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Physics: Mechanics

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Physics: Mechanics - Complete Interactive Lesson

Part 1: Kinematics & Motion

Physics: Mechanics for the MCAT

Part 1 of 7 — Kinematics

The Big 5 Kinematic Equations

$v = v_0 + at$

$\Delta x = v_0 t + \tfrac{1}{2}at^2$

$v^2 = v_0^2 + 2a\Delta x$

$\Delta x = \tfrac{1}{2}(v_0 + v)t$

$\Delta x = vt - \tfrac{1}{2}at^2$

Projectile Motion

Horizontal: $a_x = 0$ , so $v_x = v_0\cos\theta$ is constant

MCAT Tip: Free Fall

All objects fall at the same rate regardless of mass (ignoring air resistance). Use $g \approx 10\;\text{m/s}^2$ for quick calculations on the MCAT.

Relative Motion Shortcut

In one-dimensional motion, use relative velocity directly: $v_{A/B} = v_A - v_B$ . This simplifies chase and meeting-time questions.

Worked Example — Horizontal Projectile off a Cliff

A ball rolls off a $20\;\text{m}$ high table with a horizontal speed of $5\;\text{m/s}$ . Using $g = 10\;\text{m/s}^2$ , how long is it in the air and how far does it land from the base?

Step 1 — Treat the vertical motion separately. The initial vertical velocity is zero, so

Kinematics 🎯

Key Takeaways — Part 1

Use $g \approx 10\;\text{m/s}^2$ for fast MCAT calculations
Projectile motion: split into x (constant velocity) and y (constant acceleration); they are independent
Complementary angles give the same range; $45°$ gives the maximum range
Relative-velocity problems often reduce to a single subtraction when set up correctly

Part 2: Forces & Newtons Laws

Physics: Mechanics for the MCAT

Part 2 of 7 — Newton's Laws & Forces

Newton's Three Laws

Inertia: an object at rest stays at rest, and an object in motion stays in motion, unless acted on by a net force
$F = ma$ : net force equals mass times acceleration
Action-Reaction: every force has an equal and opposite force on a DIFFERENT object

Common MCAT Forces

Force	Formula	Direction
Weight	$W = mg$

Part 3: Work, Energy & Power

Physics: Mechanics for the MCAT

Part 3 of 7 — Work, Energy & Power

Work-Energy Theorem

$W_{net} = \Delta KE = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2$

Part 4: Momentum & Collisions

Physics: Mechanics for the MCAT

Part 4 of 7 — Momentum & Collisions

Linear Momentum

$\vec{p} = m\vec{v}$ — a vector quantity (units: )

Part 5: Fluids & Pressure

Physics: Mechanics for the MCAT

Part 5 of 7 — Fluids (ULTRA HIGH YIELD)

Density & Pressure

$\rho = \frac{m}{V}$ and $P = \frac{F}{A}$

Part 6: Waves & Sound

Physics: Mechanics for the MCAT

Part 6 of 7 — Torque, Equilibrium & Simple Machines

Torque

$\tau = rF\sin\theta$

$r$ = distance from the pivot (the lever arm)
$\theta$ = angle between the lever arm and the force; torque is maximum at $\theta = 90°$

Part 7: Review & MCAT Practice

Physics: Mechanics for the MCAT

Part 7 of 7 — Waves & Sound

Wave Properties

$v = f\lambda$ and period $T = \frac{1}{f}$

Vertical:

a_y = -g = -9.8\;\text{m/s}^2

Time to reach max height:

t = \frac{v_0\sin\theta}{g}

Range:

R = \frac{v_0^2\sin(2\theta)}{g}

(maximum at

45°

)

$\Delta y = \tfrac{1}{2} g t^2 \Rightarrow 20 = \tfrac{1}{2}(10) t^2 = 5 t^2$

Step 2 — Solve for the time of flight.

$t^2 = \frac{20}{5} = 4 \Rightarrow t = 2\;\text{s}$

Step 3 — Use the horizontal motion (constant velocity).

$x = v_x t = (5)(2) = 10\;\text{m}$

The ball lands $10\;\text{m}$ from the base after $2\;\text{s}$ . Key MCAT insight: the horizontal speed has NO effect on the fall time — vertical and horizontal motions are independent.

Inclined Plane (MCAT FAVORITE)

Component along the plane: $mg\sin\theta$
Component perpendicular: $mg\cos\theta$ (equals the normal force if no other vertical forces)
Friction on an incline: $f = \mu mg\cos\theta$

Force-Analysis Workflow

Isolate one object.
Draw every real force (weight, normal, tension, friction, applied).
Choose axes along the likely motion.
Write $\sum F = ma$ per axis.

Worked Example — Block on a Rough Incline

A $4\;\text{kg}$ block sits on a $30°$ incline with kinetic friction coefficient $\mu_k = 0.20$ . Using $g = 10\;\text{m/s}^2$ , find its acceleration down the slope once it is sliding.

Step 1 — Force pulling it down the plane.

$F_{\parallel} = mg\sin\theta = (4)(10)\sin 30° = 40 \times 0.5 = 20\;\text{N}$

Step 2 — Normal force and friction.

$N = mg\cos\theta = (4)(10)\cos 30° \approx 40 \times 0.87 = 34.6\;\text{N}$ , so

$f_k = \mu_k N = (0.20)(34.6) \approx 6.9\;\text{N}$

Step 3 — Net force along the plane, then Newton's second law.

$F_{net} = 20 - 6.9 = 13.1\;\text{N}$ , and $a = \frac{F_{n e t}}{m} = \frac{13.1}{4} \approx 3.3 {m/s}^{}$

Notice friction opposes the motion (subtracts), and the mass appears in both terms — when friction is absent it cancels entirely, leaving $a = g\sin\theta$ .

Forces & Newton's Laws 🎯

Key Takeaways — Part 2

$F_{net} = ma$ : always draw a free-body diagram first
Incline: $mg\sin\theta$ along the plane, $mg\cos\theta$ perpendicular
Elevator problems: apparent weight $= m(g \pm a)$
Static friction is a maximum ( $f_s \le \mu_s N$ ); kinetic friction is exact ( $f_k = \mu_k N$ )
If the speed is constant, the net force is zero even when several forces act

$W = Fd\cos\theta$ (only the force component along the displacement does work)

Conservation of Energy

$KE_i + PE_i = KE_f + PE_f$ (when no non-conservative forces act)

Kinetic energy: $KE = \tfrac{1}{2}mv^2$
Gravitational PE: $PE = mgh$
Spring PE: $PE = \tfrac{1}{2}kx^2$

$P = \frac{W}{t} = Fv$ — measured in watts (W), where $1\;\text{W} = 1\;\text{J/s}$

Conservative vs. Non-conservative Forces

Conservative (gravity, springs): path-independent work; mechanical energy is conserved
Non-conservative (friction, drag): convert mechanical energy into thermal/internal energy

When friction is present, include the non-conservative work in the energy balance.

Worked Example — Speed at the Bottom of a Ramp

A $2\;\text{kg}$ cart starts from rest at the top of a frictionless ramp $1.8\;\text{m}$ tall. Using $g = 10\;\text{m/s}^2$ , find its speed at the bottom.

Step 1 — Set up conservation of energy. All gravitational PE converts to kinetic energy:

$mgh = \tfrac{1}{2}mv^2$

Step 2 — Cancel the mass and solve for $v$ .

$v = \sqrt{2gh} = \sqrt{2(10)(1.8)} = \sqrt{36}$

Step 3 — Evaluate.

$v = 6\;\text{m/s}$

The mass dropped out, which is why $v = \sqrt{2gh}$ is worth memorizing. If friction did, say, $10\;\text{J}$ of negative work, you would instead write and solve for a smaller speed.

Work & Energy 🎯

Key Takeaways — Part 3

$W = Fd\cos\theta$ : only the force component parallel to displacement does work
Conservation of energy: $KE + PE = \text{constant}$ when no friction or drag acts
$v = \sqrt{2gh}$ for an object dropped (or sliding) from height $h$ — memorize this shortcut
Power = work / time = force $\times$ velocity
$KE \propto v^2$ : doubling speed quadruples kinetic energy
Friction does negative work and reduces mechanical energy

\text{kg}\cdot\text{m/s}

Impulse-Momentum Theorem

$\vec{J} = \vec{F}\Delta t = \Delta\vec{p}$ — a force applied over time changes momentum

Conservation of Momentum

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

Total momentum is always conserved in the absence of external forces.

Type	Momentum	Kinetic Energy
Elastic	Conserved	Conserved
Inelastic	Conserved	NOT conserved (some lost to heat/deformation)
Perfectly inelastic	Conserved	Maximum KE loss (objects stick together)

For a perfectly inelastic collision: $m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_f$

Why Increasing Collision Time Matters

From $J = F\Delta t = \Delta p$ , for a fixed momentum change, increasing $\Delta t$ lowers the average force. This principle explains airbags, padded helmets, and crumple zones.

Worked Example — Impulse Reduces Force

A $0.15\;\text{kg}$ baseball arrives at $40\;\text{m/s}$ and is caught, coming to rest. Compare the average force on the hand if the catch takes $0.01\;\text{s}$ (rigid hand) versus $0.10\;\text{s}$ (giving with the ball).

Step 1 — Find the momentum change (same for both).

$\Delta p = m\Delta v = (0.15)(0 - 40) = -6\;\text{kg}\cdot\text{m/s}$ (magnitude $6$ ).

Step 2 — Rigid catch ( $\Delta t = 0.01\;\text{s}$ ).

$F = \frac{\Delta p}{\Delta t} = \frac{6}{0.01} = 600\;\text{N}$

Step 3 — Soft catch ( $\Delta t = 0.10\;\text{s}$ ).

$F = \frac{6}{0.10} = 60\;\text{N}$

Extending the contact time tenfold cuts the average force to one-tenth. This is exactly why you pull your hands back when catching a fast ball — and why airbags save lives.

Key Takeaways — Part 4

Momentum is ALWAYS conserved in collisions (absent external forces)
KE is ONLY conserved in elastic collisions
Perfectly inelastic = objects stick together = maximum KE loss
Impulse $J = F\Delta t = \Delta p$ — extending contact time lowers the force (airbags, padding)

Hydrostatic Pressure

$P = P_0 + \rho g h$ , where $P_0$ is atmospheric pressure ( $1\;\text{atm} \approx 1.0 \times 10^5\;\text{Pa}$ )

Pressure applied to a confined fluid is transmitted equally: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ (the basis of the hydraulic lift)

Archimedes' Principle (Buoyancy)

$F_b = \rho_{fluid} \cdot V_{displaced} \cdot g$

An object floats if $\rho_{object} < \rho_{fluid}$ .

Bernoulli's Equation (energy conservation for fluids)

$P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2$

$A_1 v_1 = A_2 v_2$ — a narrower pipe forces faster flow, which (by Bernoulli) lowers pressure (the Venturi effect)

Volume flow rate $Q = Av$ ties continuity to units ( $\text{m}^3/\text{s}$ ) and to physiology passages about blood flow.

Worked Example — Buoyant Force on a Submerged Object

A solid block of volume $V = 2.0 \times 10^{-3}\;\text{m}^3$ is fully submerged in water ( $\rho_{fluid} = 1000\;\text{kg/m}^3$ ). Using $g = 10\;\text{m/s}^2$ , find the buoyant force on it.

Step 1 — Apply Archimedes' principle. Fully submerged means $V_{displaced} = V$ :

$F_b = \rho_{fluid} \cdot V_{displaced} \cdot g$

Step 2 — Substitute the values.

$F_b = (1000)(2.0 \times 10^{-3})(10)$

Step 3 — Evaluate.

$F_b = 1000 \times 0.002 \times 10 = 20\;\text{N}$

The buoyant force depends only on the displaced fluid, NOT on the block's own density. To decide whether it floats or sinks, compare this $20\;\text{N}$ buoyant force with the block's weight $mg$ : if the weight is larger, it sinks.

Key Takeaways — Part 5

Bernoulli: faster flow → lower pressure (explains aneurysms and airplane lift)
Continuity: $A_1 v_1 = A_2 v_2$ for an incompressible fluid
Buoyancy: $F_b = \rho_{fluid} V_{displaced} g$ ; an object floats when
Hydrostatic pressure rises with depth: $P = P_0 + \rho g h$
Hydraulic lift (Pascal): force multiplies by the area ratio $A_2/A_1$

Counterclockwise is positive by convention

Equilibrium Conditions

For static equilibrium: $\sum F = 0$ AND $\sum \tau = 0$

$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$

Lever: $F_1 d_1 = F_2 d_2$ (mechanical advantage)
Pulley: redirects force; compound pulleys multiply force
Inclined plane: reduces the force needed but increases the distance

Key principle: machines reduce force but NEVER reduce work ( $W = Fd$ is conserved).

Pivot Choice Strategy

Choose a pivot that eliminates an unknown force (often a support point) so the torque equation simplifies quickly.

Worked Example — Balancing a Seesaw

A child of weight $300\;\text{N}$ sits $2.0\;\text{m}$ left of a seesaw's pivot. Where must a $400\;\text{N}$ child sit on the right to balance it?

Step 1 — Write the rotational equilibrium condition. For balance the counterclockwise and clockwise torques are equal:

$\tau_{left} = \tau_{right} \Rightarrow F_1 d_1 = F_2 d_2$

Step 2 — Plug in the known values.

$(300)(2.0) = (400)(d_2)$

Step 3 — Solve for the distance.

$d_2 = \frac{600}{400} = 1.5\;\text{m}$

The heavier child sits closer to the pivot ( $1.5\;\text{m}$ vs. $2.0\;\text{m}$ ). This is the lever balance condition $F_1 d_1 = F_2 d_2$ — a staple of MCAT torque questions.

Torque & Equilibrium 🎯

Key Takeaways — Part 6

Torque $\tau = rF\sin\theta$ ; maximum when the force is perpendicular to the lever arm
Equilibrium: $\sum F = 0$ AND $\sum \tau = 0$ (you may choose any pivot point)
Lever balance: $F_1 d_1 = F_2 d_2$ — the heavier side sits closer to the pivot
Simple machines trade force for distance; work is conserved
The MCAT loves beam and seesaw problems — practice them

Transverse: oscillation perpendicular to propagation (light, waves on a string)
Longitudinal: oscillation parallel to propagation (sound)

Speed in air: $\approx 340\;\text{m/s}$ (faster in denser media such as water and solids)
Intensity: $I = \frac{P}{4\pi r^2}$ (an inverse-square law)
Decibels: $\beta = 10\log(I/I_0)$ where $I_0 = 10^{-12}\;\text{W/m}^2$
Every $10\;\text{dB}$ increase corresponds to a $10\times$ increase in intensity

$f' = f\frac{v \pm v_{observer}}{v \mp v_{source}}$

Source approaching → higher observed frequency (higher pitch)
Source receding → lower observed frequency

Both ends fixed: $\lambda_n = \frac{2L}{n}$ , $f_n = n\frac{v}{2L}$
One end open: $\lambda_n = \frac{4L}{n}$ , with only odd harmonics ( $n = 1, 3, 5, . .$ )

For a fixed source frequency, wavelength changes with the medium because $\lambda = v/f$ and the wave speed depends on the medium.

Worked Example — Decibel Change from Intensity

A sound's intensity increases from $I_1 = 10^{-6}\;\text{W/m}^2$ to $I_2 = 10^{-3}\;\text{W/m}^2$ . By how many decibels does the sound level rise?

Step 1 — Use the decibel difference formula.

$\Delta\beta = 10\log\!\left(\frac{I_2}{I_1}\right)$

Step 2 — Take the ratio.

$\frac{I_2}{I_1} = \frac{10^{-3}}{10^{-6}} = 10^{3}$

Step 3 — Evaluate the logarithm.

$\Delta\beta = 10\log(10^{3}) = 10 \times 3 = 30\;\text{dB}$

Each factor of 10 in intensity adds $10\;\text{dB}$ , so a $1000\times$ jump is $+30\;\text{dB}$ . The decibel scale is logarithmic — a useful reminder that small dB changes hide large intensity changes.

Waves & Sound 🎯

Physics Mechanics — Complete! ✅

Key relationships: $v = f\lambda$ , the Doppler effect, and the inverse-square law for intensity. Sound travels FASTER in denser media (the opposite of light). Decibels are logarithmic, so $+10\;\text{dB}$ means $10\times$ the intensity. Standing-wave harmonics depend on the boundary conditions (both ends fixed vs. one end open).

a = \frac{F _{n e t}}{m} = \frac{13.1}{4} \approx 3.3 m/s^{2}

mgh - 10 = \tfrac{1}{2}mv^2

m g h - 10 = \frac{1}{2} m v^{2}

\rho_{object} < \rho_{fluid}

Physics: Mechanics

Inclined Plane (MCAT FAVORITE)

Force-Analysis Workflow

Worked Example — Block on a Rough Incline

Key Takeaways — Part 2

Conservation of Energy

Power

Conservative vs. Non-conservative Forces

Worked Example — Speed at the Bottom of a Ramp

Key Takeaways — Part 3

Impulse-Momentum Theorem

Conservation of Momentum

Collision Types

Why Increasing Collision Time Matters

Worked Example — Impulse Reduces Force

Key Takeaways — Part 4

Hydrostatic Pressure

Pascal's Principle

Archimedes' Principle (Buoyancy)

Bernoulli's Equation (energy conservation for fluids)

Continuity Equation

Flow Rate

Worked Example — Buoyant Force on a Submerged Object

Key Takeaways — Part 5

Equilibrium Conditions

Center of Mass

Simple Machines

Pivot Choice Strategy

Worked Example — Balancing a Seesaw

Key Takeaways — Part 6

Sound

Doppler Effect

Standing Waves

Worked Example — Decibel Change from Intensity

Physics Mechanics — Complete! ✅