Motion in a Plane Class 11 Notes — Core Revision & Formulas

Understanding Vectors & 2D Kinematics

To describe motion in a plane, we must transition from single-dimension scalars to two-dimensional vectors. Unlike motion in a straight line, directional displacement in a plane requires us to decompose variables into independent horizontal ($x$) and vertical ($y$) components.

Vector Addition & Resolution

Vectors cannot be added arithmetically; they require geometric vector laws.

• Triangle Law of Vector Addition: If two vectors are represented as two sides of a triangle in order, the third closing side represents the resultant vector in reverse order.
• Parallelogram Law of Vector Addition: The resultant magnitude $R$ of two vectors $\vec{A}$ and $\vec{B}$ acting at an angle $\theta$ is:

$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$
The direction of the resultant vector with respect to vector $\vec{A}$ is given by:
$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$

• Resolution of a Vector: Any vector $\vec{A}$ in a 2D plane can be split into rectangular components:

$A_x = A\cos\theta \quad \text{and} \quad A_y = A\sin\theta$
where $\theta$ is the angle made by the vector with the positive x-axis.

Important Terms & Glossary

Scalar Quantity

A physical quantity that has magnitude only and is completely specified by a real number and a unit (e.g., mass, distance, speed).

Vector Quantity

A physical quantity that possesses both a magnitude and a specific direction, complying with the rules of vector algebra (e.g., displacement, velocity, force).

Unit Vector

A dimensionless vector of unit magnitude (value = 1) pointing in a specific direction. Represented as $\hat{a} = \vec{a} / |\vec{a}|$. The unit vectors along x, y, and z axes are denoted by $\hat{i}$, $\hat{j}$, and $\hat{k}$ respectively.

Projectile

An object thrown or projected into space, subject only to the acceleration due to gravity ($g$) and ignoring air resistance.

Trajectory

The parabolic path traced by a projectile in flight under the constant influence of gravity.

Centripetal Acceleration

The acceleration directed toward the center of a circular path that causes an object to continuously change its direction of motion while maintaining circular orbit.

Must-Remember Projectile Formulas

• Equation of Trajectory: The path of a projectile is always a parabola, mathematically represented as: y = x\tan\theta - \frac{g x^2}{2 u^2\cos^2\theta}.
• Time of Flight (T): The total duration the projectile remains in air is given by: T = \frac{2 u\sin\theta}{g}.
• Maximum Height (H): The greatest vertical distance reached above the horizontal launch plane is: H = \frac{u^2\sin^2\theta}{2 g}.
• Horizontal Range (R): The total horizontal distance covered from the launch point to landing is: R = \frac{u^2\sin 2\theta}{g}.
• Complementary Angles: A projectile has the same horizontal range for projection angles of theta and (90° - theta).
• Maximum Horizontal Range: The range is maximized when the angle of projection theta is 45°, which yields R_max = u^2 / g.
• Uniform Circular Motion Speed: Linear velocity is related to angular velocity by the vector relation: v = \omega \times r.
• Centripetal Acceleration Magnitude: The acceleration keeping an object in circular path is given by: a_c = \frac{v^2}{r} = \omega^2 r.

Projectile Motion: Horizontal vs. Vertical Component

Step-by-Step Procedure to Solve 2D Relative Velocity Problems

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Worked Revision Examples

• {"title":"Example 1: Calculating Projectile Metrics","problem":"A football is kicked with an initial velocity of 20 m/s at an angle of 30° to the horizontal. Take g = 10 m/s². Find (a) Time of Flight, (b) Max Height, and (c) Range.","solution":"Given: u = 20 m/s, theta = 30°, g = 10 m/s².\n1. Time of Flight (T) = 2 u sin(theta) / g = 2 20 sin(30°) / 10 = 40 0.5 / 10 = 2 seconds.\n2. Max Height (H) = u² sin²(theta) / (2 g) = 20² (sin(30°))² / (2 10) = 400 0.25 / 20 = 5 meters.\n3. Horizontal Range (R) = u² sin(2theta) / g = 20² sin(60°) / 10 = 400 (sqrt(3)/2) / 10 = 20 * sqrt(3) ≈ 34.64 meters."}
• {"title":"Example 2: Vector Resultant Magnitude","problem":"Two forces of magnitude 3 N and 4 N act concurrently at an angle of 90° to each other. Calculate the magnitude and direction of the resultant force.","solution":"Given: A = 3 N, B = 4 N, theta = 90°.\nUsing Parallelogram Law:\nR = sqrt(A² + B² + 2ABcos(90°))\nSince cos(90°) = 0, R = sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5 N.\nDirection (alpha) with respect to 3 N force:\ntan(alpha) = B sin(90°) / (A + B cos(90°)) = 4 1 / (3 + 0) = 4/3.\nalpha = arctan(4/3) ≈ 53.1°."}

Common Exam Traps & Scoring Keys

• The Angle Trap: Always check if the angle given in the question is with the horizontal or the vertical. Formulas like $T = \frac{2u\sin\theta}{g}$ assume $\theta$ is with the horizontal. If the question specifies $\phi$ with the vertical, substitute $\theta = 90^\circ - \phi$.
• Circular Motion Signposts: Remember that in Uniform Circular Motion, speed is constant but velocity is variable because its direction constantly changes. Therefore, acceleration is non-zero (centripetal acceleration is present, but tangential acceleration is zero).
• Horizontal Velocity Rule: Under idealized projectile conditions, horizontal acceleration $a_x = 0$. Thus, $u\cos\theta$ remains completely unchanged throughout the entire flight. Never apply gravitational acceleration to horizontal components!

Quick Revision Check Q&A

• Q: Under what conditions is the horizontal range of a projectile maximum, and what is its maximum value? A: The horizontal range is maximum when the launch angle is theta = 45 degrees. The maximum range is given by R_max = u^2 / g.
• Q: What is the angle between the velocity vector and acceleration vector at the highest point of a projectile's trajectory? A: At the highest point, velocity is completely horizontal (u_x = u cos theta) and acceleration is vertically downward (g). Hence, the angle between them is exactly 90 degrees.
• Q: Does a unit vector have any units or physical dimensions? A: No. A unit vector is dimensionless and has no unit. It is strictly used to specify a direction in 3D space.
• Q: What is the magnitude of the velocity of a projectile at its maximum height? A: At maximum height, the vertical component of velocity becomes zero. Only the horizontal component remains, so the magnitude of velocity is u * cos(theta).

Frequently Asked Questions