Resolution of a vector means expressing a single vector as the sum of two or more component vectors along chosen directions. The most common choice is a pair of mutually perpendicular axes — the rectangular components (NCERT Class 11 Physics Chapter 3, page 4).
Given a vector A making angle θ with the positive x-axis:
The original vector is recovered by A = Aₓ x̂ + Aᵧ ŷ, with magnitude A = √(Aₓ² + Aᵧ²) and direction tan θ = Aᵧ / Aₓ.
Where aspirants lose marks:
Angle-reference confusion. If the angle is given from the y-axis (or "with the vertical"), the components swap: the x-component becomes A sin θ and the y-component becomes A cos θ. Always identify the reference axis before writing cos or sin.
Sign errors in quadrants. When the vector lies in the second quadrant (90° < θ < 180°), cos θ is negative — so Aₓ is negative. Students who plug in the acute angle without thinking about sign get the wrong component.
Confusing resolution with addition. Resolution decomposes ONE vector into components; vector addition combines TWO or more vectors into a resultant. The formulas look similar but serve opposite purposes.
Non-perpendicular resolution. Components along non-orthogonal axes exist and require the sine rule or parallelogram law, but NEET questions overwhelmingly use rectangular axes. If the problem doesn't specify oblique axes, default to rectangular.
Resolution is the foundation for projectile motion (separate x and y kinematics) and inclined-plane dynamics (parallel and perpendicular to slope). Master the angle-reference check here, and those topics become mechanical.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
A force of 10 N acts at 60° above the positive x-axis. What is the x-component of this force?
A velocity vector of magnitude 20 m/s makes an angle of 30° with the vertical. What is its horizontal component?
Which of the following is the correct statement about the rectangular components of a vector?
A vector **A** has components Aₓ = −3 units and Aᵧ = 4 units. The magnitude of **A** is:
The x and y components of a vector are equal in magnitude. The angle this vector makes with the positive x-axis is:
A vector of magnitude 50 units lies in the x-y plane. Its y-component is 25 units. The angle the vector makes with the x-axis is:
A force **F** is resolved into two components along directions that are NOT perpendicular to each other. Which of the following is true?
A displacement vector has magnitude 100 m. When resolved along two perpendicular axes, the component along one axis is 60 m. The component along the other axis is:
Given
- Magnitude of displacement: d = 200 m (exact, problem-defined) - Angle with the vertical: θ_v = 60° (exact angle)
Required
Horizontal component (dₓ) and vertical component (dᵧ).
Concept
Resolution of a vector into rectangular components. The angle is stated from the vertical (y-axis), so the component along the vertical is d cos θ_v and the component along the horizontal is d sin θ_v.
Formula
- dₓ = d sin θ_v (horizontal — perpendicular to the reference axis) - dᵧ = d cos θ_v (vertical — along the reference axis)
Substitution
- dₓ = 200 × sin 60° = 200 × (√3/2) - dᵧ = 200 × cos 60° = 200 × (1/2)
Calculation
- dₓ = 200 × 0.866 = 100√3 ≈ 173.2 m - dᵧ = 200 × 0.5 = 100 m Note: 200 m, 60°, and the trigonometric values (sin 60° = √3/2, cos 60° = 1/2) are all exact in this problem. They do not limit significant figures.
Final answer
Horizontal component = 100√3 m ≈ 173.2 m; Vertical component = 100 m.
Common trap
If you read "60° with the vertical" but reflexively write dₓ = d cos 60° and dᵧ = d sin 60°, you swap the two answers: you'd get dₓ = 100 m and dᵧ = 173.2 m — exactly reversed. Always identify the reference axis first, then assign cos to the component along that axis and sin to the component perpendicular to it.
Similar NEET-style question
A force of 500 N acts at 30° with the vertical on a block resting on a horizontal surface. What is the horizontal component of the force pushing the block along the surface? *(Answer: F sin 30° = 500 × 0.5 = 250 N — the horizontal component uses sin because the angle is from the vertical.)* ---
A vector A in two dimensions can be expressed as A = Ax î + Ay ĵ where Ax = A cos θ and Ay = A sin θ are the rectangular components along the x and y axes; î and ĵ are unit vectors along x and y respectively. The magnitude is A = √(Ax² + Ay²); the direction tan θ = Ay/Ax.
-- NCERT Class 11 Physics, Ch. 3, p. 4Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant (uniform) acceleration | m/s^2 |
| t | Elapsed time | s |
Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.
| Symbol | Quantity | SI Unit |
|---|---|---|
| x | Final position | m |
| x0 | Initial position | m |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| t | Time elapsed | s |
Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | Final velocity | m/s |
| v0 | Initial velocity | m/s |
| a | Constant acceleration | m/s^2 |
| x - x0 | Displacement | m |
Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.
| Symbol | Quantity | SI Unit |
|---|---|---|
| H | Maximum height (above launch) | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle | rad/deg |
| g | Gravitational acceleration | m/s^2 |
For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R | Horizontal range | m |
| v0 | Launch speed | m/s |
| theta0 | Launch angle above horizontal | rad (or deg with sin in deg) |
| g | Gravitational acceleration | m/s^2 |
An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student answers 1:2:3:4 for distances in successive 1-second intervals (linear) instead of 1:3:5:7 (Galileo's odd numbers).
Question asks about ratios of distances traversed in successive 1-s intervals during free fall from rest.
Distance grows quadratically (y = ½ g t²); successive interval distances are y_n - y_{n-1} = ½ g (t_n² - t_{n-1}²) = ½ g (2n-1) seconds. The factor (2n-1) gives 1, 3, 5, 7, ...
Category: Graph Interpretation
Student uses sin or cos of the angle the line makes with the time axis, instead of tan, to extract velocity.
Question gives an angle the x-t line makes with the t-axis (often 30°, 45°, 60°) and asks for velocity or its ratio.
Velocity = dx/dt = slope of x-t line = tan(angle), where the angle is measured from the time axis. Always tan, not sin or cos.
Category: Overthinking
Student attempts to invert t(x) algebraically before differentiating, getting tangled in messy algebra; misses chain rule.
Question gives t as function of x (instead of x as function of t), e.g. t = x² + x.
Differentiate the given relation directly: dt/dx = (function of x). Then v = dx/dt = 1/(dt/dx). For acceleration use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.
Category: Sign Convention
Student treats a 'thrown vertically downward' problem as if the object were dropped (u = 0). The result is wrong by an additive u² term in v² = u² + 2gh. When the question explicitly states a launch speed, that speed is non-zero and CANNOT be ignored.
Question phrases: 'thrown vertically downward', 'projected with initial velocity', 'launched with speed u'. If u is given numerically, it MUST appear in the equation.
Always parse the launch verbal cue and write down u with its sign before reaching for v² = 2gh. Use the full v² = u² + 2gh (or u² - 2gh for upward motion).
Category: Similar Terms
Student plugs into v₀² sin(2θ)/g (range) when asked for maximum height, or vice versa. The two share v₀ and θ but have different sin-vs-sin² and 2g-vs-g terms.
Question mentions launch speed and angle and asks for max height (H) or range (R). Distractors include the wrong formula's answer.
Memorise BOTH formulas explicitly: H = v₀² sin² θ / (2g) (note sin²); R = v₀² sin(2θ) / g (note sin of doubled angle). Check by setting θ = 45°: max range, half max height.
Category: Sign Convention
Student plugs angle θ into v cos θ when the question states 'angle with the vertical' (which makes the horizontal component v sin θ).
Question phrases like 'thrown at angle θ with the vertical direction' or 'with horizontal'.
Always identify reference axis explicitly. From horizontal: vx = v cos θ, vy = v sin θ. From vertical: vx = v sin θ, vy = v cos θ. The two are complementary (θ_h + θ_v = 90°).
Category: Sign Convention
Student fails to distinguish between same-direction and opposite-direction relative velocities, treating both as magnitudes.
Question describes two objects moving on the same line; observer somewhere between or alongside.
Relative velocity is a VECTOR. Same direction: v_rel = v_a - v_b (smaller magnitude). Opposite direction: v_rel = v_a + v_b (larger magnitude). Use sign convention consistently along chosen axis.
Category: Overthinking
Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.
Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.
Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.
Category: Overthinking
Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.
Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.
Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.
Category: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
Question asks 'in uniform circular motion at constant speed, which is also constant?'
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Root cause: concept gap
Average velocity = total displacement / total time (vector). Average speed = total path length / total time (scalar, always >= |average velocity|). For round-trip motion, average velocity is zero; average speed is not.
Distractor offers (v1+v2)/2 instead of total-distance/total-time.
Root cause: formula misuse
The three kinematic equations require CONSTANT acceleration. For variable acceleration, use a = dv/dt and integrate, or use v dv = a dx for position-dependent acceleration. Verify constant-a before applying these formulas.
Distractor uses constant-acceleration kinematic equations on a problem where the question explicitly says acceleration changes with time or position.
Root cause: concept gap
Standard range R = v0^2 sin(2*theta)/g and H = v0^2 sin^2(theta)/(2g) assume (i) launch and landing at the same height, (ii) negligible air drag, and (iii) constant g. For asymmetric trajectories, use the full kinematic decomposition along x and y.
Distractor applies R = v0^2 sin(2*theta)/g to a projectile launched from a cliff.
Root cause: concept gap
Acceleration is the rate of change of VELOCITY (a vector), not speed. In uniform circular motion, the speed is constant but the velocity direction changes continuously, giving a centripetal acceleration of magnitude v^2/r toward the centre.
Distractor option says 'a = 0 because speed is constant'.
A particle moving with uniform speed in a circular path maintains:
The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second
uses arithmetic progression 1 2 3 4
Linear-time intuition
treats t x as explicit
Default to t-as-input thinking
treats speed ratio as distance ratio
Linear thinking: 'speed went from u to u/3, so distance traversed should be 3× the original'
uses sin or cos instead of tan
Trig confusion under time pressure
uses radius as height
Surface confusion of geometric R and projectile H
uses sin instead of sin squared
Confusing the height formula H = v²sin²θ/(2g) with the range 2v sin θ / g
uses sin instead of cos
Sign-of-angle-axis confusion
speed zero at top
True only for vertical motion (no horizontal v₀)
ignores direction relative to observer
Treating bus-passing time as absolute
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
drops initial velocity term
Student conflates 'thrown' with 'dropped' and uses v² = 2gh
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.