🧩 MODULE 4 — PROBLEM TYPES
6 Problem Types — All Fully Solved
Every question pattern in CBSE, NEET, and JEE. Examiner's intent. Concept selection. Solution. Shortcut insight.
PROBLEM TYPE 1
Direct Formula Application
CBSE boards + basic NEET. Know the formula, plug values, get marks. But you must know which formula to use.
1
Newton's 2nd Law — Direct
A body of mass 10 kg is acted upon by a force of 40 N in the positive x-direction and 30 N in the negative x-direction. Find the acceleration of the body.
Examiner tests: Whether student finds NET force correctly (not just uses one force)
Concept: F_net = ΣF (vector sum), then F_net = ma
Concept: F_net = ΣF (vector sum), then F_net = ma
COMPLETE SOLUTION
Given: m = 10 kg, F₁ = +40 N, F₂ = -30 N
Net Force: F_net = F₁ + F₂ = 40 - 30 = 10 N
Using F = ma: a = F_net / m = 10 / 10 = 1 m/s² (in +x direction)
Net Force: F_net = F₁ + F₂ = 40 - 30 = 10 N
Using F = ma: a = F_net / m = 10 / 10 = 1 m/s² (in +x direction)
Shortcut: Always add forces vectorially first. Answer = 1 m/s², NOT 7 m/s² (common error: using 40+30 = 70 N as net force by ignoring directions).
2
Atwood Machine
In an Atwood machine, masses of 4 kg and 6 kg are connected by a light string over a frictionless pulley. Find (a) acceleration of the system, (b) tension in the string. (g = 10 m/s²)
Examiner tests: FBD of each mass separately. Direction assignment for acceleration.
Concept: Atwood machine equations from FBD
Concept: Atwood machine equations from FBD
COMPLETE SOLUTION
Given: m₁ = 4 kg, m₂ = 6 kg, g = 10 m/s²
FBD of 6 kg: 6×10 - T = 6a → 60 - T = 6a ...(1)
FBD of 4 kg: T - 4×10 = 4a → T - 40 = 4a ...(2)
Adding (1) & (2): 20 = 10a → a = 2 m/s²
From (2): T = 40 + 4×2 = T = 48 N
Quick formula: a = (m₂-m₁)g/(m₁+m₂) = (6-4)×10/10 = 2 m/s² ✓
T = 2m₁m₂g/(m₁+m₂) = 2×4×6×10/10 = 48 N ✓
FBD of 6 kg: 6×10 - T = 6a → 60 - T = 6a ...(1)
FBD of 4 kg: T - 4×10 = 4a → T - 40 = 4a ...(2)
Adding (1) & (2): 20 = 10a → a = 2 m/s²
From (2): T = 40 + 4×2 = T = 48 N
Quick formula: a = (m₂-m₁)g/(m₁+m₂) = (6-4)×10/10 = 2 m/s² ✓
T = 2m₁m₂g/(m₁+m₂) = 2×4×6×10/10 = 48 N ✓
Insight: Always derive from FBD first in exams — don't just memorize formula. But memorize the formula too — it saves 30 seconds per question in NEET.
3
Friction Force Calculation
A body of mass 5 kg is placed on a rough horizontal surface (μ = 0.4). What minimum horizontal force is needed to just move the body? (g = 10 m/s²)
SOLUTION
To just move body, applied force = maximum static friction
fs(max) = μs × N = μs × mg = 0.4 × 5 × 10 = 20 N
fs(max) = μs × N = μs × mg = 0.4 × 5 × 10 = 20 N
Common error: Using mg as friction and μ as a multiplier without knowing what μ represents. μ is a dimensionless number. Friction = μ × Normal force (not μ × weight directly, unless on a horizontal surface where N = mg).
PROBLEM TYPE 2
Conceptual Questions
These test understanding, not calculation. Most NEET traps and JEE reasoning questions fall here. Memorization fails here — thinking wins.
1
Action-Reaction Identification
A book is lying on a table. Identify all action-reaction pairs. Which of the following is the correct reaction to the gravitational pull of Earth on the book?
Answer: C — Gravitational pull of book on Earth.
Newton's 3rd Law: Action-reaction pairs are of the SAME TYPE (nature), act on DIFFERENT bodies.
• Earth pulls book (gravitational) → Book pulls Earth (gravitational) — THIS is the pair.
• Normal force of table on book is a contact force — its reaction is the book pushing on table (downward).
This is a classic NEET conceptual trap.
Newton's 3rd Law: Action-reaction pairs are of the SAME TYPE (nature), act on DIFFERENT bodies.
• Earth pulls book (gravitational) → Book pulls Earth (gravitational) — THIS is the pair.
• Normal force of table on book is a contact force — its reaction is the book pushing on table (downward).
This is a classic NEET conceptual trap.
2
Friction Direction
A man stands on a stationary bus. When the driver brakes suddenly, the man lurches forward. The friction that prevents the man from sliding further acts in which direction?
Answer: B — Forward
When bus brakes, bus decelerates (backward). Man's feet tend to slide forward (relative to bus). Friction opposes relative sliding of feet → friction acts BACKWARD on feet from bus's floor.
Wait — the question asks about preventing "further sliding." If man lurches forward, his feet slide backward relative to bus. Kinetic friction on his feet acts FORWARD (opposing his feet's backward sliding).
Key: Always identify the direction of relative motion/tendency of the body (feet), then friction opposes that.
When bus brakes, bus decelerates (backward). Man's feet tend to slide forward (relative to bus). Friction opposes relative sliding of feet → friction acts BACKWARD on feet from bus's floor.
Wait — the question asks about preventing "further sliding." If man lurches forward, his feet slide backward relative to bus. Kinetic friction on his feet acts FORWARD (opposing his feet's backward sliding).
Key: Always identify the direction of relative motion/tendency of the body (feet), then friction opposes that.
Conceptual question strategy: 1) Identify which body you're analyzing. 2) Identify the tendency of relative motion. 3) Apply the principle. Don't trust your "feel" — always reason logically.
PROBLEM TYPE 3
Multi-Step Problems
This is where marks are actually won or lost. Multiple FBDs, multiple equations, multiple concepts combined. Structure your solution or lose marks.
1
Block on Rough Incline + Pulley System
A block of mass 3 kg lies on a rough inclined plane (θ = 30°, μk = 0.3). A string over a frictionless pulley connects it to a hanging mass of 5 kg. Find the acceleration of the system and tension. (g = 10 m/s²)
Examiner tests: FBD of two different bodies in different configurations. Force resolution on incline. Friction direction determination.
Multi-step: Determine motion direction → Set up FBDs → Solve system of equations
Multi-step: Determine motion direction → Set up FBDs → Solve system of equations
STEP-BY-STEP SOLUTION
Step 1: Determine direction of motion
Weight of 5 kg = 50 N (driving force for downward motion of 5 kg)
Force pulling 3 kg up incline = Tension T
Forces on 3 kg along incline: mg sinθ = 3×10×0.5 = 15 N (down the incline)
Tension T pulls it UP. Friction = μkN = 0.3×3×10×cos30° = 0.3×26 ≈ 7.79 N
If 5 kg descends: 3 kg goes up incline, friction acts DOWN the incline on 3 kg.
Step 2: FBD of 5 kg (hanging)
5g - T = 5a → 50 - T = 5a ...(1)
Step 3: FBD of 3 kg (on incline, moving up)
T - 3g sin30° - μk × 3g cos30° = 3a
T - 15 - 7.79 = 3a
T - 22.79 = 3a ...(2)
Step 4: Solve simultaneously
From (1) + (2): 50 - 22.79 = 8a → 27.21 = 8a
a ≈ 3.4 m/s²
T = 50 - 5×3.4 = T ≈ 33 N
Weight of 5 kg = 50 N (driving force for downward motion of 5 kg)
Force pulling 3 kg up incline = Tension T
Forces on 3 kg along incline: mg sinθ = 3×10×0.5 = 15 N (down the incline)
Tension T pulls it UP. Friction = μkN = 0.3×3×10×cos30° = 0.3×26 ≈ 7.79 N
If 5 kg descends: 3 kg goes up incline, friction acts DOWN the incline on 3 kg.
Step 2: FBD of 5 kg (hanging)
5g - T = 5a → 50 - T = 5a ...(1)
Step 3: FBD of 3 kg (on incline, moving up)
T - 3g sin30° - μk × 3g cos30° = 3a
T - 15 - 7.79 = 3a
T - 22.79 = 3a ...(2)
Step 4: Solve simultaneously
From (1) + (2): 50 - 22.79 = 8a → 27.21 = 8a
a ≈ 3.4 m/s²
T = 50 - 5×3.4 = T ≈ 33 N
If this step is wrong, the entire solution fails: Determining the correct direction of friction. Always ask: "Which way will the body tend to move?" Then friction opposes that tendency.
PROBLEM TYPE 4
Graph-Based Questions
Reading and interpreting graphs. JEE tests graph identification, slope analysis, and area interpretation. NEET tests basic graph reading.
Graph 1: Friction Force vs Applied Force
fs(max)
fk
f →
F_applied →
Friction force vs Applied force
- Slope of linear part = 1 (friction equals applied force)
- Peak = μs × N (maximum static friction)
- Horizontal line = μk × N (kinetic friction, constant)
- μk < μs → plateau lower than peak
JEE question: "What does the slope of the rising part represent?" Answer: 1 (or undefined if they mean the angle). The friction exactly equals applied force in static region.
Graph 2: F-t graph and Impulse
F →
t →
Area under F-t graph = Impulse = Δp
- Area = Impulse = Change in momentum
- Rectangular region: J = F × t
- Triangular region: J = ½ × F_max × t
- Net impulse = Total area (signed)
JEE Main asks: "Find change in momentum from F-t graph." Just calculate area geometrically. No complex integration needed if it's a simple shape.
PROBLEM TYPE 5
Assertion-Reason Questions
CBSE and NEET staple. Both assertion and reason must be analyzed independently, then their relationship. Don't just read one and choose.
CODES:
(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is NOT the correct explanation of A
(C) A is true, R is false
(D) A is false, R is true
(B) Both A and R are true, but R is NOT the correct explanation of A
(C) A is true, R is false
(D) A is false, R is true
1
Assertion-Reason: Friction
Assertion (A): Friction force is independent of area of contact between two surfaces.
Reason (R): Friction depends only on the normal reaction force and the nature of surfaces.
Reason (R): Friction depends only on the normal reaction force and the nature of surfaces.
Answer: A
Both assertion and reason are true, and R correctly explains A.
• A is True: Friction is independent of contact area (empirical law).
• R is True: Friction = μN, where only normal force N and μ (nature of surfaces) matter.
• R correctly explains why friction is independent of area — because the formula doesn't include area.
Both assertion and reason are true, and R correctly explains A.
• A is True: Friction is independent of contact area (empirical law).
• R is True: Friction = μN, where only normal force N and μ (nature of surfaces) matter.
• R correctly explains why friction is independent of area — because the formula doesn't include area.
2
Assertion-Reason: Action-Reaction
Assertion (A): Action and reaction forces do not cancel each other.
Reason (R): Action and reaction forces act on the same body simultaneously.
Reason (R): Action and reaction forces act on the same body simultaneously.
Answer: C — A is true, R is false
• A is True: Action and reaction forces do NOT cancel (correct).
• R is False: Action and reaction forces act on DIFFERENT bodies (NOT same body).
The trap: Students who confuse the reason immediately mark A or B. The reason "same body" is what makes them cancel if true. But since they act on different bodies, they can't cancel.
• A is True: Action and reaction forces do NOT cancel (correct).
• R is False: Action and reaction forces act on DIFFERENT bodies (NOT same body).
The trap: Students who confuse the reason immediately mark A or B. The reason "same body" is what makes them cancel if true. But since they act on different bodies, they can't cancel.
Assertion-Reason strategy: Evaluate ASSERTION first, independently. Then evaluate REASON independently. THEN check if R explains A. Don't read them together — you'll confuse yourself.
PROBLEM TYPE 6
Case-Based Questions
New in CBSE 2021+. A scenario is given, multiple questions follow. All questions are linked. Read the case carefully before solving any question.
📋 CASE STUDY
A truck is moving on a horizontal road with acceleration a = 2 m/s². A box of mass 20 kg is placed on the truck. The coefficient of static friction between box and truck floor is μs = 0.4 and kinetic friction μk = 0.3. The truck suddenly accelerates and the box may or may not slip depending on conditions. (g = 10 m/s²)
Q1 from case study:
What is the maximum acceleration of the truck for which the box does NOT slip?
Answer: B — 4 m/s²
For box not to slip: Friction ≤ fs(max)
ma_box ≤ μs × mg (friction accelerates box with truck)
a_max = μs × g = 0.4 × 10 = 4 m/s²
At a = 2 m/s² (from case), box doesn't slip since 2 < 4.
For box not to slip: Friction ≤ fs(max)
ma_box ≤ μs × mg (friction accelerates box with truck)
a_max = μs × g = 0.4 × 10 = 4 m/s²
At a = 2 m/s² (from case), box doesn't slip since 2 < 4.
Q2 from case study:
If truck acceleration increases to 6 m/s², what is the acceleration of the box?
Answer: C — 3 m/s²
When truck a = 6 m/s² > a_max = 4 m/s², box slips!
Now kinetic friction acts on box: fk = μk × mg = 0.3 × 20 × 10 = 60 N
Acceleration of box = fk/m = 60/20 = 3 m/s²
Box accelerates at 3 m/s², truck at 6 m/s² → box slides backward relative to truck.
When truck a = 6 m/s² > a_max = 4 m/s², box slips!
Now kinetic friction acts on box: fk = μk × mg = 0.3 × 20 × 10 = 60 N
Acceleration of box = fk/m = 60/20 = 3 m/s²
Box accelerates at 3 m/s², truck at 6 m/s² → box slides backward relative to truck.
Case-based questions are worth 4-5 marks in CBSE. The trick: The questions are sequential — the answer to Q1 often feeds into Q2. Read ALL questions in the case before starting. Identify what each question is actually asking.