Chapter 3: Motion in a Plane (Kinematics)

Quantities with only magnitude are scalars; those that also need a direction are vectors.

Scalars: volume, mass, speed, density, number of moles, angular frequency. Vectors: acceleration, velocity, displacement, angular velocity.

Work is a dot product ($\vec{F}\cdot \vec{d}$), a scalar. Current has a sense along the wire but does not add by the vector law, so it is a scalar.

Work and current

Impulse $=\vec{F}\Delta t=\Delta \vec{p}$, which has direction.

Impulse

(a) Not meaningful in general: only scalars of the same dimensions can be added (mass + time is meaningless).

(b) Not meaningful: a scalar and a vector cannot be added.

(c) Meaningful: e.g. $m\vec{a}=\vec{F}$; the product is a vector.

(d) Meaningful: e.g. power × time = energy; scalars of any dimensions can be multiplied.

(e) Not meaningful in general: only vectors of the same dimensions can be added (force + velocity is meaningless).

(f) A component written as a vector (say $A_x\hat{i}$) has the same dimensions as $\vec{A}$, so the sum can be formed; but NCERT counts only the operations that are meaningful for any quantities.

Only (c) and (d) are meaningful in general (NCERT answer).

(a) True. A magnitude is a pure number with units.

(b) False. A component such as $A_x\hat{i}$ is itself a vector along an axis.

(c) False. Path length ≥ |displacement|; they are equal only for motion in a straight line without turning back.

(d) True. Since path length ≥ |displacement|, dividing by the same time gives average speed ≥ |average velocity|.

(e) True. For three vectors to add to zero, they must form a closed triangle, which lies in one plane.

(a) T (b) F (c) F (d) T (e) T

Represent $\vec{a}$ and $\vec{b}$ as two sides of a triangle taken in order; then $\vec{a}+\vec{b}$ is the third side. In any triangle, one side is less than the sum of the other two and greater than their difference:

(a) $|\vec{a}+\vec{b}|<|\vec{a}|+|\vec{b}|$ (b) $|\vec{a}+\vec{b}|>||\vec{a}|-|\vec{b}||$.

Replacing $\vec{b}$ by $-\vec{b}$ (same magnitude) gives (c) $|\vec{a}-\vec{b}|<|\vec{a}|+|\vec{b}|$ and (d) $|\vec{a}-\vec{b}|>||\vec{a}|-|\vec{b}||$.

Algebraically: $|\vec{a}+\vec{b}{|}^2=a^2+b^2+2ab\cos \theta$, and $-1\le \cos \theta \le 1$ gives $(a-b{)}^2\le |\vec{a}+\vec{b}{|}^2\le (a+b{)}^2$.

The equality holds when the triangle collapses, i.e. when $\vec{a}$ and $\vec{b}$ are collinear (θ = 0° or 180°).

Triangle inequalities; equality only when $\vec{a}$ and $\vec{b}$ are collinear.

(a) Incorrect. Four non-zero vectors forming a closed quadrilateral add to zero.

(b) Correct. $\vec{a}+\vec{c}=-(\vec{b}+\vec{d})$, so their magnitudes are equal.

(c) Correct. $\vec{a}=-(\vec{b}+\vec{c}+\vec{d})$, so $|\vec{a}|=|\vec{b}+\vec{c}+\vec{d}|\le |\vec{b}|+|\vec{c}|+|\vec{d}|$.

(d) Correct. $\vec{b}+\vec{c}=-(\vec{a}+\vec{d})$, which lies in the plane of $\vec{a}$ and $\vec{d}$ (or along their line if they are collinear).

(b), (c) and (d) are correct; (a) is not.

Displacement depends only on the initial and final points. For all three it is the diameter PQ $=2\times 200=400$ m.

Girl B skates along the diameter itself, so her path length equals the displacement.

400 m for each; for girl B

(a) He returns to O, so the net displacement is zero.

(b) Average velocity = displacement ÷ time = 0.

(c) Path $=OP+\text{arc }PQ+QO=1+\frac{2\pi (1)}{4}+1=2+1.571=3.571$ km. Time $=10$ min $=\frac{1}{6}$ h.

Average speed $=\frac{3.571}{1/6}=21.4$ km h⁻¹.

(a) 0 (b) 0 (c) 21.4 km h⁻¹

Turning 60° after each 500 m makes him trace a regular hexagon of side 500 m. Take the first leg along +x; the legs point at 0°, 60°, 120°, 180°, … .

Third turn (3 legs): $\vec{d}=500[(1+\cos 60^{\circ }+\cos 120^{\circ })\hat{i}+(\sin 60^{\circ }+\sin 120^{\circ })\hat{j}]=500(\hat{i}+\sqrt{3}\hat{j})$. Magnitude $=500\times 2=1000$ m at 60° to the initial direction. Path length $=1500$ m.

Sixth turn (6 legs): he is back at the start: displacement zero. Path length $=3000$ m.

Eighth turn (8 legs = one full hexagon + 2 legs): $\vec{d}=500[(1+\cos 60^{\circ })\hat{i}+\sin 60^{\circ }\hat{j}]=500(1.5\hat{i}+0.866\hat{j})$. Magnitude $=500\sqrt{3}=866$ m at 30° to the initial direction. Path length $=4000$ m.

3rd: 1 km at 60° (path 1.5 km); 6th: zero (path 3 km); 8th: 866 m at 30° (path 4 km)

Time $=\frac{28}{60}$ h.

(a) Average speed $=\frac{23}{28/60}=49.3$ km h⁻¹. (b) |Average velocity| $=\frac{10}{28/60}=21.4$ km h⁻¹.

They are not equal: they are equal only when the path is a straight line in one direction.

(a) 49.3 km h⁻¹ (b) 21.4 km h⁻¹; not equal

The ball just touches the ceiling when $H=25$ m: $\frac{u^2{\sin }^2\theta }{2g}=25\Rightarrow {\sin }^2\theta =\frac{2\times 9.8\times 25}{1600}=0.306$, so $\sin \theta =0.553$, $\theta =33.6^{\circ }$.

$R=\frac{u^2\sin 2\theta }{g}=\frac{1600\times \sin 67.2^{\circ }}{9.8}=\frac{1600\times 0.922}{9.8}$

150.5 m

$R_{\max }=\frac{u^2}{g}=100$ m. Thrown vertically up with the same speed, the height is $H=\frac{u^2}{2g}=\frac{100}{2}$.

50 m

$\omega =2\pi \nu =2\pi \times \frac{14}{25}=3.52$ rad s⁻¹, $r=0.80$ m.

$a={\omega }^{2}r=(3.52{)}^2\times 0.80=9.91{\text{m s}}^{-2}$, directed along the string towards the centre.

9.9 m s⁻², towards the centre along the radius

$v=900\times \frac{5}{18}=250{\text{m s}}^{-1}$, $r=1000$ m.

$a_c=\frac{v^2}{r}=\frac{62500}{1000}=62.5{\text{m s}}^{-2}$; $\frac{a_c}{g}=\frac{62.5}{9.8}=6.4$.

$a_c=6.4g$

(a) False. It is true only for uniform circular motion; if the speed changes, there is also a tangential component.

(b) True. Velocity is always along the tangent to the path.

(c) True. Over a full cycle the velocity returns to its initial value, so the average acceleration $\frac{\Delta \vec{v}}{T}$ is zero.

(a) False (b) True (c) True

(a) $\vec{v}=\frac{d\vec{r}}{dt}=(3.0\hat{i}-4.0t\hat{j})$ m s⁻¹; $\vec{a}=\frac{d\vec{v}}{dt}=-4.0\hat{j}$ m s⁻².

(b) At $t=2.0$ s: $\vec{v}=3.0\hat{i}-8.0\hat{j}$, $|\vec{v}|=\sqrt{9+64}=\sqrt{73}=8.54$ m s⁻¹.

Direction: $\theta ={\tan }^{-1}\frac{8.0}{3.0}=69.4^{\circ }\approx 70^{\circ }$ below the x-axis.

(a) v = 3.0 î − 4.0t ĵ, a = −4.0 ĵ (b) 8.54 m s⁻¹, about 70° below the x-axis

$x=\frac{1}{2}(8.0)t^2=4t^2$, $y=10t+\frac{1}{2}(2.0)t^2=10t+t^2$.

(a) $4t^2=16\Rightarrow t=2$ s; $y=20+4=24$ m.

(b) $\vec{v}=8t\hat{i}+(10+2t)\hat{j}=16\hat{i}+14\hat{j}$ at t = 2 s; speed $=\sqrt{256+196}=\sqrt{452}$.

(a) 2 s, 24 m (b) 21.26 m s⁻¹

$|\hat{i}+\hat{j}|=\sqrt{2}$, at 45° to the x-axis; $|\hat{i}-\hat{j}|=\sqrt{2}$, at −45° to the x-axis.

Component of $\vec{A}$ along a unit vector $\hat{n}$ is $\vec{A}\cdot \hat{n}$:

along $\frac{\hat{i}+\hat{j}}{\sqrt{2}}$: $\frac{2+3}{\sqrt{2}}=\frac{5}{\sqrt{2}}$; along $\frac{\hat{i}-\hat{j}}{\sqrt{2}}$: $\frac{2-3}{\sqrt{2}}=-\frac{1}{\sqrt{2}}$.

√2 at 45°; √2 at −45°; components 5/√2 and −1/√2

(b) and (e) are the definitions of average velocity and average acceleration, so they always hold. (a), (c) and (d) hold only for uniform acceleration, not for arbitrary motion.

(b) and (e)

(a) False: kinetic energy (a scalar) is not conserved in an inelastic collision.

(b) False: temperature in °C or potential energy can be negative.

(c) False: mass and time are scalars with dimensions.

(d) False: temperature and gravitational potential vary from point to point.

(e) True: a scalar does not change when the axes are rotated.

Only (e) is true.

Take the observer O directly below the midpoint of the two positions A and B. Each half of AB subtends 15° at O at height 3400 m:

$AB=2\times 3400\tan 15^{\circ }=2\times 3400\times 0.2679=1822$ m.

Speed $=\frac{1822}{10.0}$

182 m s⁻¹