Motion in a Plane

In this chapter: scalars and vectors, multiplying a vector by a number, triangle and parallelogram laws, unit vectors and resolution, analytical addition, position, velocity and acceleration in two dimensions, motion with constant acceleration, projectile motion and uniform circular motion.

Scalars and vectors

A scalar has only magnitude: mass, time, temperature, work, energy. Scalars add by ordinary arithmetic. A vector has magnitude and direction and obeys the triangle (or parallelogram) law of addition: displacement, velocity, acceleration, force, momentum. In print a vector is written in bold or with an arrow, A, and its magnitude as |A| or A.

• Two vectors are equal only if they have the same magnitude and the same direction. A vector may be shifted parallel to itself without changing it.
• Multiplying A by a positive number λ changes its magnitude to λA and keeps its direction. A negative number reverses the direction.
• The null vector 0 has zero magnitude: A + 0 = A and A − A = 0. The displacement of a body that returns to its starting point is a null vector.

Adding and subtracting graphically

Triangle law: place the tail of B at the head of A; the vector from the tail of A to the head of B is A + B. Parallelogram law: draw both from a common tail and complete the parallelogram; the diagonal from the common tail is the resultant. Vector addition is commutative (A + B = B + A) and associative. Subtraction is addition of the negative: A − B = A + (−B).

Resolution of vectors

A unit vector has magnitude 1 and only shows direction: n̂ = A/A. The unit vectors along the x, y and z axes are î, ĵ and k̂. Any vector in the x-y plane can be written as the sum of its components:

Adding vectors analytically

With components, just add like components: if R = A + B, then R_x = A_x + B_x and R_y = A_y + B_y. For two vectors with an angle θ between them, the law of cosines gives

Worked example: Two forces of 3 N and 4 N act on a body with an angle of 60° between them. Find the magnitude of the resultant.
Solution: R = √(9 + 16 + 2 × 3 × 4 × cos 60°) = √(25 + 12) = √37 ≈ 6.1 N. If the angle were 90°, R would be exactly 5 N; at 0° it is 7 N and at 180° it is 1 N.

Position, velocity and acceleration in a plane

The position vector of a particle is r = x î + y ĵ. Its velocity and acceleration are

The velocity at any point is always along the tangent to the path, in the direction of motion. Acceleration need not be along the path; in circular motion it is perpendicular to it.

Constant acceleration

The important idea is that motion in a plane can be treated as two separate straight-line motions along x and y, linked only by the common time t. Solve each direction with the Chapter 2 equations and join them through t.

Projectile motion

A body thrown into the air and moving under gravity alone (air resistance neglected) is a projectile. Take the launch point as origin, x horizontal and y vertically up, with speed u at angle θ above the horizontal. Then a_x = 0 and a_y = −g.

• The range is maximum at θ = 45°: R_max = u²/g.
• Angles θ and (90° − θ) give the same range but different heights and times of flight.
• At the highest point the velocity is u cos θ (horizontal) and the acceleration is still g downward, so the two are at 90°. The velocity is never zero unless θ = 90°.
• A body thrown horizontally with speed u from a height h takes t = √(2h/g) to land, the same time as a body simply dropped from that height, and lands a horizontal distance u√(2h/g) away.

Worked example: A ball is thrown with a speed of 20 m s⁻¹ at 30° above the horizontal. Take g = 10 m s⁻². Find its time of flight, maximum height and range.
Solution: T = 2(20)(sin 30°)/10 = 2 × 20 × 0.5/10 = 2 s. H = (20)²(sin 30°)²/(2 × 10) = 400 × 0.25/20 = 5 m. R = (20)² sin 60°/10 = 40 × 0.866 ≈ 34.6 m. A throw at 60° with the same speed would give the same range but a height of 15 m.

Uniform circular motion

A body moving on a circle of radius R at constant speed v is in uniform circular motion. The speed is constant but the direction of velocity keeps changing, so the body is accelerated. This acceleration, the centripetal acceleration, is always directed towards the centre.

Worked example: An insect trapped in a circular groove of radius 12 cm moves along it steadily and completes 7 revolutions in 100 s. Find its angular speed, linear speed and acceleration.
Solution: ν = 7/100 = 0.07 rev s⁻¹, so ω = 2πν = 2 × 3.14 × 0.07 ≈ 0.44 rad s⁻¹. v = ωR = 0.44 × 12 ≈ 5.3 cm s⁻¹. a = ω²R = (0.44)² × 12 ≈ 2.3 cm s⁻², directed towards the centre of the groove.

Relative velocity in two dimensions

As in one dimension, v_BA = v_B − v_A, but now the subtraction is vector subtraction. The classic example is rain: a person walking through vertically falling rain sees it coming at an angle from the front, and should tilt the umbrella forward by the angle whose tangent is (walking speed)/(rain speed).

Common mistakes: (1) Saying the velocity at the top of a projectile is zero; only the vertical component is zero. (2) Using R = u² sin 2θ/g when the landing point is not at the launch level. (3) Adding magnitudes of vectors directly (3 N + 4 N is not 7 N unless they are parallel). (4) Calling uniform circular motion "unaccelerated" because the speed is constant. (5) Mixing rev s⁻¹ and rad s⁻¹; multiply the frequency by 2π to get ω.

JEE and NEET focus

• Resultant of two vectors, its range, and the angle it makes with one of them.
• T, H, R of a projectile; complementary angles; relation between R and H (R = 4H cot θ).
• Horizontal projection from a height, and finding velocity at a given time or height.
• Centripetal acceleration from frequency, period or speed.
• Two-dimensional motion given as x(t) and y(t), finding velocity and acceleration vectors by differentiation.

Practice questions