Motion in 2D

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Motion in 2D — Key Facts Projectile motion: motion under gravity with constant horizontal velocity Range $R = \frac{u^2 \sin 2\theta}{g}$; Maximum height $H = \frac{u^2 \sin^2\theta}{2g}$; Time of flight $T = \frac{2u\sin\theta}{g}$ Horizontal range is maximum at $\theta = 45°$; same range at complementary angles ($\theta$ and $90°-\theta$) ⚡ Exam tip: At the highest point, velocity is purely horizontal ($v_y = 0$), but acceleration remains $g$ downward

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of 2D motion kinematics.

Motion in 2D — CUET Physics Study Guide

Motion in two dimensions involves vectors in perpendicular directions (typically $x$ and $y$) that can be analysed independently using the principle of superposition. The horizontal and vertical components of motion in projectile motion are completely independent — the horizontal motion has zero acceleration (ignoring air resistance), while the vertical motion has constant acceleration $g$ downward.

Trajectory Equation: Eliminating time $t$ from $x = u\cos\theta \cdot t$ and $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ gives: $$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$$

This is the equation of a parabola. The shape depends on $u$ and $\theta$. For a given initial speed $u$, the maximum range occurs at $\theta = 45°$, giving $R_{\max} = u^2/g$.

Relative Motion in 2D: The velocity of A relative to B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. A classic problem is a boat crossing a river: if the river flows with speed $v_r$ and the boat’s speed in still water is $v_b$ perpendicular to the current, the actual path is diagonal. Time to cross $= \frac{d}{v_b}$ (where $d$ is river width), while downstream drift $= v_r \times \frac{d}{v_b}$.

Circular Motion Basics: Angular displacement $\theta$ (radians), angular velocity $\omega = \frac{d\theta}{dt}$, tangential speed $v = \omega r$, centripetal acceleration $a_c = \frac{v^2}{r} = \omega^2 r$ directed toward the centre.

Example: A ball thrown at 20 m/s at 37° from horizontal. Find range and max height.

• $u_x = 20 \cos 37° = 20 \times 0.8 = 16$ m/s; $u_y = 20 \sin 37° = 20 \times 0.6 = 12$ m/s
• Time of flight $T = \frac{2 \times 12}{10} = 2.4$ s; Range $R = 16 \times 2.4 = 38.4$ m
• Max height $H = \frac{12^2}{20} = 7.2$ m

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious preparation with derivations and applications.

Motion in 2D — Complete CUET Physics Notes

Projectile Motion from a Height: When a projectile is launched from height $h$ above ground (instead of ground level), the equations modify. Time to hit ground: solve $h + u\sin\theta \cdot t - \frac{1}{2}gt^2 = 0$. The range depends on whether the projectile lands above or below launch point. If launched downward from height $h$ with speed $u$ at angle $\theta$, the range is reduced compared to ground-level launch.

Projectile Motion on an Inclined Plane: When the target plane is inclined at angle $\phi$ to the horizontal, resolve axes along and perpendicular to the plane. The range along the inclined plane differs from the horizontal range formula. For a projectile launched up an incline, both the angle of projection and incline angle affect the range.

Non-Uniform Circular Motion: When tangential acceleration exists (e.g., a car speeding up on a circular track), total acceleration is the vector sum of centripetal acceleration $a_c = v^2/r$ radially inward and tangential acceleration $a_t = \frac{dv}{dt}$ along the tangent. The magnitude is $a = \sqrt{a_c^2 + a_t^2}$.

Banking of Roads: For a vehicle on a circular road of radius $r$ banked at angle $\theta$ without friction, $\tan\theta = \frac{v^2}{rg}$. This allows safe turning at speed $v$ without relying on friction. The ideal speed is $v = \sqrt{rg\tan\theta}$. On level roads, friction provides the centripetal force: $f = \frac{mv^2}{r}$.

Conical Pendulum: A mass suspended by a string swinging in a horizontal circle, with the string making constant angle $\theta$ with the vertical. Tension $T = \frac{mg}{\cos\theta}$, and the horizontal component provides centripetal force: $T\sin\theta = \frac{mv^2}{r}$. Time period $T = 2\pi\sqrt{\frac{l\cos\theta}{g}}$.

Relative Velocity in 2D: The general formula $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ applies. For rainfall problems: if rain appears vertical to a moving observer, $\vec{v}_r - \vec{v}_o = 0$ for no deflection. The angle of rain apparent velocity gives the solution.

CUET Exam Patterns (2022–2024):

• River-boat problems appeared in CUET 2023 (1 mark)
• Banking of roads is frequently tested in Section B
• Projectile from height combined with horizontal launch (2 marks)
• Conical pendulum not directly tested but related concepts appear
• Common mistakes: using $g = 10$ m/s² inconsistently, wrong sign for $g$, mixing up components

⚡ Key insight: In projectile problems, always resolve initial velocity into components first. Then treat horizontal ($a=0$) and vertical ($a=-g$) independently. Time is the common link between $x$ and $y$ equations.

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