they cannot cancel. Cancellation requires both forces on the same object.
Many students think action-reaction forces cancel each other because they are equal and opposite. Remember: they act on two different objects, so they cannot cancel. Cancellation requires both forces on the same object.
Q4. [Theoretical] Explain the force of friction with the help of Activity 6.1 (rubber band and coins experiment). What conclusion can be drawn about friction on different surfaces? (5 Marks)
Ans:
- Objective: To investigate how the force of friction varies with the nature of surfaces.
- Setup: A stack of 4 coins (secured with adhesive tape) is launched using a stretched rubber band. The rubber band is stretched to the same mark (point C) every time. This ensures the same initial force each time.
- Procedure: The experiment is repeated on different surfaces — wooden table top, laminated table top, and polished marble/tile floor. The distance the coin stack travels before stopping is recorded.
- Observation: The stack travels the shortest distance on the wooden surface (rough) and the longest distance on the polished marble (smooth).
- Explanation: On rough surfaces, the force of friction is greater — it reduces the coin stack’s velocity quickly, so it stops in a shorter distance. On smooth surfaces, friction is smaller, so the velocity decreases more slowly and the coins travel farther.
- Conclusion: The force of friction depends on the nature of surfaces in contact — rougher surfaces produce greater friction. A spring balance can be used to directly measure the friction force on each surface, confirming the conclusion.
Extended Idea (Thought Experiment)
If the surface were perfectly smooth (zero friction), the coin stack would never stop after being released — it would continue moving forever. This is the condition described in Newton’s First Law.
If the surface were perfectly smooth (zero friction), the coin stack would never stop after being released — it would continue moving forever. This is the condition described in Newton’s First Law.
⚡ Numerical Problems — Application (Q5–Q8)
Q5. [Practical/Application] A student tries to push a stationary block of 25 kg on a horizontal floor. The maximum force of friction is 50 N. Find the displacement of the block in 2 seconds when the student pushes it with (i) 50 N and (ii) 55 N. (5 Marks)
Ans: Using Newton’s Second Law (F = ma) and kinematic equation (s = ut + ½at²):
Given: mass (m) = 25 kg, max friction = 50 N, time (t) = 2 s, initial velocity (u) = 0
Case (i): Applied force = 50 N
Net Force = 50 − 50 = 0 N
Since net force = 0, acceleration = 0, block remains stationary.
∴ Displacement = 0 m (block does not move)
Case (ii): Applied force = 55 N
Net Force = 55 − 50 = 5 N
Acceleration: a = F/m = 5/25 = 0.2 m s⁻²
Displacement: s = ut + ½at² = 0 + ½ × 0.2 × (2)² = 0.4 m
∴ Displacement = 0.4 m in the forward direction
Case (i): Applied force = 50 N
Net Force = 50 − 50 = 0 N
Since net force = 0, acceleration = 0, block remains stationary.
∴ Displacement = 0 m (block does not move)
Case (ii): Applied force = 55 N
Net Force = 55 − 50 = 5 N
Acceleration: a = F/m = 5/25 = 0.2 m s⁻²
Displacement: s = ut + ½at² = 0 + ½ × 0.2 × (2)² = 0.4 m
∴ Displacement = 0.4 m in the forward direction
Marks Tip
For numericals: always write (1) Given values, (2) Formula, (3) Substitution, (4) Final answer with units. Each step carries marks.
For numericals: always write (1) Given values, (2) Formula, (3) Substitution, (4) Final answer with units. Each step carries marks.
Q6. [Practical/Application] A sports car of mass 1500 kg moves east. Its velocity-time graph shows: 0–5 s (velocity increases from 0 to 10 m/s), 5–10 s (constant at 10 m/s), 10–15 s (velocity decreases from 10 to 0 m/s). Calculate the force during each time interval. (5 Marks)
Ans: Using v = u + at to find acceleration, then F = ma:
Given: mass (m) = 1500 kg
Interval 1: 0 s to 5 s
u = 0 m/s, v = 10 m/s, t = 5 s
a₁ = (v − u)/t = (10 − 0)/5 = 2 m s⁻²
F₁ = 1500 × 2 = 3000 N (eastward)
Interval 2: 5 s to 10 s
Constant velocity → acceleration = 0
F₂ = 1500 × 0 = 0 N (no net force)
Interval 3: 10 s to 15 s
u = 10 m/s, v = 0 m/s, t = 5 s
a₃ = (0 − 10)/5 = −2 m s⁻²
F₃ = 1500 × (−2) = −3000 N
∴ F₁ = 3000 N (east), F₂ = 0 N, F₃ = 3000 N (west, opposing motion)
Interval 1: 0 s to 5 s
u = 0 m/s, v = 10 m/s, t = 5 s
a₁ = (v − u)/t = (10 − 0)/5 = 2 m s⁻²
F₁ = 1500 × 2 = 3000 N (eastward)
Interval 2: 5 s to 10 s
Constant velocity → acceleration = 0
F₂ = 1500 × 0 = 0 N (no net force)
Interval 3: 10 s to 15 s
u = 10 m/s, v = 0 m/s, t = 5 s
a₃ = (0 − 10)/5 = −2 m s⁻²
F₃ = 1500 × (−2) = −3000 N
∴ F₁ = 3000 N (east), F₂ = 0 N, F₃ = 3000 N (west, opposing motion)
Q7. [Practical/Application] A bullet of mass 50 g moving with a speed of 100 m/s enters a stationary wooden block and stops after penetrating 50 cm. Estimate the stopping force acting on the bullet (assume constant acceleration). (5 Marks)
Ans: Using kinematics to find acceleration, then Newton’s Second Law:
Given: mass (m) = 50 g = 0.05 kg, initial velocity (u) = 100 m/s, final velocity (v) = 0 m/s, distance (s) = 50 cm = 0.5 m
Step 1 — Find deceleration using v² = u² + 2as:
0 = (100)² + 2 × a × 0.5
0 = 10000 + a
a = −10000 m s⁻²
Step 2 — Find stopping force:
F = m × |a| = 0.05 × 10000
∴ Stopping Force = 500 N (acting opposite to motion)
Step 1 — Find deceleration using v² = u² + 2as:
0 = (100)² + 2 × a × 0.5
0 = 10000 + a
a = −10000 m s⁻²
Step 2 — Find stopping force:
F = m × |a| = 0.05 × 10000
∴ Stopping Force = 500 N (acting opposite to motion)
Common Mistake
Always convert units first: grams to kilograms (÷1000), cm to metres (÷100). Forgetting unit conversion is one of the most common errors in numericals.
Always convert units first: grams to kilograms (÷1000), cm to metres (÷100). Forgetting unit conversion is one of the most common errors in numericals.
Q8. [Practical/Application] When a 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, the gun recoils. Calculate the magnitudes of initial accelerations of the bullet and the gun. Why are they different even though the force is the same? (5 Marks)
Ans: By Newton’s Third Law, the recoil force on the gun equals the force on the bullet (2 N), but in opposite direction.
Given: mass of bullet = 0.1 kg, mass of gun = 5 kg, force = 2 N
Acceleration of bullet:
a_bullet = F / m_bullet = 2 / 0.1 = 20 m s⁻²
Acceleration of gun:
a_gun = F / m_gun = 2 / 5 = 0.4 m s⁻²
∴ Bullet: 20 m s⁻², Gun: 0.4 m s⁻² (in opposite directions)
Acceleration of bullet:
a_bullet = F / m_bullet = 2 / 0.1 = 20 m s⁻²
Acceleration of gun:
a_gun = F / m_gun = 2 / 5 = 0.4 m s⁻²
∴ Bullet: 20 m s⁻², Gun: 0.4 m s⁻² (in opposite directions)
Why are the accelerations different? Even though the forces are equal (Newton’s Third Law), the masses are very different. By a = F/m, the lighter bullet gains much greater acceleration than the heavy gun. The gun “recoils” slowly while the bullet shoots forward fast.
Important Concept
Equal forces ≠ Equal accelerations. Acceleration also depends on mass: a = F/m. The larger the mass, the smaller the acceleration for the same force.
Equal forces ≠ Equal accelerations. Acceleration also depends on mass: a = F/m. The larger the mass, the smaller the acceleration for the same force.
⚡ Systems & Application Questions (Q9–Q10)
Q9. [Practical/Application] An object of mass 2 kg moves at 10 m/s. It enters a rough patch where friction = 7 N and an additional opposing force of 3 N is applied. How far does it travel before stopping? (5 Marks)
Ans: Using Newton’s Second Law to find deceleration, then kinematics to find distance:
Given: m = 2 kg, u = 10 m/s, v = 0, friction force = 7 N, additional force = 3 N (both opposing motion)
Step 1 — Total opposing force:
F_net = 7 + 3 = 10 N (opposing motion)
Step 2 — Deceleration (negative acceleration):
a = −F/m = −10/2 = −5 m s⁻²
Step 3 — Distance using v² = u² + 2as:
0 = (10)² + 2 × (−5) × s
0 = 100 − 10s
s = 100/10 = 10 m
∴ The object travels 10 m before coming to rest.
Step 1 — Total opposing force:
F_net = 7 + 3 = 10 N (opposing motion)
Step 2 — Deceleration (negative acceleration):
a = −F/m = −10/2 = −5 m s⁻²
Step 3 — Distance using v² = u² + 2as:
0 = (10)² + 2 × (−5) × s
0 = 100 − 10s
s = 100/10 = 10 m
∴ The object travels 10 m before coming to rest.
Q10. [Theoretical] Two boxes of masses m₁ and m₂ are connected by a string on a frictionless surface. A force F is applied on Box 1. Explain the concept of treating them as a system and derive the expression for the acceleration of the system. (5 Marks)
Ans:
- Setup: Box 1 (mass m₁) and Box 2 (mass m₂) are connected by a string on a frictionless horizontal surface. A horizontal force F is applied on Box 1 to the right.
- Tension as Internal Force: Box 1 pulls Box 2 forward through the string with tension T. By Newton’s Third Law, Box 2 pulls Box 1 backward with the same tension T. T is an internal force within the two-box system.
- Treating as a System: When we consider both boxes together as a single system, internal forces (T on Box 1 and T on Box 2) cancel each other out. Only external forces matter for the system’s overall acceleration.
- Applying Newton’s Second Law to the System: The only external horizontal force is F. Total mass of system = m₁ + m₂.
- Derivation: a = F / (m₁ + m₂)
- Physical Meaning: The system of two connected boxes accelerates exactly like a single object of mass (m₁ + m₂) under force F. This approach greatly simplifies analysis of connected objects.
a = F / (m₁ + m₂)
Real-Life Connection
When you walk, your arms, legs, and body all move differently. But your overall motion can be analysed by treating your entire body as a single object. This is the power of the “system” approach in Newton’s Laws!
When you walk, your arms, legs, and body all move differently. But your overall motion can be analysed by treating your entire body as a single object. This is the power of the “system” approach in Newton’s Laws!
Formula & Key Terms Quick Reference
⚡ Key Formulas
Newton’s 2nd Law: F = m × a
Acceleration: a = F / m
Gravitational Force (Weight): W = m × g (g = 9.8 m s⁻² ≈ 10 m s⁻²)
Kinematic Equation 1: v = u + at
Kinematic Equation 2: s = ut + ½at²
Kinematic Equation 3: v² = u² + 2as
System Acceleration: a = F / (m₁ + m₂)
📖 Key Terms
| Term | Definition |
|---|---|
| Force (बल) | A push or pull that changes or tends to change the state of rest or motion of an object. SI unit: newton (N). |
| Net Force | The vector sum of all forces acting on an object. Determines if the object accelerates. |
| Balanced Forces | Forces with zero net force; object remains at rest or moves with constant velocity. |
| Inertia (जड़त्व) | Tendency of an object to resist change in its state of rest or uniform motion. |
| Friction (घर्षण) | Force opposing relative motion between surfaces in contact; acts opposite to direction of motion. |
| Newton (N) | SI unit of force; 1 N = force producing 1 m s⁻² acceleration in 1 kg mass. |
| Action-Reaction Pair | Equal and opposite forces acting on two different interacting objects (Newton’s 3rd Law). |
| Normal Force | Force exerted by a surface perpendicular to the surface on an object resting on it; balanced by weight. |
Common Exam Mistakes to Avoid
Mistake 1 — Action-Reaction forces cancelling each other
Students often write “action and reaction forces cancel each other.” This is WRONG. Action-reaction forces act on two different objects and cannot cancel. Forces cancel only when they act on the same object.
Students often write “action and reaction forces cancel each other.” This is WRONG. Action-reaction forces act on two different objects and cannot cancel. Forces cancel only when they act on the same object.
Mistake 2 — Forgetting unit conversions in numericals
Always convert: grams → kg (÷1000), cm → m (÷100), km/h → m/s (÷3.6). Not converting units before calculation is the most common reason for losing marks in numericals.
Always convert: grams → kg (÷1000), cm → m (÷100), km/h → m/s (÷3.6). Not converting units before calculation is the most common reason for losing marks in numericals.
Mistake 3 — Writing “Newton” with capital N for the unit
The unit is spelled “newton” (lowercase n) when written in full, but its symbol is “N” (uppercase). For example, write “5 N” or “five newtons” — NOT “five Newtons.”
The unit is spelled “newton” (lowercase n) when written in full, but its symbol is “N” (uppercase). For example, write “5 N” or “five newtons” — NOT “five Newtons.”
Mistake 4 — Confusing balanced forces with no forces
“Balanced forces” means net force = 0, but it does NOT mean no forces act. Multiple forces can act on an object and still be balanced (e.g., a book on a table has gravity and normal force both acting — they are balanced).
“Balanced forces” means net force = 0, but it does NOT mean no forces act. Multiple forces can act on an object and still be balanced (e.g., a book on a table has gravity and normal force both acting — they are balanced).
Mistake 5 — Not mentioning direction in force/net force answers
Force is a vector quantity — always state direction. Write “3 N towards the right” not just “3 N.” Especially important in net force calculations.
Force is a vector quantity — always state direction. Write “3 N towards the right” not just “3 N.” Especially important in net force calculations.
Mistake 6 — Assuming equal forces produce equal accelerations
Equal and opposite forces (Newton’s 3rd Law) do NOT produce equal accelerations unless masses are equal. Acceleration = F/m depends on mass. Bullet and gun feel same force but have very different accelerations.
Equal and opposite forces (Newton’s 3rd Law) do NOT produce equal accelerations unless masses are equal. Acceleration = F/m depends on mass. Bullet and gun feel same force but have very different accelerations.
Mistake 7 — Omitting the law statement before explaining/applying it
In questions about Newton’s Laws, always begin by stating the law in full. Many students jump to explanation directly. The law statement itself carries 1–2 marks.
In questions about Newton’s Laws, always begin by stating the law in full. Many students jump to explanation directly. The law statement itself carries 1–2 marks.
🗂️ Quick Revision Summary
Force (बल)A push or pull; SI unit = newton (N). Has both magnitude and direction.
Balanced vs Unbalanced ForcesBalanced → net force = 0 → no acceleration. Unbalanced → net force ≠ 0 → acceleration.
Friction (घर्षण)Always opposes motion; depends on surface type. Smaller friction → object travels farther.
Newton’s 1st LawObject stays at rest or in uniform motion unless a net force acts. Explains inertia.
Newton’s 2nd LawF = ma. Greater force → greater acceleration. Greater mass → smaller acceleration.
Newton’s 3rd LawEvery action has an equal and opposite reaction. Forces act on different objects — don’t cancel.
g = 9.8 m s⁻²Acceleration due to gravity is constant near Earth’s surface. Independent of object’s mass.
System of ObjectsInternal forces cancel; only external force matters. a = F / (m₁ + m₂).
Real-Life: Airbag / Catching BallIncreasing stopping time → smaller acceleration → smaller force → less injury (Newton’s 2nd Law).
Real-Life: Rocket LaunchGas expelled downward (action) → equal force pushes rocket upward (reaction) — Newton’s 3rd Law.
Final Exam Strategy
For this chapter: (1) Memorise all three Newton’s Law statements word-for-word — they carry guaranteed marks. (2) In every numerical, write Given → Formula → Substitution → Answer with units. (3) For application questions (airbag, cricket, rocket), always name the specific Newton’s Law before explaining. (4) Remember: Friction is your friend in walking, but an enemy when sliding! Good luck! 🌟
For this chapter: (1) Memorise all three Newton’s Law statements word-for-word — they carry guaranteed marks. (2) In every numerical, write Given → Formula → Substitution → Answer with units. (3) For application questions (airbag, cricket, rocket), always name the specific Newton’s Law before explaining. (4) Remember: Friction is your friend in walking, but an enemy when sliding! Good luck! 🌟

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