This chapter connects the cause (Force) with the effect (Motion). AMU loves asking conceptual questions on the 3 laws and numericals on conservation of momentum — that's the star of this chapter. Recoil of gun, explosion of a bomb, collisions, friction, free body diagrams — all covered here. 15+ PYQs and practice exercises included. Get ready.
1) What is Force?
A force is a push or pull that can change or tend to change the state of rest or of uniform motion of a body. It can also change the shape or size of a body.
Force (F) = mass (m) × acceleration (a)
SI unit: Newton (N) = kg m/s²
CGS unit: dyne = g cm/s²
1 N = 10⁵ dyne
Effects of Force
| Effect | Example |
|---|---|
| Starts motion | Pushing a stationary car |
| Stops motion | Applying brakes |
| Changes speed | Accelerating a bike |
| Changes direction | Turning a cycle handle |
| Changes shape | Squeezing a sponge |
Types of Forces
| Category | Type | Description |
|---|---|---|
| Contact forces | Muscular force | Exerted by muscles of animals/humans (pulling, pushing, lifting) |
| Frictional force | Opposes relative motion between two surfaces in contact | |
| Normal reaction | Force exerted by a surface perpendicular to it (supports weight) | |
| Tension force | Force transmitted through a string/rope when pulled | |
| Non-contact forces | Gravitational force | Force of attraction between any two masses |
| Electrostatic force | Force between charged particles | |
| Magnetic force | Force between magnets or magnetic materials |
🔥 AMU PYQ: Which of the following is a contact force?
(A) Gravitational force (B) Magnetic force (C) Frictional force (D) Electrostatic force
Ans: C (Friction requires contact between surfaces)
(A) Gravitational force (B) Magnetic force (C) Frictional force (D) Electrostatic force
Ans: C (Friction requires contact between surfaces)
2) Balanced vs Unbalanced Forces
| Feature | Balanced forces | Unbalanced forces |
|---|---|---|
| Net force | Zero | Non-zero |
| Effect on motion | No change (body at rest or uniform motion) | Change in state of motion (acceleration produced) |
| Example | Book lying on a table (gravity = normal reaction) | Pushing a box that moves |
🟡 Senior's Tip: Balanced forces mean net force = 0, so by Newton's 1st law, the body continues in its current state (rest or uniform motion). Unbalanced forces cause acceleration (Newton's 2nd law).
3) Newton's First Law of Motion (Law of Inertia)
Statement: Every body continues in its state of rest or of uniform motion in a straight line unless compelled by an external unbalanced force to change that state.
Galileo's Experiment
Galileo demonstrated that a ball rolling down one incline will roll up an opposite incline to nearly the same height. As the slope is decreased, the ball travels farther. In the ideal case of no friction, the ball would roll forever — thus a body in motion stays in motion unless acted upon by a force.
Inertia
The natural tendency of a body to resist any change in its state of rest or motion is called inertia.
Inertia ∝ Mass
Mass is the measure of inertia. Heavier objects have more inertia → harder to start or stop. A truck has more inertia than a bicycle.
Types of Inertia
| Type | Description | Example |
|---|---|---|
| Inertia of rest | Body resists being set into motion | When a bus suddenly starts, passengers fall backward. Dust falls off a carpet when beaten with a stick. |
| Inertia of motion | Body resists being stopped | When a bus suddenly stops, passengers fall forward. A person alighting from a moving train runs forward. |
| Inertia of direction | Body resists change in direction | A cyclist leans while turning. Mud flies off a rotating tire tangentially. A passenger falls sideways when a car turns sharply. |
🔥 AMU PYQ: The law of inertia states that an object at rest will remain at rest and an object in motion will remain in motion in a straight line unless acted upon by an external force.
Ans: True
🔥 AMU PYQ: When a bus suddenly starts, passengers fall backward. This is due to:
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Friction
Ans: A
🔥 AMU PYQ: Which of the following has the highest inertia?
(A) A bicycle (B) A car (C) A truck (D) A motorcycle
Ans: C (inertia ∝ mass, truck has most mass)
Ans: True
🔥 AMU PYQ: When a bus suddenly starts, passengers fall backward. This is due to:
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Friction
Ans: A
🔥 AMU PYQ: Which of the following has the highest inertia?
(A) A bicycle (B) A car (C) A truck (D) A motorcycle
Ans: C (inertia ∝ mass, truck has most mass)
4) Newton's Second Law of Motion
Statement: The rate of change of momentum of a body is directly proportional to the applied unbalanced force and takes place in the direction of the force.
Derivation of F = ma
From 2nd law: F ∝ (mv - mu) / t
F ∝ m(v - u) / t
F ∝ m × a
F = k × m × a (where k = 1 in SI units)
→ F = m × a
F ∝ m(v - u) / t
F ∝ m × a
F = k × m × a (where k = 1 in SI units)
→ F = m × a
Momentum (p)
Momentum is the quantity of motion contained in a body. It is a vector quantity.
p = m × v
SI unit: kg m/s
CGS unit: g cm/s
Numerical Example:
A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 25 s. Find force.
F = m(v-u)/t = 3(3.5-2)/25 = 3×1.5/25 = 0.18 N
A constant force acting on a body of mass 3 kg changes its speed from 2 m/s to 3.5 m/s in 25 s. Find force.
F = m(v-u)/t = 3(3.5-2)/25 = 3×1.5/25 = 0.18 N
🔥 AMU PYQ: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Ans: True (this is the 2nd law!)
🔥 AMU PYQ: A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Magnitude of acceleration is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (Fnet = √(64+36) = 10 N, a = 10/5 = 2 m/s²)
Ans: True (this is the 2nd law!)
🔥 AMU PYQ: A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Magnitude of acceleration is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (Fnet = √(64+36) = 10 N, a = 10/5 = 2 m/s²)
5) Newton's Third Law of Motion
Statement: To every action, there is an equal and opposite reaction.
FAB = - FBA
Key Points — Action-Reaction Pairs:
- Action and reaction act on different bodies, so they never cancel each other.
- They are equal in magnitude and opposite in direction.
- Both act simultaneously.
Why Action-Reaction Don't Cancel
Many students ask: "If action and reaction are equal and opposite, why don't they cancel out and produce zero net force?" The answer: They act on different bodies! For example:
- When you push a wall (action on wall), the wall pushes you back (reaction on you).
- The action (you on wall) affects the wall. The reaction (wall on you) affects YOU.
- Since they act on different objects, they can't cancel each other.
- If you push the wall hard enough, the wall doesn't move (it's fixed), but YOU move backward!
Examples of Action-Reaction Pairs:
| Situation | Action | Reaction |
|---|---|---|
| Walking | Foot pushes ground backward | Ground pushes foot forward |
| Gun firing | Bullet pushed forward | Gun recoils backward |
| Rocket launch | Hot gases pushed downward | Gases push rocket upward |
| Swimming | Hands push water backward | Water pushes swimmer forward |
| Book on table | Book exerts weight on table | Table exerts normal force on book |
🔥 AMU PYQ: The third law of motion states that for every action, there is an equal and opposite reaction.
Ans: True
🔥 AMU PYQ: Action and reaction forces act on:
(A) Same body (B) Different bodies (C) Both (D) None
Ans: B
Ans: True
🔥 AMU PYQ: Action and reaction forces act on:
(A) Same body (B) Different bodies (C) Both (D) None
Ans: B
6) Conservation of Linear Momentum (⭐ Most Important)
This is the most important topic in this chapter for AMU. Expect at least 1-2 questions every year.
Statement
In the absence of an external unbalanced force, the total momentum of an isolated system remains conserved (constant).
Total initial momentum = Total final momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Derivation from Newton's Laws
From Newton's 2nd: F = dp/dt
If no external force (F = 0): dp/dt = 0 → p = constant
For two interacting bodies A and B:
FAB = -FBA (by Newton's 3rd law)
dpA/dt = -dpB/dt
d(pA + pB)/dt = 0
→ Total momentum of the system is conserved!
If no external force (F = 0): dp/dt = 0 → p = constant
For two interacting bodies A and B:
FAB = -FBA (by Newton's 3rd law)
dpA/dt = -dpB/dt
d(pA + pB)/dt = 0
→ Total momentum of the system is conserved!
Real-Life Applications (All Asked in AMU):
| Application | How It Works |
|---|---|
| Recoil of a gun | Before: p=0. After: mbvb + mgvg = 0 → vg = -(mbvb)/mg |
| Rocket propulsion | Exhaust gases get downward momentum → rocket gets upward momentum |
| Explosion of a bomb | Initially at rest (p=0). After explosion, fragments fly in different directions — vector sum = 0 |
| Collision of balls | Total momentum before = total after (if no external force) |
| Man jumping from boat | Man jumps forward → boat moves backward (conservation of momentum) |
| Two ice skaters pushing off | Both move in opposite directions with momenta equal in magnitude |
🔥🔥🔥 AMU PYQ — Most asked question:
A shell of mass 0.020 kg is fired from a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?
Solution:
Before firing: p = 0
After: (0.020 × 80) + (100 × vg) = 0
1.6 + 100 vg = 0
vg = -1.6/100 = -0.016 m/s (negative = backward)
Recoil speed = 0.016 m/s
A shell of mass 0.020 kg is fired from a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, what is the recoil speed of the gun?
Solution:
Before firing: p = 0
After: (0.020 × 80) + (100 × vg) = 0
1.6 + 100 vg = 0
vg = -1.6/100 = -0.016 m/s (negative = backward)
Recoil speed = 0.016 m/s
🟡 Senior's Pro Tip for Conservation of Momentum:
• Always draw a diagram (before / after)
• Write pinitial = pfinal
• For recoil: m₁v₁ + m₂v₂ = 0 → v₂ = -m₁v₁/m₂
• For collisions: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
• If bodies stick together after collision: v₁ = v₂ = v (common velocity)
• If a body is initially at rest and explodes: total initial p = 0
• Always draw a diagram (before / after)
• Write pinitial = pfinal
• For recoil: m₁v₁ + m₂v₂ = 0 → v₂ = -m₁v₁/m₂
• For collisions: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
• If bodies stick together after collision: v₁ = v₂ = v (common velocity)
• If a body is initially at rest and explodes: total initial p = 0
7) Impulse and Impulse-Momentum Theorem
When a large force acts for a very short time, it's called an impulsive force. The product of force and the time for which it acts is called impulse.
Impulse (J) = F × t = Change in momentum = mv - mu
SI unit: N s (same as kg m/s)
Impulse-Momentum Theorem: J = Δp
Examples of Impulse:
- Cricket fielder moves hands backward while catching a ball — this increases time (t), so force (F) decreases for same impulse (F×t = constant).
- Hammer striking a nail — large force applied for very short time to drive the nail.
- A karate expert breaks a pile of bricks by applying a large impulsive force.
- A glass falls on a carpet vs hard floor — carpet increases stopping time, reducing force, so glass doesn't break.
Numerical Example:
A 0.15 kg cricket ball moving at 30 m/s is hit by a bat and returns at 40 m/s. If contact time is 0.01 s, find the impulse and average force.
Impulse = Δp = mv - mu = 0.15×40 - 0.15×(-30) = 6 + 4.5 = 10.5 N s
F = Impulse / t = 10.5 / 0.01 = 1050 N
A 0.15 kg cricket ball moving at 30 m/s is hit by a bat and returns at 40 m/s. If contact time is 0.01 s, find the impulse and average force.
Impulse = Δp = mv - mu = 0.15×40 - 0.15×(-30) = 6 + 4.5 = 10.5 N s
F = Impulse / t = 10.5 / 0.01 = 1050 N
🔥 AMU PYQ: When a batsman hits a cricket ball, the impulse is equal to the ______.
Ans: Change in momentum
Ans: Change in momentum
8) Friction — Types and Properties
Friction is the force that opposes relative motion between two surfaces in contact. It is a contact force.
Frictional force (f) = μ × N
μ = coefficient of friction (depends on materials)
N = normal reaction (force perpendicular to surface)
Types of Friction:
| Type | Definition | Key point |
|---|---|---|
| Static friction (fs) | Opposes the start of motion. Acts when body is at rest relative to surface. | Maximum value is called limiting friction: fs(max) = μsN |
| Kinetic / Sliding friction (fk) | Opposes motion when surfaces are already sliding. | Kinetic friction < Static friction. fk = μkN |
| Rolling friction (fr) | Opposes rolling motion (ball, wheel, roller). | Rolling friction << Sliding friction (about 100× less) |
🟡 Order of Friction (smallest to largest):
Rolling friction < Kinetic friction < Static friction
That's why wheels are used instead of sliding — rolling friction is the least!
Rolling friction < Kinetic friction < Static friction
That's why wheels are used instead of sliding — rolling friction is the least!
Factors Affecting Friction:
- Nature of surfaces: Rough surfaces → more friction. Smooth surfaces → less friction.
- Weight / Normal force: More weight → more friction (f ∝ N).
- Surface area: Friction does NOT depend on area of contact (surprising but true!).
Advantages of Friction:
- Walking is possible because of friction between shoes and ground.
- Writing with a pen/pencil requires friction.
- Brakes work because of friction between brake pads and wheels.
- Vehicles move because tires grip the road through friction.
- Lighting a matchstick uses friction.
Disadvantages of Friction:
- Causes wear and tear of machinery parts.
- Wastes energy (heat is produced).
- Reduces efficiency of machines.
Methods to Reduce Friction:
- Using lubricants (oil, grease) between surfaces.
- Using wheels / ball bearings.
- Polishing surfaces to make them smooth.
- Streamlining (reducing air resistance for fast-moving objects).
Methods to Increase Friction:
- Treads on tires (increase roughness).
- Using sand/gravel on icy roads.
- Gymnasts use chalk powder on their hands for better grip.
- Spikes on sports shoes.
🔥 AMU PYQ: Which type of friction is the smallest?
(A) Static (B) Sliding (C) Rolling (D) Fluid
Ans: C (Rolling friction is the smallest)
🔥 AMU PYQ: Which factor does NOT affect friction?
(A) Nature of surface (B) Normal force (C) Area of contact (D) Roughness
Ans: C (Friction is independent of area of contact)
(A) Static (B) Sliding (C) Rolling (D) Fluid
Ans: C (Rolling friction is the smallest)
🔥 AMU PYQ: Which factor does NOT affect friction?
(A) Nature of surface (B) Normal force (C) Area of contact (D) Roughness
Ans: C (Friction is independent of area of contact)
9) Free Body Diagrams (NEW)
A free body diagram (FBD) is a diagram that shows all the forces acting on a body. It helps analyze the net force and motion.
How to Draw an FBD:
- Draw the body as a simple dot or box.
- Identify all forces acting on the body (not forces exerted by the body on others).
- Draw arrows from the centre of the body representing each force.
- Label each force clearly.
- Choose a coordinate system (usually x-y axes).
- Resolve forces into components if needed.
Common Forces in FBD:
| Force | Direction | Notation |
|---|---|---|
| Weight (gravity) | Downward (towards Earth's centre) | W = mg |
| Normal reaction | Perpendicular to surface, upward | N |
| Tension | Along the string, away from body | T |
| Friction | Opposite to direction of motion | f or Ff |
| Applied force | In the direction of push/pull | F |
Example — Book on a table:
Forces on book:
• Weight (mg) → downward
• Normal (N) → upward
Since book is at rest: N = mg (balanced forces)
Forces on book:
• Weight (mg) → downward
• Normal (N) → upward
Since book is at rest: N = mg (balanced forces)
Example — Box pulled on rough surface:
Forces on box:
• Applied force (F) → horizontal right
• Weight (mg) → downward
• Normal (N) → upward
• Friction (f) → horizontal left (opposes motion)
Net horizontal: Fnet = F - f
Net vertical: N - mg = 0 → N = mg
Forces on box:
• Applied force (F) → horizontal right
• Weight (mg) → downward
• Normal (N) → upward
• Friction (f) → horizontal left (opposes motion)
Net horizontal: Fnet = F - f
Net vertical: N - mg = 0 → N = mg
10) Motion of Connected Bodies (Pulley Problems — Simple)
When two bodies are connected by a string over a pulley, they move together with the same acceleration (assuming string is inextensible and massless).
Atwood Machine (simple vertical pulley):
Two masses m₁ and m₂ (m₁ > m₂) connected over a frictionless pulley.
Two masses m₁ and m₂ (m₁ > m₂) connected over a frictionless pulley.
Net force = m₁g - m₂g = (m₁ + m₂)a
Acceleration: a = (m₁ - m₂)g / (m₁ + m₂)
Tension: T = (2m₁m₂ g) / (m₁ + m₂)
Numerical Example:
Two masses 5 kg and 3 kg are connected by a string over a pulley. Find acceleration and tension.
a = (5-3)g/(5+3) = 2×9.8/8 = 2.45 m/s²
T = 2×5×3×9.8/(5+3) = 294/8 = 36.75 N
Two masses 5 kg and 3 kg are connected by a string over a pulley. Find acceleration and tension.
a = (5-3)g/(5+3) = 2×9.8/8 = 2.45 m/s²
T = 2×5×3×9.8/(5+3) = 294/8 = 36.75 N
🟡 Senior's Trick for Pulley Problems:
• Assume the heavier mass moves down
• The acceleration is the same for both masses
• The tension is the same throughout the string (ideal string)
• Write F = ma for each mass separately, then solve
• Assume the heavier mass moves down
• The acceleration is the same for both masses
• The tension is the same throughout the string (ideal string)
• Write F = ma for each mass separately, then solve
11) Types of Collisions and Momentum Conservation
Collisions are classified into elastic and inelastic based on whether kinetic energy is conserved.
| Feature | Elastic collision | Inelastic collision |
|---|---|---|
| Kinetic energy conserved? | Yes (KE before = KE after) | No (some KE converted to heat/sound) |
| Momentum conserved? | Yes | Yes |
| Bodies stick together? | No (they bounce off) | Yes (perfectly inelastic) |
| Examples | Billiard balls, ideal gas molecules | Two cars colliding, a bullet embedding in wood |
Perfectly Inelastic Collision (bodies stick together):
m₁u₁ + m₂u₂ = (m₁ + m₂)v
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Common velocity: v = (m₁u₁ + m₂u₂) / (m₁ + m₂)
Numerical Example:
A 2 kg block moving at 4 m/s collides with a stationary 3 kg block. After collision they stick together. Find common velocity.
v = (2×4 + 3×0) / (2+3) = 8/5 = 1.6 m/s
A 2 kg block moving at 4 m/s collides with a stationary 3 kg block. After collision they stick together. Find common velocity.
v = (2×4 + 3×0) / (2+3) = 8/5 = 1.6 m/s
🟡 Senior's Tip — Collision Shortcuts:
• For perfectly inelastic collision (one body at rest): v = m₁u₁/(m₁+m₂)
• For elastic collision of equal masses: velocities get exchanged! (v₁ = u₂, v₂ = u₁)
• For elastic collision, relative speed of approach = relative speed of separation
• For perfectly inelastic collision (one body at rest): v = m₁u₁/(m₁+m₂)
• For elastic collision of equal masses: velocities get exchanged! (v₁ = u₂, v₂ = u₁)
• For elastic collision, relative speed of approach = relative speed of separation
🔥 AMU PYQ: A bullet of mass 50 g moving at 100 m/s strikes a 2 kg wooden block at rest and gets embedded. Find the velocity of the block-bullet system.
Ans: v = (0.05×100)/(0.05+2) = 5/2.05 = 2.44 m/s
Ans: v = (0.05×100)/(0.05+2) = 5/2.05 = 2.44 m/s
12) Applications of Newton's Laws in Daily Life
1st Law Applications (Inertia):
- Seat belts in cars — when brakes are applied suddenly, your body tends to continue moving forward (inertia of motion). Seat belts provide the external force to stop you.
- Dusting a carpet — beat the carpet with a stick. The dust particles fall off due to inertia of rest (they were at rest and tend to remain at rest).
- Shaking a tree — branches move but fruits fall off (inertia of rest of fruits).
- Headrest in cars — prevents whiplash when car is hit from behind (inertia of rest of your head).
2nd Law Applications (F = ma):
- Design of vehicles — more powerful engines produce more force, hence greater acceleration.
- Bowling in cricket — a fast bowler applies more force → ball accelerates more → higher speed.
- Weightlifting — heavier weights require more force to lift (F ∝ m).
3rd Law Applications (Action-Reaction):
- Walking — you push the ground backward (action), ground pushes you forward (reaction).
- Rowing a boat — oars push water backward, water pushes the boat forward.
- Helicopter — rotor blades push air downward, air pushes the helicopter upward.
- Balloon — air rushes out backward, balloon moves forward.
- Recoil of a gun — bullet forward, gun backward.
🔥 AMU PYQ: Which law explains the recoil of a gun?
(A) 1st law (B) 2nd law (C) 3rd law (D) Law of gravitation
Ans: C (3rd law — action-reaction)
(A) 1st law (B) 2nd law (C) 3rd law (D) Law of gravitation
Ans: C (3rd law — action-reaction)
13) Common Mistakes Students Make in Force & Laws
⚠️ Mistake 1: Thinking action-reaction forces cancel each other.
Fix: They act on different bodies, so they DON'T cancel!
⚠️ Mistake 2: Forgetting to include all forces in FBD.
Fix: Always list weight, normal, tension, friction, and applied forces.
⚠️ Mistake 3: Using F = ma without checking units.
Fix: Ensure mass in kg, acceleration in m/s² → force in N.
⚠️ Mistake 4: Confusing mass and weight.
Fix: Mass (kg) is constant. Weight (N) = mg, varies with g.
⚠️ Mistake 5: Forgetting that impulse = change in momentum (not just momentum).
Fix: J = mv - mu, not just mv.
⚠️ Mistake 6: Thinking friction always opposes motion (it opposes relative motion).
Fix: When you walk, friction pushes you forward — it's not always opposing!
Fix: They act on different bodies, so they DON'T cancel!
⚠️ Mistake 2: Forgetting to include all forces in FBD.
Fix: Always list weight, normal, tension, friction, and applied forces.
⚠️ Mistake 3: Using F = ma without checking units.
Fix: Ensure mass in kg, acceleration in m/s² → force in N.
⚠️ Mistake 4: Confusing mass and weight.
Fix: Mass (kg) is constant. Weight (N) = mg, varies with g.
⚠️ Mistake 5: Forgetting that impulse = change in momentum (not just momentum).
Fix: J = mv - mu, not just mv.
⚠️ Mistake 6: Thinking friction always opposes motion (it opposes relative motion).
Fix: When you walk, friction pushes you forward — it's not always opposing!
12) Important Formulae — At a Glance
| Quantity | Formula | SI unit |
|---|---|---|
| Force (Newton's 2nd law) | F = ma | N (kg m/s²) |
| Momentum | p = mv | kg m/s |
| Impulse | J = FΔt = Δp = mv - mu | N s |
| Weight | W = mg | N |
| Friction (kinetic) | f = μN | N |
| Conservation of momentum | m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ | — |
| Recoil velocity | V = -(m v) / M | m/s |
| Pulley acceleration | a = (m₁ - m₂)g / (m₁ + m₂) | m/s² |
| Pulley tension | T = 2m₁m₂g / (m₁ + m₂) | N |
| Newton's 3rd law | FAB = -FBA | N |
| 1 N in CGS | 1 N = 10⁵ dyne | — |
13) PYQ Battle — 16 AMU Past Questions on Force & Laws
Q1. The SI unit of force is:
(A) Dyne (B) Newton (C) Joule (D) Pascal
Ans: B
(A) Dyne (B) Newton (C) Joule (D) Pascal
Ans: B
Q2. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. The magnitude of acceleration is:
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (Fnet = √(64+36) = 10 N, a = 10/5 = 2 m/s²)
(A) 1 m/s² (B) 2 m/s² (C) 3 m/s² (D) 4 m/s²
Ans: B (Fnet = √(64+36) = 10 N, a = 10/5 = 2 m/s²)
Q3. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially at 15 m/s. How long does it take to stop?
Ans: a = 50/20 = 2.5 m/s², v = u - at → 0 = 15 - 2.5t → t = 6 s
Ans: a = 50/20 = 2.5 m/s², v = u - at → 0 = 15 - 2.5t → t = 6 s
Q4. A gun of mass 100 kg fires a bullet of mass 0.02 kg with velocity 80 m/s. Recoil velocity of the gun is:
(A) 0.016 m/s (B) 0.16 m/s (C) 1.6 m/s (D) 16 m/s
Ans: A
(A) 0.016 m/s (B) 0.16 m/s (C) 1.6 m/s (D) 16 m/s
Ans: A
Q5. Which has the highest inertia?
(A) Bicycle (B) Car (C) Truck (D) Motorcycle
Ans: C (inertia ∝ mass)
(A) Bicycle (B) Car (C) Truck (D) Motorcycle
Ans: C (inertia ∝ mass)
Q6. When a bus suddenly starts, passengers fall backward due to:
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Friction
Ans: A
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Friction
Ans: A
Q7. When a bus suddenly stops, passengers fall forward due to:
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Gravity
Ans: B
(A) Inertia of rest (B) Inertia of motion (C) Inertia of direction (D) Gravity
Ans: B
Q8. Action and reaction forces act on:
(A) Same body (B) Different bodies (C) Both (D) None
Ans: B
(A) Same body (B) Different bodies (C) Both (D) None
Ans: B
Q9. Momentum of a body of mass 5 kg moving with velocity 2 m/s is:
(A) 5 (B) 10 (C) 2.5 (D) 7 kg m/s
Ans: B (p = 5×2 = 10 kg m/s)
(A) 5 (B) 10 (C) 2.5 (D) 7 kg m/s
Ans: B (p = 5×2 = 10 kg m/s)
Q10. A bomb at rest explodes into three fragments. Two fragments of equal mass fly perpendicular with equal velocity v. The third (mass = 2×) moves with:
(A) v/2 (B) v/√2 (C) v (D) √2 v
Ans: B (momentum conservation)
(A) v/2 (B) v/√2 (C) v (D) √2 v
Ans: B (momentum conservation)
Q11. In the absence of external force, the total momentum of a system is:
(A) Zero (B) Constant (C) Increasing (D) Decreasing
Ans: B (Conserved)
(A) Zero (B) Constant (C) Increasing (D) Decreasing
Ans: B (Conserved)
Q12. 1 N = ______ dyne.
(A) 10³ (B) 10⁴ (C) 10⁵ (D) 10⁶
Ans: C
(A) 10³ (B) 10⁴ (C) 10⁵ (D) 10⁶
Ans: C
Q13. Rocket propulsion is based on:
(A) Newton's 1st law (B) Newton's 2nd law (C) Conservation of momentum (D) Both B and C
Ans: D
(A) Newton's 1st law (B) Newton's 2nd law (C) Conservation of momentum (D) Both B and C
Ans: D
Q14. A 50 g bullet hits a 2 kg block of wood and becomes embedded. If bullet's initial speed was 100 m/s, find common speed.
Ans: m₁u₁ = (m₁+m₂)v → 0.05×100 = 2.05v → v = 5/2.05 = 2.44 m/s
Ans: m₁u₁ = (m₁+m₂)v → 0.05×100 = 2.05v → v = 5/2.05 = 2.44 m/s
Q15. Which is the weakest among the following?
(A) Static friction (B) Sliding friction (C) Rolling friction (D) All are equal
Ans: C
(A) Static friction (B) Sliding friction (C) Rolling friction (D) All are equal
Ans: C
Q16. Two blocks of masses 4 kg and 2 kg are connected by a string over a pulley. Find acceleration.
Ans: a = (4-2)g/(4+2) = 2×9.8/6 = 3.27 m/s²
Ans: a = (4-2)g/(4+2) = 2×9.8/6 = 3.27 m/s²
14) Practice Exercise — 15 Problems with Answers
Try these on your own before looking at the solutions.
1. Find force if mass = 10 kg and acceleration = 4 m/s².
Answer
F = 10×4 = 40 N
2. A 6 kg object accelerates at 3 m/s². Find net force.
Answer
F = 6×3 = 18 N
3. Momentum of a 500 g ball moving at 10 m/s.
Answer
p = 0.5×10 = 5 kg m/s
4. A 150 g bullet is fired from a 5 kg gun at 300 m/s. Find recoil speed of gun.
Answer
0.15×300 + 5×v = 0 → v = -45/5 = -9 m/s (recoil speed = 9 m/s)
5. A 0.2 kg ball moving at 15 m/s is caught by a fielder. If stopping time is 0.05 s, find average force.
Answer
Impulse = 0.2×15 = 3 N s. F = 3/0.05 = 60 N
6. A car of mass 1000 kg moving at 20 m/s is brought to rest in 5 s. Find retarding force.
Answer
a = (0-20)/5 = -4 m/s². F = 1000×(-4) = -4000 N (retarding force = 4000 N)
7. Two bodies of masses 8 kg and 2 kg have equal momentum. Find ratio of their KE.
Answer
p₁=p₂ → m₁v₁=m₂v₂ → v₁/v₂ = 2/8 = 1/4. KE₁/KE₂ = (½m₁v₁²)/(½m₂v₂²) = (8×1)/(2×16) = 8/32 = 1:4
8. A block of mass 5 kg is pulled on a rough surface with force 30 N. If μ = 0.2, find acceleration. (g = 10 m/s²)
Answer
N = mg = 50 N. f = 0.2×50 = 10 N. F_net = 30 - 10 = 20 N. a = 20/5 = 4 m/s²
9. Find tension when a 10 kg mass hangs from a vertical string at rest.
Answer
T = mg = 10×9.8 = 98 N (since at rest, net force = 0)
10. A 60 kg man jumps from a 2 kg boat at 3 m/s. Find boat's recoil speed.
Answer
60×3 + 2×v = 0 → v = -180/2 = -90 m/s (Boat moves backward at 90 m/s)
11. A force of 20 N acts on a body of mass 4 kg for 3 s. Find change in momentum.
Answer
Δp = F×t = 20×3 = 60 kg m/s
12. Two balls of masses 2 kg (v=3 m/s) and 3 kg (v=2 m/s) move towards each other. After collision they stick together. Find common velocity.
Answer
2×3 + 3×(-2) = (2+3)v → 6-6 = 5v → v = 0 m/s (they come to rest)
13. A body of mass 10 kg at rest explodes into two pieces of masses 4 kg and 6 kg. If the 4 kg piece moves at 15 m/s, find velocity of 6 kg piece.
Answer
4×15 + 6×v = 0 → v = -60/6 = -10 m/s (opposite direction)
14. Why does a fielder pull his hands backward while catching a fast ball?
Answer
To increase the stopping time, which reduces the average force (impulse = F×t = constant = Δp). Longer t → smaller F.
15. A box weighing 100 N is pushed with 40 N force. If μ = 0.3, will the box move?
Answer
fs(max) = 0.3×100 = 30 N. Pushing force (40 N) > 30 N, so YES, the box will move.15) Quick Revision — One-liners
- Newton's 1st law → Law of Inertia (body resists change in state)
- Inertia ∝ Mass (more mass = more inertia)
- Newton's 2nd law → F = ma (force causes acceleration)
- Newton's 3rd law → Action = -Reaction (on different bodies)
- Momentum p = mv, SI unit: kg m/s (vector quantity)
- Impulse = F × t = Δp (change in momentum)
- Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (when Fext = 0)
- Recoil velocity: V = -(m v) / M
- 1 N = 10⁵ dyne
- Friction f = μN; Rolling < Kinetic < Static friction
- Friction does NOT depend on area of contact
- Balanced forces → no acceleration; Unbalanced → acceleration
- Pulley acceleration: a = (m₁-m₂)g/(m₁+m₂); T = 2m₁m₂g/(m₁+m₂)
- Free body diagram shows all forces acting on a body
- Action-reaction pairs act on different bodies — they never cancel
✅ Force & Newton's Laws done. Master the conservation of momentum numericals — that's where AMU awards marks. Next: Gravitation!