Motion In A Plane: Complete Chapter Revision Notes For Class 11 NEET by PW

Questions from Motion In A Plane often require you to break motion into horizontal and vertical components before applying the right equation. This makes vectors an important starting point, while applications such as projectile motion and relative velocity add another layer to the chapter. A clear understanding of these basics is useful for handling numerical questions in NEET Physics.

The PW Motion In A Plane: Complete Chapter One-Shot Video For Class 11 NEET 2027 covers the concepts and important results needed to revise the chapter, from vector representation and resolution to projectile motion, relative velocity, river crossing, and rain-man applications. Use these notes to refresh the concepts and formulas before practising questions.

Velocity–Acceleration Angle And Nature Of Motion

The angle between velocity and acceleration gives information about the nature of motion. When velocity and acceleration are parallel or antiparallel, the angle between them is 0° or 180°, and the motion is along a straight line.

If the angle between velocity and acceleration is neither 0° nor 180°, the path is curved.

Projectile motion generally has a changing angle between velocity and acceleration, so its path is curved.

Vector Kinematic Equations

For constant acceleration, the equations of motion can be written in vector form:

→v = →u + →a t

→s = →u t + ½ →a t²

For motion in two dimensions, resolve the motion into x- and y-components and solve them separately.

If the coordinates of a particle are x = x(t) and y = y(t):

vₓ = dx/dt

vᵧ = dy/dt

→v = vₓ →i + vᵧ →j

aₓ = dvₓ/dt

aᵧ = dvᵧ/dt

→a = aₓ →i + aᵧ →j

Memory Tip: Position → Velocity → Acceleration, with each step obtained by differentiation with respect to time.

For a vector A = a →i + b →j, its magnitude is:

|A| = √(a² + b²)

Projectile Motion: Definition And Assumptions

When a body is projected into the air, and only gravity acts on it after projection, it is called a projectile. The motion of the body is called projectile motion, and the path followed by it is called its trajectory.

The standard assumptions are:

1. Air resistance is neglected.

2. The ground is considered flat.

3. Acceleration due to gravity, g, is constant.

4. The curvature of the Earth is neglected.

Projectile motion is uniformly accelerated motion. In oblique projection, the initial velocity has both horizontal and vertical components, while acceleration acts vertically downward.

Oblique Projectile Motion

A projectile is launched from the ground with an initial speed u at an angle θ to the horizontal.

The initial velocity components are:

uₓ = u cosθ

uᵧ = u sinθ

The trajectory of an oblique projectile is a downward-opening parabola.

Memory Tip: In the standard projectile formulas, θ is measured from the horizontal. If the angle is measured from the vertical as φ, then θ = 90° − φ.

Independence Of Horizontal And Vertical Motion

Horizontal and vertical motions are independent of each other.

• Horizontal acceleration: aₓ = 0

• Horizontal velocity: vₓ = u cosθ

• Vertical acceleration: aᵧ = −g

• Vertical velocity: vᵧ = u sinθ − gt

Standard Projectile Formulas

For a projectile projected and landing at the same level:

Time of flight:

T = 2u sinθ/g

Maximum height:

Hₘₐₓ = u² sin²θ/(2g)

Horizontal range:

R = u² sin2θ/g

In terms of the initial velocity components:

T = 2uᵧ/g

Hₘₐₓ = uᵧ²/(2g)

R = 2uₓuᵧ/g

The range is maximum when:

θ = 45°

The maximum range is:

Rₘₐₓ = u²/g

For maximum range:

Hₘₐₓ = Rₘₐₓ/4

This relation applies specifically when θ = 45° and the range is maximum.

General Relation Between Maximum Height And Range

For a projectile launched and landing at the same level:

Hₘₐₓ/R = tanθ/4

Therefore:

4Hₘₐₓ = R tanθ

This relation is valid for any angle of projection under the standard projectile assumptions.

If the maximum height is equal to the range:

Hₘₐₓ = R

Using 4Hₘₐₓ = R tanθ:

4R = R tanθ

Therefore:

tanθ = 4

θ = tan⁻¹4 ≈ 76°

So, θ = tan⁻¹4 is a special result obtained from the condition Hₘₐₓ = R, and is not a general projectile relation.

Complementary Angles

Two projectiles launched with the same speed at complementary angles θ and (90° − θ) have the same horizontal range.

R₁ = R₂

Their times of flight and maximum heights are generally different.

For θ and (90° − θ):

T₁/T₂ = tanθ

H₁/H₂ = tan²θ

The product of their times of flight is:

T₁T₂ = 2R/g

The product of their maximum heights is:

H₁H₂ = R²/16

Velocity, Energy, And Momentum

At any time t, the velocity of an oblique projectile is:

→v = u cosθ →i + (u sinθ − gt) →j

The speed is:

v = √(u² + g²t² − 2ugt sinθ)

At the highest point:

vᵧ = 0

vₓ = u cosθ

Therefore, the velocity of the projectile is not zero at the highest point. Its speed is minimum and is given by:

vₘᵢₙ = u cosθ

At the highest point, velocity and acceleration are perpendicular, so the angle between them is 90°.

At the same horizontal level, the projectile has the same speed while moving upward and downward.

For a projectile landing at the same level from which it was launched:

Initial velocity:

→u = u cosθ →i + u sinθ →j

Final velocity:

→v = u cosθ →i − u sinθ →j

The initial and final speeds are equal, so their kinetic energies are also equal. However, their velocity vectors are different.

Change in velocity:

Δ→v = −2u sinθ →j

Magnitude of change in momentum:

|Δ→p| = 2mu sinθ

However, the change in the magnitude of momentum is:

Δ|→p| = 0

Memory Tip: Do not confuse the magnitude of the change in momentum, |Δ→p|, with the change in the magnitude of momentum, Δ|→p|.

Equation Of Trajectory

For a projectile launched from the origin:

x = u cosθ t

y = u sinθ t − ½gt²

Eliminating t gives the equation of trajectory:

y = x tanθ − gx²/(2u² cos²θ)

This represents a downward-opening parabola.

The trajectory equation can also be written in terms of range:

y = x tanθ (1 − x/R)

For a point (a, b), the minimum speed required for a projectile to pass through the point is:

vₘᵢₙ = √[g(b + √(a² + b²))]

Memory Tip: When a question directly involves the range, the range form of the trajectory equation can make the calculation easier.

Horizontal Projectile

For horizontal projection from a height h, the initial vertical velocity is zero.

The time taken to reach the ground is:

t = √(2h/g)

The horizontal range is:

R = u√(2h/g)

At time t:

→v = u →i − gt →j

The speed is:

v = √(u² + g²t²)

The trajectory is:

y = −gx²/(2u²)

A body dropped from rest and a body projected horizontally from the same height reach the ground simultaneously because their vertical motions are identical.

Relative Motion

The velocity of A with respect to B is:

→v_AB = →v_A − →v_B

Or, using a common reference frame:

→v_AB = →v_AG − →v_BG

The relative velocities satisfy:

→v_AB = −→v_BA

Therefore, they have equal magnitudes but opposite directions.

For two trains moving in opposite directions:

t = (L₁ + L₂)/(v₁ + v₂)

For two trains moving in the same direction, when the faster train overtakes the slower train:

t = (L₁ + L₂)/(v₁ − v₂)

For a train crossing a bridge:

t = (L_train + L_bridge)/v

For a train crossing a person or a bird, only the train length is considered:

t = L_train/v_relative

The relative equations of motion are:

v_rel = u_rel + a_rel t

s_rel = u_rel t + ½a_rel t²

Collision Condition For Two Particles

Two particles collide when they occupy the same position at the same time.

Let their initial position vectors be →R_A and →R_B, and their constant velocity vectors be →V_A and →V_B.

Their positions after time t are:

→r_A = →R_A + →V_A t

→r_B = →R_B + →V_B t

For collision:

→r_A = →r_B

Therefore:

→R_A − →R_B = (→V_B − →V_A)t

This gives the collision condition:

→R_A − →R_B is parallel to →V_B − →V_A

The time of collision, when the direction condition is satisfied, is:

t = |→R_A − →R_B| / |→V_B − →V_A|

The particles must also be moving toward the same meeting point. Parallel relative position and relative velocity alone are not sufficient if their directions do not lead to an intersection at t > 0.

Collision Of Two Projectiles

Consider two projectiles launched simultaneously from different points at the same vertical level, with the same downward acceleration g.

For the projectiles to collide at a later time, their vertical positions must be equal.

Since both projectiles have the same downward acceleration, their vertical displacement difference depends only on their initial vertical velocity components.

Therefore, for a collision:

u₁ sinθ₁ = u₂ sinθ₂

This means their initial vertical components must be equal.

Once this condition is satisfied, their vertical separation remains zero. The collision time is then determined by their horizontal motion.

If their initial horizontal positions are x₁ and x₂, and their horizontal velocity components are u₁ cosθ₁ and u₂ cosθ₂:

x₁ + u₁ cosθ₁ t = x₂ + u₂ cosθ₂ t

Therefore:

t = (x₂ − x₁)/(u₁ cosθ₁ − u₂ cosθ₂)

The value of t must be positive for a physical collision after launch.

Important: The equal vertical-component condition applies to projectiles launched simultaneously from the same vertical level under the same gravitational acceleration. If the initial heights are different or the launch times are different, the condition has to be modified accordingly.

River Swimmer And Rain-Man Problems

For a swimmer crossing a river:

→V_SG = →V_SR + →V_RG

where:

• →V_SG = velocity of swimmer with respect to ground

• →V_SR = velocity of swimmer with respect to river

• →V_RG = velocity of river with respect to ground

For a river of width d, if the swimmer moves at an angle θ such that the component across the river is V_SR cosθ:

t = d/(V_SR cosθ)

Minimum time occurs when the swimmer moves perpendicular to the river:

tₘᵢₙ = d/V_SR

For the shortest path, the downstream drift must be cancelled:

V_SR sinθ = V_RG

The corresponding time is:

t_shortest = d/√(V_SR² − V_RG²)

This condition is possible only when:

V_SR > V_RG

For rain-man problems, the velocity of rain relative to the man is:

→V_RM = →V_RG − →V_MG

The umbrella should be held along the direction of the apparent rain velocity to protect against the rain.

Revise Motion In A Plane Class 11 NEET to improve your understanding of vectors and their applications in different types of two-dimensional motion. The PW Motion In A Plane: Complete Chapter One-Shot Video For Class 11 NEET 2027 can help you refresh key concepts and formulas before practising NEET Physics questions.

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