Motion in a Plane — Class 11 Physics

1. Scalars and Vectors

A scalar has magnitude only, such as mass, time, distance and speed. A vector has both magnitude and direction, such as displacement, velocity, acceleration and force.

Vectors are represented by arrows. The length represents magnitude and the arrowhead represents direction.

2. Vector Addition

Vectors can be added using the triangle or parallelogram law. If two vectors A and B make an angle θ, the magnitude of their resultant is R=√(A²+B²+2AB cosθ).

For perpendicular vectors, θ=90°, so R=√(A²+B²).

3. Components of a Vector

A vector A making angle θ with the positive x-axis can be resolved into rectangular components: A_x=A cosθ and A_y=A sinθ.

The magnitude is recovered using A=√(A_x²+A_y²), and its direction satisfies tanθ=A_y/A_x, with the correct quadrant considered.

4. Motion in Two Dimensions

Position, velocity and acceleration can be treated as vectors. In Cartesian components, r=x i + y j, v=v_xi+v_yj, and a=a_xi+a_yj.

5. Projectile Motion

For a projectile launched with speed u at angle θ to the horizontal, neglecting air resistance:

Horizontal velocity: u cosθ
Vertical initial velocity: u sinθ
Time of flight: T=2u sinθ/g
Maximum height: H=u²sin²θ/(2g)
Horizontal range: R=u²sin2θ/g

These standard results assume launch and landing occur at the same level and g is constant.

6. Uniform Circular Motion

In uniform circular motion, speed is constant but velocity continuously changes direction, so acceleration is non-zero. The centripetal acceleration is a_c=v²/r=ω²r and is directed toward the centre.

Worked Numerical 1 — Resultant of Perpendicular Vectors

Question: A student walks 6 m east and then 8 m north. Find the magnitude and direction of displacement.

Solution: Resultant = √(6²+8²)=√100=10 m.

tanθ=8/6=4/3, so θ≈53.1° north of east.

Worked Numerical 2 — Vector Components

Question: Resolve a 20 N force acting at 30° above the horizontal.

F_x=20 cos30°=10√3 N≈17.32 N.

F_y=20 sin30°=10 N.

Worked Numerical 3 — Projectile Range

Question: A ball is projected at 20 m/s at 30° to the horizontal. Take g=10 m/s². Find the time of flight and range.

T=2u sin30°/g=2(20)(1/2)/10=2 s.

R=u²sin60°/g=400(√3/2)/10=20√3 m≈34.6 m.

Worked Numerical 4 — Maximum Height

Question: A projectile is fired with speed 20 m/s at 30°. Take g=10 m/s². Find maximum height.

H=u²sin²30°/(2g)=400(1/4)/20=5 m.

Worked Numerical 5 — Centripetal Acceleration

Question: A car moves around a circular track of radius 50 m at 10 m/s. Find its centripetal acceleration.

a_c=v²/r=100/50=2 m/s², directed toward the centre.

Explained MCQs

Which of the following is a vector?
A) speed B) distance C) displacement D) mass
Answer: C. Displacement requires both magnitude and direction.
Two perpendicular vectors have magnitudes 3 and 4. Their resultant is:
Answer: 5. √(3²+4²)=5.
At the highest point of a projectile, which component of velocity is zero?
Answer: Vertical component. The horizontal component remains u cosθ when air resistance is neglected.
At the highest point of projectile motion, is acceleration zero?
Answer: No. Acceleration remains g downward.
In uniform circular motion, which quantity changes continuously?
Answer: Velocity. Speed remains constant but direction of velocity changes.
For a given launch speed on level ground, maximum projectile range occurs at:
Answer: 45°. Since sin2θ is maximum when 2θ=90°.

Competency Question

A person walks 10 m east and then 10 m north. Explain why the total distance is 20 m but the magnitude of displacement is only 10√2 m. Draw the vector triangle representing the motion.

HOTS

Two projectiles are launched with the same speed at angles θ and (90°−θ) on level ground. Show why they have the same horizontal range. Do they necessarily have the same maximum height? Explain.

Common Mistakes

Adding vector magnitudes without considering direction.
Confusing speed with velocity.
Assuming projectile acceleration becomes zero at the highest point.
Using projectile formulas when launch and landing heights are different without modification.
For circular motion, assuming constant speed means zero acceleration.

Quick Revision

A_x=A cosθ, A_y=A sinθ
Resultant: √(A²+B²+2AB cosθ)
Projectile T=2u sinθ/g
Projectile H=u²sin²θ/(2g)
Projectile R=u²sin2θ/g
Centripetal acceleration=v²/r=ω²r
At projectile maximum height: v_y=0, but a=g downward

Class 11 Physics – Motion in a Plane