Physics - Mechanics and Properties of Matter - How to Solve: Motion in a Plane (Projectile, Horizontal/Vertical Circles, River-Boat) – NEET UG Physics Guide

How to Solve: Motion in a Plane (Projectile, Horizontal/Vertical Circles, River-Boat) – NEET UG Physics Guide

Introduction

Mastering Motion in a Plane unlocks 5-7 direct NEET questions (18-25 marks) every year—enough to boost your rank by 10,000+ places. Whether it’s a projectile clearing a wall, a car on a banked curve, or a boat crossing a river, these problems test vector resolution, time of flight, and circular motion—skills that also apply to chemistry (molecular motion) and biology (blood flow in arteries).

WHAT YOU NEED TO KNOW FIRST

1. Vector Resolution: Splitting a vector into horizontal (x) and vertical (y) components.
2. Equations of Motion: For constant acceleration (given on NEET sheet).
3. Circular Motion Basics: Centripetal force, angular velocity, and radius relationships.

KEY TERMS & FORMULAS

1. Projectile Motion

Key Terms: - Range (R): Horizontal distance covered. - Time of Flight (T): Total time in air. - Maximum Height (H): Peak vertical displacement.

Formulas:
1. Time of Flight (T) [ T = \frac{2u \sin \theta}{g} ] - (u) = initial velocity - (\theta) = launch angle - (g) = acceleration due to gravity (MEMORISE THIS)

1. Range (R) [ R = \frac{u^2 \sin 2\theta}{g} ] (MEMORISE THIS – Maximum range at (\theta = 45^\circ))

2. Maximum Height (H) [ H = \frac{u^2 \sin^2 \theta}{2g} ] (MEMORISE THIS)

3. Equation of Trajectory [ y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta} ] (Given on NEET sheet – use for "height at a given distance" problems)

2. Horizontal Circular Motion (e.g., Car on a Curve)

Key Terms: - Centripetal Force (F_c): Inward force keeping object in circle. - Banking Angle ((\theta)): Angle of road tilt to prevent skidding.

Formulas:
1. Centripetal Force [ F_c = \frac{mv^2}{r} ] - (m) = mass - (v) = velocity - (r) = radius (MEMORISE THIS)

1. Maximum Speed Without Skidding (Unbanked Curve) [ v_{\text{max}} = \sqrt{\mu r g} ]

2. (\mu) = coefficient of friction (MEMORISE THIS)

3. Banking Angle (No Friction) [ \tan \theta = \frac{v^2}{r g} ] (MEMORISE THIS)

3. Vertical Circular Motion (e.g., Roller Coaster Loop)

Key Terms: - Critical Velocity (v_c): Minimum speed at top to complete loop. - Tension (T): Force in string/rope.

Formulas:
1. Critical Velocity at Top [ v_c = \sqrt{r g} ] (MEMORISE THIS)

1. Tension at Bottom [ T_{\text{bottom}} = \frac{mv^2}{r} + m g ] (MEMORISE THIS)

2. Tension at Top [ T_{\text{top}} = \frac{mv^2}{r} - m g ] (MEMORISE THIS)

4. River-Boat Problems (Relative Motion)

Key Terms: - Resultant Velocity (v_r): Boat’s velocity relative to ground. - Drift (x): Horizontal displacement due to river flow.

Formulas:
1. Resultant Velocity [ v_r = \sqrt{v_b^2 + v_r^2 + 2 v_b v_r \cos \theta} ] - (v_b) = boat velocity relative to water - (v_r) = river velocity - (\theta) = angle between boat and river flow (MEMORISE THIS)

1. Time to Cross River [ t = \frac{d}{v_b \sin \theta} ]

2. (d) = width of river (MEMORISE THIS)

3. Drift (x) [ x = (v_r + v_b \cos \theta) \times t ] (MEMORISE THIS)

STEP-BY-STEP METHOD

For Projectile Motion:

1. Resolve Initial Velocity into (u_x = u \cos \theta) and (u_y = u \sin \theta).
2. Time of Flight (T): Use (T = \frac{2u_y}{g}).
3. Range (R): Use (R = u_x \times T).
4. Maximum Height (H): Use (H = \frac{u_y^2}{2g}).
5. Trajectory Equation: If asked for height at a distance, plug (x) into (y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}).

For Horizontal Circular Motion:

1. Identify Forces: Centripetal force = friction or normal reaction.
2. Write Force Equation: (\frac{mv^2}{r} = \text{Net inward force}).
3. Solve for Unknown: Speed, radius, or banking angle.

For Vertical Circular Motion:

1. At Top: (T + mg = \frac{mv^2}{r}) (if tension is involved).
2. At Bottom: (T - mg = \frac{mv^2}{r}).
3. Critical Velocity: (v_{\text{min}} = \sqrt{r g}) at top.

For River-Boat Problems:

1. Resolve Boat Velocity: (v_b) relative to water.
2. Resultant Velocity: Use vector addition.
3. Time to Cross: (t = \frac{d}{v_b \sin \theta}).
4. Drift: (x = v_r \times t).

WORKED EXAMPLES

Example 1 – Basic Projectile

Question: A ball is thrown at (20 \, \text{m/s}) at (30^\circ). Find: a) Time of flight b) Range c) Maximum height

Solution:
1. Resolve Velocity: - (u_x = 20 \cos 30^\circ = 20 \times 0.866 = 17.32 \, \text{m/s}) - (u_y = 20 \sin 30^\circ = 20 \times 0.5 = 10 \, \text{m/s})

1. Time of Flight: [ T = \frac{2u_y}{g} = \frac{2 \times 10}{9.8} = 2.04 \, \text{s} ]

2. Range: [ R = u_x \times T = 17.32 \times 2.04 = 35.33 \, \text{m} ]

3. Maximum Height: [ H = \frac{u_y^2}{2g} = \frac{10^2}{2 \times 9.8} = 5.1 \, \text{m} ]

What we did and why: - Split velocity into x and y components to separate motion. - Used time of flight formula for vertical motion. - Calculated range using horizontal velocity × time. - Found max height using vertical motion equations.

Example 2 – Medium (Banked Curve)

Question: A car takes a turn of radius (50 \, \text{m}) at (15 \, \text{m/s}). If the road is banked at (30^\circ), find the coefficient of friction needed to prevent skidding.

Solution:
1. Forces Involved: - Normal force ((N)) perpendicular to road. - Friction ((f = \mu N)) parallel to road.

1. Resolve Forces:
2. Vertical: (N \cos \theta = mg + f \sin \theta)

3. Horizontal: (N \sin \theta + f \cos \theta = \frac{mv^2}{r})

4. Substitute (f = \mu N): [ N \cos \theta = mg + \mu N \sin \theta \quad \text{(1)} ] [ N \sin \theta + \mu N \cos \theta = \frac{mv^2}{r} \quad \text{(2)} ]

5. Divide (2) by (1): [ \frac{\sin \theta + \mu \cos \theta}{\cos \theta - \mu \sin \theta} = \frac{v^2}{r g} ] [ \frac{0.5 + \mu \times 0.866}{0.866 - \mu \times 0.5} = \frac{15^2}{50 \times 9.8} = 0.459 ]

6. Solve for (\mu): [ 0.5 + 0.866 \mu = 0.459 (0.866 - 0.5 \mu) ] [ 0.5 + 0.866 \mu = 0.397 - 0.23 \mu ] [ 1.096 \mu = -0.103 \quad \text{(No solution? Check signs!)} ]

Correction: Friction acts down the slope (opposing motion). - Equation (1): (N \cos \theta + \mu N \sin \theta = mg) - Equation (2): (N \sin \theta - \mu N \cos \theta = \frac{mv^2}{r})

Now solve: [ \frac{\sin \theta - \mu \cos \theta}{\cos \theta + \mu \sin \theta} = \frac{v^2}{r g} ] [ \frac{0.5 - 0.866 \mu}{0.866 + 0.5 \mu} = 0.459 ] [ 0.5 - 0.866 \mu = 0.459 (0.866 + 0.5 \mu) ] [ 0.5 - 0.866 \mu = 0.397 + 0.23 \mu ] [ 0.103 = 1.096 \mu ] [ \mu = 0.094 ]

What we did and why: - Drew free-body diagram to identify forces. - Resolved forces parallel and perpendicular to the slope. - Used centripetal force equation for circular motion. - Corrected friction direction (common mistake!).

Example 3 – Exam-Style (River-Boat with Angle)

Question: A boat crosses a (100 \, \text{m}) wide river flowing at (3 \, \text{m/s}). The boat’s speed relative to water is (5 \, \text{m/s}) at (60^\circ) upstream. Find: a) Time to cross b) Drift downstream

Solution:
1. Resolve Boat Velocity: - (v_{bx} = 5 \cos 60^\circ = 2.5 \, \text{m/s}) (upstream) - (v_{by} = 5 \sin 60^\circ = 4.33 \, \text{m/s}) (across river)

1. Time to Cross: [ t = \frac{d}{v_{by}} = \frac{100}{4.33} = 23.09 \, \text{s} ]

2. Drift Downstream:

3. Net horizontal velocity = (v_r - v_{bx} = 3 - 2.5 = 0.5 \, \text{m/s}) [ x = 0.5 \times 23.09 = 11.55 \, \text{m} ]

What we did and why: - Split boat velocity into x and y components. - Used perpendicular component to find crossing time. - Calculated drift using net horizontal velocity.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for Motion in a Plane!

1. Projectile Motion:
2. Split velocity into x and y.
3. Time of flight: (T = \frac{2u \sin \theta}{g}).
4. Range: (R = \frac{u^2 \sin 2\theta}{g}).

5. Max height: (H = \frac{u^2 \sin^2 \theta}{2g}).

6. Circular Motion:

7. Centripetal force: (\frac{mv^2}{r}).
8. Banked curve: (\tan \theta = \frac{v^2}{r g}).

9. Vertical loop: Critical speed at top = (\sqrt{r g}).

10. River-Boat:

11. Time to cross: (t = \frac{d}{v_b \sin \theta}).
12. Drift: (x = v_r \times t).

Pro Tips: - Always draw a diagram. - Check units (m/s, m, s). - For river problems, resolve boat velocity relative to water. - NEET loves projectile from height—use (y = u_y t - \frac{1}{2} g t^2).

You’ve got this! Now go crush those 5-7 questions tomorrow!