Motion in a Plane for NEET UG
In two dimensions, motion is handled by splitting vectors into perpendicular parts. This one idea explains projectiles, river crossings and circular motion.
PhysicsClass 1167 MCQs3 lessons11 flashcards3 memory tricks
Practise Motion in a Plane free
What you'll learn
- Vectors: adding and resolving
Add two vectors at an angle and split a vector into x and y components. - Projectile motion
Find the time of flight, maximum height and range of a projectile. - Uniform circular motion
Explain why a body moving in a circle at constant speed is accelerating, and calculate that acceleration.
Key formulas
| Resultant of A and B at angle θ | R = √(A² + B² + 2AB cos θ) |
|---|---|
| Direction of R from A | tan α = B sin θ / (A + B cos θ) |
| Components of A at angle θ to x-axis | Aₓ = A cos θ, A_y = A sin θ |
| Time of flight | T = 2u sin θ / g |
| Maximum height | H = u² sin²θ / 2g |
| Horizontal range | R = u² sin 2θ / g |
| Path | y = x tan θ − gx² / (2u² cos²θ), a parabola |
| Angular speed | ω = θ/t = 2π/T = 2πν |
Memory tricks
R = u² sin 2θ / g, so θ and (90° − θ) give the same range. Range is maximum at 45°. The higher angle keeps the ball in the air longer.
Rain's velocity relative to you = v_rain − v_you, so it seems to come at your face. Tilt the umbrella forward by an angle θ from the vertical, where tan θ = v_you / v_rain.
Speed is constant but velocity changes, so there is a centripetal acceleration v²/r = ω²r pointing towards the centre.
Try 3 NEET-style questions
Q1. A swimmer crosses a 100 m wide river by swimming perpendicular to the current at 2 m/s relative to the water. The river flows at 1.5 m/s. How far downstream does the swimmer land?
- 50 m
- 100 m
- 125 m
- 75 m
Show answer
Answer: (4) 75 m
Crossing time = 100 m / 2 m/s = 50 s. Drift = 1.5 m/s × 50 s = 75 m.
Q2. A ball is thrown horizontally at 10 m/s from the top of a 45 m tall cliff. Taking g = 10 m/s², it lands at a horizontal distance of
- 15 m
- 30 m
- 45 m
- 90 m
Show answer
Answer: (2) 30 m
Time to fall: t = √(2h/g) = √9 = 3 s. Horizontal distance = 10 × 3 = 30 m.
Q3. At the highest point of an oblique projectile (launched at angle θ with speed u), its velocity is
- zero
- horizontal, equal to u cos θ
- vertical, equal to u sin θ
- equal to u
Show answer
Answer: (2) horizontal, equal to u cos θ
At the top the vertical component of velocity is zero, and the horizontal component u cos θ remains. So the velocity is horizontal with magnitude u cos θ.
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