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Motion in a Plane – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 3. Notes, key terms, important questions with answers and MCQs.

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Chapter overview

Motion in two dimensions uses vectors. Projectile motion and uniform circular motion are the key examples.

Vectors

Vectors add by the triangle or parallelogram law: R = √(A² + B² + 2AB cos θ).
Resolution: Aₓ = A cos θ, Aᵧ = A sin θ.
Unit vectors î, ĵ, k̂ along axes.

Projectile and circular motion

Projectile at angle θ: time of flight T = 2u sin θ/g, max height H = u² sin²θ/2g, range R = u² sin 2θ/g (max at 45°).
Path of a projectile is a parabola; horizontal velocity stays constant.
Uniform circular motion: centripetal acceleration a = v²/r = ω²r, directed towards the centre; v = ωr.

Key terms

Projectile: object moving under gravity alone after being thrown
Centripetal acceleration: acceleration towards the centre in circular motion
Angular velocity: rate of change of angular displacement, ω = dθ/dt; v = ωr

Formulas

R = u² sin 2θ/g
H = u² sin²θ/2g
a = v²/r

Important questions with answers

Show that the maximum range of a projectile is four times its maximum height when projected at 45°.2 marks

At 45°: R = u²/g and H = u² sin²45°/2g = u²/4g. So R = 4H.

A man walks 4 km east and then 3 km north. Find his displacement in magnitude and direction.3 marks

Magnitude = √(16 + 9) = 5 km. Direction: tan θ = 3/4, θ ≈ 37° north of east.

Derive the equation of the path of a projectile and show it is a parabola. Also derive the time of flight and range.5 marks

Projected with speed u at angle θ: x = (u cos θ)t and y = (u sin θ)t − ½gt². Eliminating t = x/(u cos θ): y = x tan θ − gx²/(2u² cos²θ), which is of the form y = ax − bx², a parabola. Time of flight: y = 0 gives t = 2u sin θ/g. Range R = (u cos θ) × T = u² × 2 sin θ cos θ/g = u² sin 2θ/g, maximum at θ = 45°.

Quick revision

Max range at 45°.
Horizontal velocity constant in projectile.
Speed constant but velocity changes in uniform circular motion.

Exam tips

Angles θ and (90° − θ) give the same range.
Resolve projectile motion into independent horizontal (uniform velocity) and vertical (uniform acceleration g) motions.

Practice MCQs

Range of a projectile is maximum at
1. 30°
2. 45°
3. 60°
4. 90°
Show answer
Answer: (b) 45°. sin 90° = 1.
Path of a projectile is a
1. parabola
2. circle
3. straight line
4. ellipse
Show answer
Answer: (a) parabola. Parabola.
Resultant of two perpendicular forces 3 N and 4 N is
1. 7 N
2. 1 N
3. 12 N
4. 5 N
Show answer
Answer: (d) 5 N. √(9 + 16).
A body moves in a circle of radius 2 m at 4 m/s. Centripetal acceleration is
1. 2 m/s²
2. 8 m/s²
3. 16 m/s²
4. 4 m/s²
Show answer
Answer: (b) 8 m/s². 16/2.
A ball is thrown horizontally at 10 m/s from a 20 m high cliff (g = 10 m/s²). Horizontal distance covered is
1. 10 m
2. 20 m
3. 40 m
4. 200 m
Show answer
Answer: (b) 20 m. t = 2 s.
A projectile is thrown at 20 m/s at 30° (g = 10 m/s²). Time of flight is
1. 1 s
2. 2 s
3. 4 s
4. 3 s
Show answer
Answer: (b) 2 s. 2 × 20 × 0.5/10.
Two projectiles at 30° and 60° with the same speed have
1. different ranges
2. same height
3. same time
4. same range
Show answer
Answer: (d) same range. Complementary angles.
A wheel makes 120 revolutions per minute. Its angular speed is
1. 4π rad/s
2. 2π rad/s
3. 120 rad/s
4. 60π rad/s
Show answer
Answer: (a) 4π rad/s. 2 rev/s × 2π.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.