IB Physics How to Solve: IB Physics – Mechanics (Projectile, Circular, Energy, Momentum, Collisions)

How to Solve: IB Physics – Mechanics (Projectile, Circular, Energy, Momentum, Collisions)

Complete Guide

Introduction

"Mastering IB Physics Mechanics unlocks 20–25% of your Paper 1 and Paper 2 marks—enough to boost your grade by a full level. Whether it’s calculating the range of a projectile, the tension in a rollercoaster loop, or the velocity after a collision, these skills apply to real-world problems like rocket launches, car crashes, and even sports physics."

WHAT YOU NEED TO KNOW FIRST

1. Kinematic equations (for projectile motion)
2. Newton’s laws of motion (for circular motion and collisions)
3. Conservation principles (energy and momentum)

KEY TERMS & FORMULAS

1. Projectile Motion

Key Terms: - Range (R): Horizontal distance traveled - Time of flight (t): Total time in the air - Maximum height (H): Peak vertical displacement

Formulas: 1. Time of flight (symmetrical launch & land heights)
[
t = \frac{2u \sin \theta}{g}
]
- ( u ) = initial velocity (m/s)
- ( \theta ) = launch angle (°)
- ( g ) = acceleration due to gravity (9.81 m/s²) → MEMORISE THIS

1. Range (R)
   [
   R = \frac{u^2 \sin 2\theta}{g}
   ]
   → MEMORISE THIS

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

3. Horizontal velocity (constant)
   [
   v_x = u \cos \theta
   ]
   → MEMORISE THIS

4. Vertical velocity at time t
   [
   v_y = u \sin \theta - gt
   ]
   → MEMORISE THIS

2. Circular Motion

Key Terms: - Centripetal force (F_c): Net force toward the center - Centripetal acceleration (a_c): Acceleration toward the center - Angular velocity (ω): Radians per second

Formulas: 1. Centripetal force
[
F_c = \frac{mv^2}{r} = m \omega^2 r
]
- ( m ) = mass (kg)
- ( v ) = linear velocity (m/s)
- ( r ) = radius (m)
- ( \omega ) = angular velocity (rad/s) → MEMORISE THIS

1. Centripetal acceleration
   [
   a_c = \frac{v^2}{r} = \omega^2 r
   ]
   → MEMORISE THIS

2. Angular velocity
   [
   \omega = \frac{2\pi}{T} = 2\pi f
   ]

3. ( T ) = period (s)
4. ( f ) = frequency (Hz) → MEMORISE THIS

3. Energy (Work, Power, Conservation)

Key Terms: - Work (W): Energy transferred by a force - Kinetic energy (KE): Energy of motion - Potential energy (PE): Stored energy (gravitational, elastic) - Power (P): Rate of energy transfer

Formulas: 1. Work done by a constant force
[
W = F \cdot d \cdot \cos \theta
]
- ( F ) = force (N)
- ( d ) = displacement (m)
- ( \theta ) = angle between force and displacement → MEMORISE THIS

1. Kinetic energy
   [
   KE = \frac{1}{2} m v^2
   ]
   → MEMORISE THIS

2. Gravitational potential energy (near Earth’s surface)
   [
   PE = mgh
   ]
   → MEMORISE THIS

3. Elastic potential energy (spring)
   [
   PE_{elastic} = \frac{1}{2} k x^2
   ]

4. ( k ) = spring constant (N/m)

5. ( x ) = extension/compression (m) → MEMORISE THIS

6. Power
   [
   P = \frac{W}{t} = F \cdot v
   ]
   → MEMORISE THIS

7. Conservation of mechanical energy (no friction)
   [
   KE_i + PE_i = KE_f + PE_f
   ]
   → MEMORISE THIS

4. Momentum & Collisions

Key Terms: - Momentum (p): Mass × velocity - Impulse (J): Change in momentum - Elastic collision: KE conserved - Inelastic collision: KE not conserved (objects stick together)

Formulas: 1. Momentum
[
p = mv
]
→ MEMORISE THIS

1. Impulse
   [
   J = \Delta p = F \Delta t
   ]
   → MEMORISE THIS

2. Conservation of momentum (closed system)
   [
   m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
   ]
   → MEMORISE THIS

3. Elastic collision (1D, final velocities)
   [
   v_1 = \frac{(m_1 - m_2) u_1 + 2 m_2 u_2}{m_1 + m_2}
   ]
   [
   v_2 = \frac{(m_2 - m_1) u_2 + 2 m_1 u_1}{m_1 + m_2}
   ]
   → Given on exam sheet (but know how to use it)

4. Perfectly inelastic collision (objects stick together)
   [
   m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f
   ]
   → MEMORISE THIS

STEP-BY-STEP METHOD

1. Projectile Motion (Range, Time, Height)

Steps: 1. Resolve initial velocity into horizontal (( v_x = u \cos \theta )) and vertical (( v_y = u \sin \theta )) components. 2. Time of flight: Use ( t = \frac{2 u \sin \theta}{g} ) (if launch and land heights are equal). 3. Range: Use ( R = \frac{u^2 \sin 2\theta}{g} ). 4. Maximum height: Use ( H = \frac{u^2 \sin^2 \theta}{2g} ). 5. Check units: Ensure ( u ) is in m/s, ( g = 9.81 ) m/s².

2. Circular Motion (Centripetal Force, Acceleration)

Steps: 1. Identify the force causing circular motion (e.g., tension, friction, gravity). 2. Write ( F_c = \frac{mv^2}{r} ) and set it equal to the net force. 3. Solve for the unknown (e.g., velocity, radius, mass). 4. If angular velocity is given, use ( F_c = m \omega^2 r ). 5. Check direction: Centripetal force always points toward the center.

3. Energy Problems (Conservation, Work, Power)

Steps: 1. Identify initial and final states (e.g., top of a hill vs. bottom). 2. Write conservation equation: ( KE_i + PE_i = KE_f + PE_f ). 3. Substitute formulas (( \frac{1}{2} m v^2 ), ( mgh ), ( \frac{1}{2} k x^2 )). 4. Solve for the unknown (e.g., velocity, height, spring constant). 5. If work is done by a non-conservative force (e.g., friction), add ( W_{nc} ) to the equation.

4. Momentum & Collisions

Steps: 1. Draw a before/after diagram (label masses and velocities). 2. Write conservation of momentum equation: ( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 ). 3. If elastic collision, use the given formulas for ( v_1 ) and ( v_2 ). 4. If inelastic collision, objects stick together: ( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f ). 5. Solve for the unknown velocity. 6. Check if KE is conserved (only for elastic collisions).

WORKED EXAMPLES

Example 1 – Basic Projectile Motion

Question: A ball is kicked at 20 m/s at 30° above the horizontal. Find: a) Time of flight b) Range c) Maximum height

Solution: 1. Resolve velocity:
- ( v_x = 20 \cos 30° = 17.3 ) m/s
- ( v_y = 20 \sin 30° = 10 ) m/s

1. Time of flight:
   [
   t = \frac{2 \times 10}{9.81} = 2.04 \text{ s}
   ]

2. Range:
   [
   R = \frac{20^2 \sin 60°}{9.81} = \frac{400 \times 0.866}{9.81} = 35.3 \text{ m}
   ]

3. Maximum height:
   [
   H = \frac{10^2}{2 \times 9.81} = 5.1 \text{ m}
   ]

What we did and why: - Split velocity into components to separate horizontal (constant) and vertical (accelerated) motion. - Used time of flight formula for symmetrical launch/land heights. - Applied range and height formulas directly.

Example 2 – Medium Circular Motion

Question: A 1.2 kg ball is attached to a 0.8 m string and swung in a vertical circle at 4 m/s. Find the tension at the bottom of the circle.

Solution: 1. Forces at the bottom: Tension (up) and weight (down). 2. Net force = centripetal force:
[
F_c = T - mg = \frac{mv^2}{r}
] 3. Substitute values:
[
T - (1.2 \times 9.81) = \frac{1.2 \times 4^2}{0.8}
]
[
T - 11.77 = 24
]
[
T = 35.8 \text{ N}
]

What we did and why: - Identified that tension and weight both act at the bottom. - Set net force equal to centripetal force. - Solved for tension, ensuring units were consistent.

Example 3 – Exam-Style Energy Problem

Question: A 500 g block slides down a 2 m frictionless ramp inclined at 30°. What is its speed at the bottom?

Solution: 1. Initial energy: Only PE at the top.
[
PE_i = mgh = 0.5 \times 9.81 \times (2 \sin 30°) = 4.905 \text{ J}
] 2. Final energy: Only KE at the bottom.
[
KE_f = \frac{1}{2} m v^2
] 3. Conservation of energy:
[
PE_i = KE_f \implies 4.905 = \frac{1}{2} \times 0.5 \times v^2
] 4. Solve for ( v ):
[
v^2 = \frac{4.905 \times 2}{0.5} = 19.62
]
[
v = 4.43 \text{ m/s}
]

What we did and why: - Used ( h = d \sin \theta ) to find vertical height. - Applied conservation of energy (no friction). - Solved for velocity using KE formula.

Example 4 – Exam-Style Collision

Question: A 3 kg cart moving at 4 m/s collides with a stationary 2 kg cart. After the collision, the 3 kg cart moves at 2 m/s. Is the collision elastic?

Solution: 1. Conservation of momentum:
[
(3 \times 4) + (2 \times 0) = (3 \times 2) + (2 \times v_2)
]
[
12 = 6 + 2 v_2 \implies v_2 = 3 \text{ m/s}
] 2. Check KE before and after:
- Before: ( KE_i = \frac{1}{2} \times 3 \times 4^2 = 24 ) J
- After: ( KE_f = \frac{1}{2} \times 3 \times 2^2 + \frac{1}{2} \times 2 \times 3^2 = 6 + 9 = 15 ) J 3. KE not conserved → inelastic collision.

What we did and why: - Used momentum conservation to find the second velocity. - Compared initial and final KE to determine collision type.

COMMON MISTAKES

1. MISTAKE: Forgetting to resolve velocity into components in projectile motion.
   WHY IT HAPPENS: Students treat velocity as a single vector.
   CORRECT APPROACH: Always split into ( v_x ) and ( v_y ).

2. MISTAKE: Using ( F_c = \frac{mv}{r} ) instead of ( \frac{mv^2}{r} ).
   WHY IT HAPPENS: Confusing centripetal force with linear momentum.
   CORRECT APPROACH: Remember ( F_c ) depends on ( v^2 ).

3. MISTAKE: Ignoring the direction of forces in circular motion.
   WHY IT HAPPENS: Not drawing free-body diagrams.
   CORRECT APPROACH: Always draw forces and label directions.

4. MISTAKE: Assuming all collisions are elastic.
   WHY IT HAPPENS: Overlooking KE conservation.
   CORRECT APPROACH: Check if KE is conserved before using elastic formulas.

5. MISTAKE: Mixing up ( g = 9.81 ) m/s² with 10 m/s².
   WHY IT HAPPENS: Using approximate values when exact is needed.
   CORRECT APPROACH: Use 9.81 unless the question specifies 10.

EXAM TRAPS

1. TRAP: Projectile motion with unequal launch/land heights.
   HOW TO SPOT IT: Question mentions "launched from a cliff" or "lands on a platform."
   HOW TO AVOID IT: Use ( t = \frac{v_y \pm \sqrt{v_y^2 + 2gh}}{g} ) for time of flight.

2. TRAP: Circular motion with changing speed (e.g., rollercoaster loops).
   HOW TO SPOT IT: Question asks for minimum speed at the top.
   HOW TO AVOID IT: Set ( F_c = mg ) at the top (minimum speed condition).

3. TRAP: Energy problems with friction.
   HOW TO SPOT IT: Question mentions "rough surface" or "work done against friction."
   HOW TO AVOID IT: Add ( W_{nc} = F_f \cdot d ) to the energy equation.

1-MINUTE RECAP

"Here’s what you need to remember tonight: 1. Projectile motion: Split velocity into ( v_x ) and ( v_y ). Use ( t = \frac{2u \sin \theta}{g} ), ( R = \frac{u^2 \sin 2\theta}{g} ), and ( H = \frac{u^2 \sin^2 \theta}{2g} ). 2. Circular motion: ( F_c = \frac{mv^2}{r} ). Always point toward the center. 3. Energy: ( KE + PE = \text{constant} ) (no friction). Work = ( F \cdot d \cdot \cos \theta ). 4. Momentum: ( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 ). Elastic = KE conserved, inelastic = not. 5. Exam traps: Watch for unequal heights, friction, and minimum speed conditions. Now go practice—you’ve got this!