Motion in a Plane

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 3.

How does a fielder know where the catch will land?

A skied cricket ball is 30 m away and 25 m up. The fielder does no arithmetic. He simply runs at steady speed to a spot and waits. He can do that because the ball's horizontal motion and its vertical motion are completely independent: one is steady, the other is the familiar up-and-down of free fall. Splitting a motion into two independent straight-line motions is the big idea of this chapter, and vectors are the language that makes it possible.

Scalars and vectors

A scalar has only magnitude: mass, time, temperature, density, work, energy, distance. A vector has magnitude and direction and combines by the triangle or parallelogram law: displacement, velocity, acceleration, force, momentum.

The position vector r points from a chosen origin to the particle. The displacement vector points from the starting position to the final one and depends only on those two points, not on the route. Its magnitude is never larger than the path length. Two vectors are equal only if they have the same magnitude and the same direction, wherever they are drawn.

Beyond the textbook. For the exam: a vector is a quantity with magnitude and direction. In reality: that is necessary but not sufficient. Electric current has a direction along the wire and a magnitude, yet two currents meeting at a junction add as ordinary numbers, not by the parallelogram law, so current is a scalar. Why it matters: the real test of a vector is how it adds, which is why a "direction arrow" alone does not make a quantity a vector.

Combining vectors

Multiplying by a number. Multiplying a vector A by a positive number λ stretches it by λ and keeps its direction. A negative λ reverses it. Multiplying by zero gives the null vector, which has zero magnitude and no definite direction. It appears whenever a body returns to its start.

Adding. In the triangle method, 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. In the parallelogram method, draw both from a common point and take the diagonal. Both give the same answer. Vector addition is commutative ($A + B = B + A$) and associative ($(A + B) + C = A + (B + C)$). Subtracting is adding the negative: $A - B = A + (-B)$.

Resolving. Any vector in a plane can be split into two perpendicular components. If A makes angle θ with the x-axis,

$$A_x = A\cos\theta, \qquad A_y = A\sin\theta$$

with magnitude $A = \sqrt{A_x^2 + A_y^2}$ and direction $\tan \theta = A_y/A_x$. The numbers $A_x$ and $A_y$ are scalars (they can be negative), and $A_x\hat{i}$ is the vector. Here $\hat{i}$, $\hat{j}$ and k̂ are unit vectors along x, y and z: they have magnitude 1, no unit and no dimension, and only point the way. So $A = A_x\hat{i} + A_y\hat{j}$.

Adding with components is the method that always works: add the x-components, add the y-components, then rebuild the magnitude and angle. For two vectors of magnitudes A and B at angle θ to each other, the law of cosines gives the resultant directly:

$$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$

$$\tan \alpha = \dfrac{B\sin\theta}{A + B\cos\theta}$$

where α is the angle R makes with A. The special cases are worth remembering.

For the difference A − B, the formula is the same with the sign inside the root changed: the magnitude of $A - B$ is $\sqrt{A^2 + B^2 - 2AB\cos\theta}$.

Worked example: two forces at 60°. Forces of 3 N and 5 N act on a body with 60° between them. Find the resultant and its direction relative to the 3 N force.

The common wrong step: writing R = 3 + 5 = 8 N, which is true only when the forces point the same way, or using a minus sign inside the root, which gives $\sqrt{19} \approx 4.4$ N (the magnitude of A − B) instead.

Worked example: components. Given $A = 3\hat{i} + 4\hat{j}$ and $B = \hat{i} - 2\hat{j}$, find $A + 2B$ and the magnitude of $A - B$.

The common wrong step: adding an x-component to a y-component (3 + 4). The two axes never mix. The vector $5\hat{i}$ has no y part at all.

Velocity and acceleration in a plane

With position $r = x\hat{i} + y\hat{j}$, the average velocity is $\Delta r/\Delta t$, and the instantaneous velocity is

$$v = \dfrac{dr}{dt} = v_x\hat{i} + v_y\hat{j}$$

It always points along the tangent to the path, because as Δt shrinks the chord Δr lines up with the curve. Acceleration is $a = dv/dt = a_x\hat{i} + a_y\hat{j}$. In a plane, a can make any angle with v, which is new compared with a straight line, where a was either along v or against it.

If the acceleration is constant, the vector equations look just like the 1-D ones:

$$v = v_0 + at, \qquad r = r_0 + v_0 t + \dfrac{1}{2}at^2$$

Each splits into two independent 1-D problems, one along x and one along y.

Worked example: constant acceleration in a plane. A particle starts at the origin at t = 0 with velocity $8\hat{j}$ m s⁻¹ and moves in the x–y plane with constant acceleration $(4\hat{i} + 2\hat{j})$ m s⁻². When its x-coordinate is 18 m, find its y-coordinate and its speed.

The common wrong step: using the speed 8 m s⁻¹ along x as well. The initial velocity has no x-component, so the x-motion starts from rest.

Projectile motion

A projectile is a body thrown into the air and moving under gravity alone. Fire it with speed $v_0$ at angle $\theta_0$ above the horizontal. Horizontally there is no force, so $a_x = 0$. Vertically gravity pulls down, so $a_y = -g$. Taking the launch point as the origin,

Why the path is a parabola. Solve the first position equation for t and put it into the second:

$$y = (\tan \theta_0)\,x - \dfrac{g x^2}{2 v_0^2 \cos^2 \theta_0}$$

This has the form $y = ax - bx^2$, a parabola.

The results you must know (launch and landing at the same level):

Four consequences follow from these formulas. The range is largest at $\theta_0 = 45^\circ$, where $\sin 2\theta_0 = 1$, and then $R_{\max} = v_0^2/g$. Angles that add to 90° (such as 30° and 60°) give the same range, because $\sin 2\theta$ is the same, but the higher angle goes higher and stays longer in the air. The height is largest at 90°, and $R/H = 4/\tan\theta_0$, so at 45° the range is four times the height. At the top of the path, $v_y = 0$ but $v_x = v_0 \cos \theta_0$ is unchanged, so the speed is not zero there, and the acceleration is still g downward.

Worked example: a ball on level ground. A ball is kicked at 20 m s⁻¹ at 30° above the horizontal (g = 10 m s⁻²). Find the maximum height, time of flight, range and speed at the top.

The common wrong step: using the speed at the top as zero, or using 20 m s⁻¹ as the horizontal velocity. The 20 is the speed along the slant, and only its horizontal part, $20 \cos 30^\circ$, drives the range.

Pause and predict: one ball is dropped from a tall cliff, and at the same instant another is thrown horizontally from the same cliff. Which one hits the ground first?

Check your answer: they land together. Both start with zero vertical velocity and both accelerate downward at g. The horizontal motion changes only where the second ball lands, not when. This is the independence of the two motions in its simplest form.

Horizontal projection. When a body is launched horizontally from height h at speed u, the vertical motion is a plain drop from rest, so $t = \sqrt{2h/g}$, the range is $u\sqrt{2h/g}$, and the landing speed has the horizontal part $u$ and a vertical part $gt$.

Worked example: a stone thrown from a cliff. A stone is thrown horizontally at 15 m s⁻¹ from a cliff 80 m high (g = 10 m s⁻²). Find the time of fall, the horizontal distance covered and the velocity on landing.

The common wrong step: using the range formula $v_0^2 \sin 2\theta_0/g$ here. It assumes the landing is at the launch level, which is false for a cliff.

Beyond the textbook. For the exam: the path is a parabola and 45° gives the longest range. In reality: air drag makes the descent steeper than the climb, so the path is no longer a parabola, and the best angle for a ball in air falls below 45°. Even without drag, if the ball leaves the hand above the ground, as in the shot put, the best angle is below 45°. Why it matters: the clean formulas are the vacuum case from level ground, and every sporting situation bends those assumptions a little.

Uniform circular motion

A body moving on a circle at constant speed is in uniform circular motion (UCM). Its speed is constant, but its velocity is not, because the direction keeps changing. A change of velocity is an acceleration, so the body is accelerating even though the speedometer reads steady.

Why the acceleration points to the centre. The velocity is along the tangent. Over a short time the velocity vector rotates by a small angle, and the change in velocity points toward the centre. This centripetal acceleration has magnitude

$$a_c = \dfrac{v^2}{R} = \omega^2 R$$

The angular speed ω is measured in rad s⁻¹, and the other quantities are tied to it by

$$v = \omega R, \qquad T = \dfrac{2\pi}{\omega} = \dfrac{2\pi R}{v}, \qquad \nu = \dfrac{1}{T}, \qquad \omega = 2\pi\nu$$

which gives $a_c = 4\pi^2\nu^2 R = 4\pi^2 R/T^2$. The velocity (tangent) and the acceleration (radius) are perpendicular at every instant. The magnitude of a_c is constant, but its direction keeps rotating, so a_c is not a constant vector. That is why the equations of constant acceleration cannot be applied to UCM.

Worked example: a spinning flywheel. A flywheel of radius 20 cm turns at 300 revolutions per minute. Find its period, the speed of a point on the rim and the centripetal acceleration there.

The common wrong step: substituting 300 as ω. Revolutions per minute must be converted to radians per second, multiplying by $2\pi$ and dividing by 60.

Pause and predict: a stone is whirled in a horizontal circle on a string. The string snaps. In which direction does the stone fly off?

Check your answer: along the tangent, in a straight line, with the speed it had at that instant. With the string gone nothing supplies the inward acceleration, so the stone follows the first law and keeps its velocity.

Beyond the textbook. For the exam: $a_c = \omega^2 R$ is the centripetal acceleration of any body in UCM. In reality: you are in UCM right now because the Earth turns once in about 24 hours. A person standing on the equator has $\omega \approx 7.3 \times 10^{-5}$ rad s⁻¹ and $R \approx 6.4 \times 10^{6}$ m, so $a_c = \omega^2 R \approx 0.034$ m s⁻², roughly 0.3 % of g. That tiny inward acceleration is a part of why weighing scales read very slightly less at the equator than at the poles. Why it matters: it shows how small ω²R is for everyday rotation, compared with the 197 m s⁻² of the flywheel.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Motion in a Straight Line supplies the kinematic equations that you apply twice here, once along each axis.
• Next: Laws of Motion shows which real force provides the centripetal acceleration (friction, tension, the normal reaction) and uses the vector resolution you practised here.
• Later in Class XI: Work, Energy and Power multiplies vectors to define work, and Gravitation uses circular motion to explain orbits.