Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 3 · MECHANICS

DETAILED NOTES

Motion in a Plane

The complete chapter, section by section: vectors, resolving and adding them, motion with constant acceleration, projectile motion, and uniform circular motion, explained in plain language with original worked examples. For the interactive projectile simulation, see the concept page.

3.1 Vectors and Scalars

The last chapter described motion along a single straight line, where a plain ++ or −- sign was enough to capture direction, since only two directions were ever possible. Motion in a plane or in space needs more than a sign: it needs vectors.

A scalar is a quantity with magnitude only, completely specified by a single number and a unit, distance, mass, temperature, time. Scalars combine by ordinary algebra. A vector has both a magnitude and a direction, and combines by a different rule entirely, the triangle law or the parallelogram law of addition, not plain arithmetic. Displacement, velocity, acceleration, and force are all vectors.

3.1.1 Position and displacement vectors

To locate an object moving in a plane, pick a fixed origin OO. The straight line from OO to the object’s location PP is its position vector, r\mathbf{r}. If the object later sits at P′P', the vector from PP to P′P' is the displacement. Crucially, that displacement vector is the same regardless of which path the object actually travelled between PP and P′P', a straight walk and a long, winding detour between the same two points have identical displacement, even though the winding path is obviously longer. This is why the total path length travelled is always greater than or equal to the magnitude of the displacement, and equal only when the object never changes direction.

3.1.2 Equality of vectors

Two vectors are equal only if they share both the same magnitude and the same direction. Two arrows of identical length pointing in different directions are not equal, no matter how similar they look. Because a vector in this course carries no fixed location, sliding it parallel to itself never changes it, so equality can always be checked by shifting one vector until its tail meets the other’s and comparing tips.

3.2 Multiplying and Combining Vectors

Multiplying a vector A\mathbf{A} by a positive number λ\lambda scales its magnitude by λ\lambda and leaves its direction unchanged. Multiplying by a negative number scales the magnitude by ∣λ∣|\lambda| and flips the direction. Multiplying a constant velocity vector by a duration of time, for instance, produces a displacement vector, and the dimensions multiply along with the numbers.

Adding two vectors A\mathbf{A} and B\mathbf{B} graphically means placing B\mathbf{B}’s tail at A\mathbf{A}’s head and drawing the vector from A\mathbf{A}’s tail to B\mathbf{B}’s head, the head-to-tail or triangle method. Equivalently, bringing both tails to a common origin and completing a parallelogram gives the sum along its diagonal, the parallelogram method; the two methods always agree. Vector addition is commutative and associative:

A+B=B+A,(A+B)+C=A+(B+C)\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}, \qquad (\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})

A vector added to its own negative gives the null vector 0\mathbf{0}, which has zero magnitude and, since there’s nothing left to point anywhere, no defined direction. Subtraction is just addition of the negative: A−B=A+(−B)\mathbf{A}-\mathbf{B}=\mathbf{A}+(-\mathbf{B}).

3.3 Resolving a Vector into Components

Just as a vector can be built up from two others, it can be taken apart into two component vectors along any pair of non-parallel directions in the same plane. The most useful choice is a rectangular coordinate system with unit vectors i^\hat{\mathbf{i}} and j^\hat{\mathbf{j}} along the x- and y-axes, each of magnitude exactly 1 and carrying no dimension of its own, they only specify direction. Any vector A\mathbf{A} in the plane can then be written as:

A=Axi^+Ayj^,Ax=Acos⁡θ,    Ay=Asin⁡θ\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}, \qquad A_x = A\cos\theta,\;\; A_y = A\sin\theta

where θ\theta is the angle A\mathbf{A} makes with the x-axis. A component like AxA_x is a plain number, not a vector, it can come out positive, negative, or zero depending on θ\theta. Given the components instead of the magnitude and angle, the reverse conversion is just Pythagoras and inverse tangent:

A=Ax2+Ay2,θ=tan⁡−1 ⁣(AyAx)A = \sqrt{A_x^2+A_y^2}, \qquad \theta = \tan^{-1}\!\left(\dfrac{A_y}{A_x}\right)

3.4 Adding Vectors Analytically

The graphical method is intuitive but tedious and only ever as accurate as the drawing. Once every vector is written in component form, addition becomes ordinary arithmetic done twice, once per axis:

R=A+B    ⇒    Rx=Ax+Bx,Ry=Ay+By\mathbf{R}=\mathbf{A}+\mathbf{B} \;\;\Rightarrow\;\; R_x = A_x+B_x, \qquad R_y = A_y+B_y

The same component-by-component rule extends to any number of vectors, in three dimensions as easily as two, and to subtraction by simply flipping a sign before adding.

Worked example

Relative velocity by components

A cyclist moves at vc=6i^+8j^\mathbf{v}_c = 6\hat{\mathbf{i}} + 8\hat{\mathbf{j}} km/h relative to the ground, while the wind blows at vw=−2i^−3j^\mathbf{v}_w = -2\hat{\mathbf{i}} - 3\hat{\mathbf{j}} km/h relative to the ground. What is the cyclist’s velocity relative to the wind?

Relative velocity is a subtraction, done component by component:

vc/w=vc−vw=(6−(−2))i^+(8−(−3))j^=8i^+11j^ km/h\mathbf{v}_{c/w} = \mathbf{v}_c - \mathbf{v}_w = (6-(-2))\hat{\mathbf{i}} + (8-(-3))\hat{\mathbf{j}} = 8\hat{\mathbf{i}} + 11\hat{\mathbf{j}}\text{ km/h}

with magnitude 82+112≈13.6\sqrt{8^2+11^2} \approx 13.6 km/h. Take care to distinguish this from the resultant of two velocities acting on one object, v=v1+v2\mathbf{v}=\mathbf{v}_1+\mathbf{v}_2 (addition): relative velocity of object 1 with respect to object 2 is always v1−v2\mathbf{v}_1-\mathbf{v}_2, a subtraction, and the two are easy to mix up.

3.5 Position, Velocity and Acceleration in a Plane

Everything from one-dimensional kinematics carries over once position, velocity, and acceleration are written as vectors. The position vector is r=xi^+yj^\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}, and a displacement over some interval is Δr=Δx i^+Δy j^\Delta\mathbf{r} = \Delta x\,\hat{\mathbf{i}} + \Delta y\,\hat{\mathbf{j}}.

Average velocity is displacement divided by the time interval, vˉ=Δr/Δt\bar{\mathbf{v}} = \Delta\mathbf{r}/\Delta t, and points along Δr\Delta\mathbf{r}. Shrinking Δt\Delta t toward zero turns this into the instantaneous velocity:

v=drdt=vxi^+vyj^,vx=dxdt,    vy=dydt\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}}, \qquad v_x = \dfrac{dx}{dt},\;\; v_y = \dfrac{dy}{dt}

As the time interval shrinks, the displacement direction rotates smoothly until it lies exactly along the curve itself, so velocity at any point is always tangent to the path, pointing the way the object is actually moving. This is worth holding onto: in one dimension, velocity and acceleration are always along the same line, but for motion in a plane the angle between them can be anywhere from 0∘0^\circ to 180∘180^\circ.

Acceleration follows the identical pattern one derivative up:

a=dvdt=axi^+ayj^,ax=dvxdt,    ay=dvydt\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}}, \qquad a_x = \dfrac{dv_x}{dt},\;\; a_y = \dfrac{dv_y}{dt}

3.6 Motion with Constant Acceleration

When acceleration a\mathbf{a} stays constant, the same equations from straight-line motion apply, just written as vectors, and consequently apply separately and independently to each axis:

v=v0+atr=r0+v0t+12at2\begin{gathered} \mathbf{v} = \mathbf{v}_0 + \mathbf{a}t \\[6px] \mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2}\mathbf{a}t^2 \end{gathered}

In components, this is really two separate one-dimensional problems running at once:

vx=v0x+axt,x=x0+v0xt+12axt2vy=v0y+ayt,y=y0+v0yt+12ayt2\begin{gathered} v_x = v_{0x} + a_x t, \qquad x = x_0 + v_{0x}t + \tfrac{1}{2}a_x t^2 \\[6px] v_y = v_{0y} + a_y t, \qquad y = y_0 + v_{0y}t + \tfrac{1}{2}a_y t^2 \end{gathered}

That independence of the two axes is the single most useful idea in this chapter: motion in a plane is nothing but two ordinary one-dimensional motions, along two perpendicular directions, happening at the same time without interfering with each other. Projectile motion, next, is the clearest possible illustration.

3.7 Projectile Motion

An object launched into the air and then left to gravity alone, a ball, a shell, a shot put, is a projectile. Its motion is exactly the superposition described above: horizontally, nothing acts on it (ignoring air resistance), so that component of velocity never changes; vertically, gravity provides constant acceleration −g-g. It was Galileo who first argued, in 1632, that these two components could be treated as independent.

Launching at speed v0v_0 and angle θ0\theta_0 from the origin gives:

x(t)=(v0cos⁡θ0) t,y(t)=(v0sin⁡θ0) t−12gt2x(t) = (v_0\cos\theta_0)\,t, \qquad y(t) = (v_0\sin\theta_0)\,t - \tfrac{1}{2}gt^2

Eliminating tt between the two shows the path is a parabola, and standard bookkeeping on the same two equations gives the time of flight, maximum height, and horizontal range:

Tf=2v0sin⁡θ0g,hm=v02sin⁡2θ02gR=v02sin⁡2θ0g\begin{gathered} T_f = \dfrac{2v_0\sin\theta_0}{g}, \qquad h_m = \dfrac{v_0^2\sin^2\theta_0}{2g} \\[8px] R = \dfrac{v_0^2\sin 2\theta_0}{g} \end{gathered}

For a fixed launch speed, RR is largest when sin⁡2θ0\sin 2\theta_0 is largest, at θ0=45∘\theta_0=45^\circ. The full derivation, an interactive simulation, and exactly where the familiar “time up equals time down” and “45° is always best” shortcuts quietly assume equal launch and landing heights, live on the concept page.

Worked example

The angle where range equals maximum height

For what launch angle θ0\theta_0 does a projectile’s range exactly equal its maximum height?

Setting R=hmR = h_m and cancelling the common factor of v02/gv_0^2/g:

sin⁡2θ0=12sin⁡2θ0    ⇒    2sin⁡θ0cos⁡θ0=12sin⁡2θ0\sin 2\theta_0 = \tfrac{1}{2}\sin^2\theta_0 \;\;\Rightarrow\;\; 2\sin\theta_0\cos\theta_0 = \tfrac{1}{2}\sin^2\theta_0

Dividing both sides by sin⁡θ0\sin\theta_0 (valid since θ0≠0\theta_0 \ne 0) and rearranging:

4cos⁡θ0=sin⁡θ0    ⇒    tan⁡θ0=4    ⇒    θ0=tan⁡−1(4)≈76∘4\cos\theta_0 = \sin\theta_0 \;\;\Rightarrow\;\; \tan\theta_0 = 4 \;\;\Rightarrow\;\; \theta_0 = \tan^{-1}(4) \approx \textcolor{#e08a1e}{76^\circ}

A much steeper launch than the 45° that maximises range, which makes sense: reaching that height at all needs most of the velocity pointed upward.

3.8 Uniform Circular Motion

An object moving at constant speed around a circle is in uniform circular motion. “Uniform” describes the speed only, the velocity itself is constantly changing direction, tangent to the circle at every instant, and a changing velocity means the object is accelerating even though its speed never changes.

Comparing the velocity vectors at two nearby points and taking the limit as they merge shows that this acceleration always points toward the centre of the circle, which is why it is called centripetal (“centre-seeking”) acceleration, a term Newton proposed. Its magnitude works out to:

Centripetal acceleration
ac=v2R\textcolor{#e08a1e}{a_c} = \dfrac{v^2}{R}

A second, equivalent description uses the angular speed ω\omega, the rate at which the radius vector sweeps out angle, ω=Δθ/Δt\omega = \Delta\theta/\Delta t. Since the arc length covered is R ΔθR\,\Delta\theta, the linear speed is v=ωRv=\omega R, and substituting back:

ac=ω2Ra_c = \omega^2 R

In terms of the time period TT (one full revolution) or frequency ν=1/T\nu = 1/T: ω=2πν\omega = 2\pi\nu, v=2πRνv = 2\pi R\nu, and ac=4π2ν2Ra_c = 4\pi^2\nu^2 R. Because the direction of ac\mathbf{a}_c keeps changing even though its magnitude doesn’t, centripetal acceleration is not a constant vector, and the ordinary constant-acceleration equations from §3.6 do not apply to circular motion at all.

— From the NCERT exercises

A few of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 3.9

A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference (a quarter of the circle) to a point Q, and returns to the centre along QO. If the round trip takes 10 minutes, find (a) the net displacement, (b) the average velocity, and (c) the average speed of the cyclist.

Solution

(a) The cyclist starts and ends at the same point O, so the net displacement is zero.

(b) Average velocity is net displacement divided by time, and with zero displacement the average velocity is zero too, regardless of how far the cyclist actually rode.

(c) Average speed uses the total path length instead:

path=OP⏟1 km+14(2πR)⏟PQ,  ≈1.57 km+QO⏟1 km≈3.57 km\text{path} = \underbrace{OP}_{1\text{ km}} + \underbrace{\tfrac{1}{4}(2\pi R)}_{PQ,\;\approx 1.57\text{ km}} + \underbrace{QO}_{1\text{ km}} \approx 3.57\text{ km}average speed=3.57 km10/60 h≈21.4 km/h\text{average speed} = \dfrac{3.57\text{ km}}{10/60\text{ h}} \approx \textcolor{#e08a1e}{21.4\text{ km/h}}

A clean illustration of the chapter’s own warning: average speed and the magnitude of average velocity agree only when the path never doubles back on itself.

NCERT Exercise 3.11

A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 minutes. Find (a) the average speed of the taxi, (b) the magnitude of the average velocity. Are the two equal?

Solution

Both use the same time interval, 28 min = 7/15 h:

average speed=23 km7/15 h≈49.3 km/h\text{average speed} = \dfrac{23\text{ km}}{7/15\text{ h}} \approx \textcolor{#e08a1e}{49.3\text{ km/h}}∣average velocity∣=10 km7/15 h≈21.4 km/h|\text{average velocity}| = \dfrac{10\text{ km}}{7/15\text{ h}} \approx \textcolor{#e08a1e}{21.4\text{ km/h}}

Not equal, and by a wide margin. The gap between them is exactly the padding the dishonest route added: 23 km travelled to cover what is, in a straight line, only 10 km.

NCERT Exercise 3.14

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?

Solution

The angular speed comes first:

ω=2π×1425≈3.52 rad/s\omega = \dfrac{2\pi \times 14}{25} \approx 3.52\text{ rad/s}

and then the centripetal acceleration, using §3.8’s angular form:

ac=ω2R=(3.52)2×0.80≈9.9 m/s2, directed toward the centre of the circle at every instanta_c = \omega^2 R = (3.52)^2 \times 0.80 \approx \textcolor{#e08a1e}{9.9\text{ m/s}^2}\text{, directed toward the centre of the circle at every instant}