Lesson 10 of 10 · 15 min
Chapter review
Watch a class
The whole chapter on YouTube
Complete chapter walkthrough with NCERT coverage
Prashant Kirad 11th & 12th · Hinglish · Whole chapter · Open on YouTube
One-shot revision with derivations
PW Class 11 Science · Hinglish · Whole chapter · Open on YouTube
Must-know facts
16 facts
- 1Displacement magnitude ≤ path length; equal only for straight-line motion in one direction.
- 2Unit vectors î, ĵ, k̂ have magnitude 1 and no unit or dimension.
- 3Ax = A cos θ, Ay = A sin θ, where θ is measured from the x-axis.
- 4R² = A² + B² + 2AB cos θ; tan α = B sin θ / (A + B cos θ).
- 5|A − B| ≤ |A + B| ≤ A + B.
- 6A − B = A + (−B); A + (−A) = 0, the null vector.
- 7Instantaneous velocity is tangent to the path.
- 8In a plane the angle between v and a can be anything from 0° to 180°.
- 9Constant a: v = v₀ + at, r = r₀ + v₀t + ½at², component by component.
- 10Projectile: horizontal velocity v₀ cos θ₀ constant; vertical acceleration −g.
- 11T_f = 2v₀ sin θ₀/g; hₘ = v₀² sin² θ₀/2g; R = v₀² sin 2θ₀/g.
- 12Maximum range v₀²/g at 45°; θ and 90° − θ give equal ranges.
- 13The trajectory of a projectile is a parabola: y = x tan θ₀ − gx²/(2v₀² cos² θ₀).
- 14Uniform circular motion: a_c = v²/R = ω²R = 4π²ν²R, towards the centre.
- 15v = Rω; ω = 2πν = 2π/T.
- 16Average speed ≥ magnitude of average velocity.
Common traps
Where marks are lost
Adding the magnitudes: 40 m and 30 m give 70 m whatever the angle.
Taking Ax = A sin θ out of habit.
Saying the velocity of a projectile is zero at the highest point.
Thinking uniform circular motion has no acceleration because the speed is constant.
Using v = u + at along the circle for uniform circular motion.
Believing a ball thrown horizontally takes longer to fall than one dropped from the same height.
Assuming a higher launch angle always sends a projectile farther.
Giving a unit vector a unit, such as î metres.
Formulas
13 to know
Components
Ax = A cos θ, Ay = A sin θ
θ measured from the x-axis.
Magnitude and direction
A = √(Ax² + Ay²), tan θ = Ay/Ax
In 3D add Az² under the root.
Resultant of two vectors
R² = A² + B² + 2AB cos θ
Law of cosines; θ is the angle between A and B.
Direction of the resultant
tan α = B sin θ / (A + B cos θ)
α measured from A.
Law of sines
R/sin θ = A/sin β = B/sin α
Angles of the vector triangle.
Constant acceleration
v = v₀ + at, r = r₀ + v₀t + ½at²
Applies to each component separately.
Projectile position
x = v₀ cos θ₀ t, y = v₀ sin θ₀ t − ½gt²
Launch from the origin; g = 9.8 m/s².
Projectile path
y = x tan θ₀ − gx² / (2v₀² cos² θ₀)
A parabola.
Time of flight
T_f = 2v₀ sin θ₀ / g
Time to the top is half of it.
Maximum height
hₘ = v₀² sin² θ₀ / 2g
Reached when vᵧ = 0.
Horizontal range
R = v₀² sin 2θ₀ / g
Maximum v₀²/g at 45°.
Centripetal acceleration
a_c = v²/R = ω²R = 4π²ν²R
Directed towards the centre.
Angular speed
v = Rω, ω = 2πν = 2π/T
ω in rad/s, ν in s⁻¹ (Hz).
Key terms
13 terms
- Scalar
- A quantity fully given by a number and a unit.
- Vector
- A quantity with magnitude and direction that adds by the triangle law.
- Position vector
- The arrow from the chosen origin to the point.
- Displacement
- The change in position vector, from the starting point to the end point.
- Null vector
- The vector of zero magnitude, the sum of a vector and its reverse.
- Unit vector
- A dimensionless vector of magnitude 1 that only marks a direction.
- Resolution
- Splitting a vector into components along chosen directions.
- Projectile
- A body moving under gravity alone after being launched.
- Trajectory
- The path a body follows; for a projectile, a parabola.
- Horizontal range
- The horizontal distance from launch to return to the launch level.
- Centripetal acceleration
- The acceleration towards the centre of a circle, v²/R in size.
- Angular speed
- The rate at which the angle swept about the centre changes, in rad/s.
- Frequency
- Number of revolutions per second, the reciprocal of the time period.