Newton’s First Law, Inertia Statement An object at rest remains at rest, and an object in motion continues with constant velocity, unless Acted upon by a net external force.
Inertia Inertia is the tendency of an object to resist changes in its state of motion. It is measured by Mass, the greater the mass, the greater the inertia.
Inertial Frames of Reference A frame of reference in which Newton’s first law holds is called an inertial frame . Accelerating Frames (like a car rounding a corner) are non-inertial.
Applications Passengers lurch forward when a bus brakes suddenly. Tablecloth can be pulled out from under dishes if done quickly (dishes have inertia). A spacecraft in deep space continues at constant velocity without thrust. Newton’s Second Law Explore how net force, mass, and acceleration are related by applying forces to objects. Experiment with friction, applied forces, and different masses to see how they affect motion.
Statement The net force acting on an object equals the rate of change of momentum:
{ ⃗ F } { n e t } } = { d p ⃗ } { d t \vec\{F\}_{\mathrm\{net\}\} = \frac\{d\vec{p}\}\{dt} { F } { n e t }} = d { p } { d t For constant mass:
{ ⃗ F } { { n e t } } = m { ⃗ a } \vec\{F\}_\{\mathrm\{net\}\} = m\vec\{a\} { F } { { n e t }} = m { a } Key Points Force is a vector quantity, direction matters. F ⃗ n e t \vec{F}_{\mathrm{net}} F net resultant force ).The acceleration is always in the direction of the net force. SI unit: newton (N \mathrm N N 1 N = 1 k g ⋅ m / s 2 1\mathrm{ N} = 1\mathrm{ kg} \cdot \mathrm{m/s}^2 1 N = 1 kg ⋅ m/s 2 Free-Body Diagrams A free-body diagram shows all forces acting ON an object:
Isolate the object. Draw all forces as arrows (label each force). Do NOT include forces exerted BY the object. Include the weight (m g mg m g N N N f f f Component Resolution For forces at angles, resolve into components:
F x = F cos θ , F y = F sin θ F_x = F\cos\theta, \quad F_y = F\sin\theta F x = F cos θ , F y = F sin θ Apply Newton’s second law in each direction:
∑ F x = m a x , ∑ F y = m a y \sum F_x = ma_x, \quad \sum F_y = ma_y ∑ F x = m a x , ∑ F y = m a y
A 5 k g 5\mathrm{ kg} 5 kg 30 N 30\mathrm{ N} 30 N 30 ° 30\degree 30° 0.3 0.3 0.3
Vertical :
N + 30 sin 30 ° = m g N + 30\sin 30\degree = mg N + 30 sin 30° = m g N = 5 ( 9.81 ) − 30 ( 0.5 ) = 49.05 − 15 = 34.05 { N } N = 5(9.81) - 30(0.5) = 49.05 - 15 = 34.05\mathrm\{ N\} N = 5 ( 9.81 ) − 30 ( 0.5 ) = 49.05 − 15 = 34.05 { N } Friction : f k = μ k N = 0.3 ( 34.05 ) = 10.22 N f_k = \mu_k N = 0.3(34.05) = 10.22\mathrm{ N} f k = μ k N = 0.3 ( 34.05 ) = 10.22 N
Horizontal :
30 cos 30 ° − f k = m a 30\cos 30\degree - f_k = ma 30 cos 30° − f k = ma 25.98 − 10.22 = 5 a 25.98 - 10.22 = 5a 25.98 − 10.22 = 5 a A = { 1 5.76 } { 5 } = 3.15 { m / s } 2 A = \frac\{15.76\}\{5\} = 3.15\mathrm\{ m/s\}^2 A = 1 { 5.76 } { 5 } = 3.15 { m / s } 2 Connected Bodies For systems of connected objects (e.g., pulley systems):
Draw a separate free-body diagram for each object. Apply F ⃗ = m a ⃗ \vec{F} = m\vec{a} F = m a Use the constraint that connected objects share the same acceleration magnitude. The tension in an ideal (massless, frictionless) string is the same throughout.
Two masses m 1 = 3 k g m_1 = 3\mathrm{ kg} m 1 = 3 kg m 2 = 5 k g m_2 = 5\mathrm{ kg} m 2 = 5 kg m 1 m_1 m 1 m 2 m_2 m 2 μ k = 0.4 \mu_k = 0.4 μ k = 0.4
For m 1 m_1 m 1 m 1 g − T = m 1 a m_1 g - T = m_1 a m 1 g − T = m 1 a
For m 2 m_2 m 2 T − μ k m 2 g = m 2 a T - \mu_k m_2 g = m_2 a T − μ k m 2 g = m 2 a
Adding: m 1 g − μ k m 2 g = ( m 1 + m 2 ) a m_1 g - \mu_k m_2 g = (m_1 + m_2)a m 1 g − μ k m 2 g = ( m 1 + m 2 ) a
A = { m 1 g − μ k m 2 g } { m 1 + m 2 } = { 3 ( 9.81 ) − 0.4 ( 5 ) ( 9.81 ) } { 8 } = { 2 9.43 − 19.62 } { 8 } = { 9 .81 } { 8 } = 1.23 { m / s } 2 A = \frac\{m_1 g - \mu_k m_2 g\}\{m_1 + m_2\} = \frac\{3(9.81) - 0.4(5)(9.81)\}\{8\} = \frac\{29.43 - 19.62\}\{8\} = \frac\{9.81\}\{8\} = 1.23\mathrm\{ m/s\}^2 A = m { 1 g − μ k m 2 g } { m 1 + m 2 } = 3 { ( 9.81 ) − 0.4 ( 5 ) ( 9.81 )} { 8 } = 2 { 9.43 − 19.62 } { 8 } = 9 { .81 } { 8 } = 1.23 { m / s } 2 T = m 1 ( g − a ) = 3 ( 9.81 − 1.23 ) = 25.74 { N } T = m_1(g - a) = 3(9.81 - 1.23) = 25.74\mathrm\{ N\} T = m 1 ( g − a ) = 3 ( 9.81 − 1.23 ) = 25.74 { N } Apparent Weight and Elevators In an elevator accelerating upward at a a a
N − m g = m a ⟹ N = m ( g + a ) N - mg = ma \implies N = m(g + a) N − m g = ma ⟹ N = m ( g + a ) The apparent weight is N = m ( g + a ) N = m(g + a) N = m ( g + a )
In an elevator accelerating downward at a a a
M g − N = m a ⟹ N = m ( g − a ) Mg - N = ma \implies N = m(g - a) M g − N = ma ⟹ N = m ( g − a ) The apparent weight is N = m ( g − a ) N = m(g - a) N = m ( g − a )
Newton’s Third Law Statement For every action, there is an equal and opposite reaction:
{ ⃗ F } { A { o n } B } = − { ⃗ F } { B { o n } A } \vec\{F\}_\{A \mathrm\{ on \} B\} = -\vec\{F\}_\{B \mathrm\{ on \} A\} { F } { A { o n } B } = − { F } { B { o n } A } Key Points Action-reaction pairs act on different objects . They are equal in magnitude and opposite in direction. They are the same type of force (both gravitational, both normal, etc.). Action-reaction forces do NOT cancel because they act on different bodies. Common Action-Reaction Pairs Action Reaction Book pushes down on table Table pushes up on book (normal force) Earth pulls on apple (weight) Apple pulls on Earth Foot pushes backward on ground Ground pushes forward on foot Rocket pushes exhaust gases down Exhaust gases push rocket up
A common mistake is to confuse Newton’s third law pairs with balanced forces on the same object. The Weight and normal force on a stationary book are NOT an action-reaction pair, they act on the SAME Object and balance. The reaction to the book’s weight is the book pulling the Earth upward.
Friction Static Friction Static friction prevents a stationary object from starting to move. It varies from 0 0 0
F s ≤ μ s N F_s \le \mu_s N F s ≤ μ s N Where μ s \mu_s μ s coefficient of static friction and N N N
Kinetic Friction Kinetic friction opposes the motion of a sliding object:
F k = μ k N F_k = \mu_k N F k = μ k N Where μ k \mu_k μ k coefficient of kinetic friction .
Key Relationships μ s > μ k \mu_s \gt \mu_k μ s > μ k Friction is independent of the area of contact. Friction is proportional to the normal force. Kinetic friction is approximately constant (independent of speed). Angle of Repose The angle at which an object on an inclined plane begins to slide:
tan θ = μ s \tan\theta = \mu_s tan θ = μ s
A block rests on an inclined plane at 35 ° 35\degree 35° μ s = 0.6 \mu_s = 0.6 μ s = 0.6
tan 35 ° ≈ 0.700 \tan 35\degree \approx 0.700 tan 35° ≈ 0.700 Since tan 35 ° > μ s = 0.6 \tan 35\degree \gt \mu_s = 0.6 tan 35° > μ s = 0.6
Friction on an Inclined Plane For a block on an incline at angle θ \theta θ
Force Expression Component of weight along plane m g sin θ mg\sin\theta m g sin θ Component of weight perpendicular to plane m g cos θ mg\cos\theta m g cos θ Normal force N = m g cos θ N = mg\cos\theta N = m g cos θ Friction force f = μ N = μ m g cos θ f = \mu N = \mu mg\cos\theta f = μ N = μ m g cos θ
The block slides down if m g sin θ > μ s m g cos θ mg\sin\theta \gt \mu_s mg\cos\theta m g sin θ > μ s m g cos θ tan θ > μ s \tan\theta \gt \mu_s tan θ > μ s
Momentum and Impulse Linear Momentum { ⃗ p } = m { ⃗ v } \vec\{p\} = m\vec\{v\} { p } = m { v } Momentum is a vector quantity with SI unit k g ⋅ m / s \mathrm{kg} \cdot \mathrm{m/s} kg ⋅ m/s
{ ⃗ F } { n e t } } = { d p ⃗ } { d t \vec\{F\}_{\mathrm\{net\}\} = \frac\{d\vec{p}\}\{dt} { F } { n e t }} = d { p } { d t Impulse Impulse equals the change in momentum:
{ ⃗ J } = { ⃗ F } { { n e t } } ⋅ Δ t = Δ { ⃗ p } = m { ⃗ v } − m { ⃗ u } \vec\{J\} = \vec\{F\}_\{\mathrm\{net\}\} \cdot \Delta t = \Delta\vec\{p\} = m\vec\{v\} - m\vec\{u\} { J } = { F } { { n e t }} ⋅ Δ t = Δ { p } = m { v } − m { u } For a constant force: J = F Δ t J = F \Delta t J = F Δ t
The unit of impulse is N ⋅ s = k g ⋅ m / s \mathrm N \cdot \mathrm s = \mathrm{kg} \cdot \mathrm{m/s} N ⋅ s = kg ⋅ m/s
Impulse and Force-Time Graphs The area under a force-time graph equals the impulse.
A 0.15 k g 0.15\mathrm{ kg} 0.15 kg 30 m / s 30\mathrm{ m/s} 30 m/s 20 m / s 20\mathrm{ m/s} 20 m/s 0.01 s 0.01\mathrm{ s} 0.01 s
J = Δ p = 0.15 ( 20 − ( − 30 ) ) = 0.15 × 50 = 7.5 { N } ⋅ s J = \Delta p = 0.15(20 - (-30)) = 0.15 \times 50 = 7.5\mathrm\{ N\}\cdot\mathrm s J = Δ p = 0.15 ( 20 − ( − 30 )) = 0.15 × 50 = 7.5 { N } ⋅ s F = { J } { Δ t } = { 7 .5 } { 0.01 } = 750 { N } F = \frac\{J\}\{\Delta t\} = \frac\{7.5\}\{0.01\} = 750\mathrm\{ N\} F = J { } { Δ t } = 7 { .5 } { 0.01 } = 750 { N } Conservation of Linear Momentum In a closed system (no external forces), the total momentum is conserved:
M 1 { ⃗ u } 1 + m 2 { ⃗ u } 2 = m 1 { ⃗ v } 1 + m 2 { ⃗ v } 2 M_1\vec\{u\}_1 + m_2\vec\{u\}_2 = m_1\vec\{v\}_1 + m_2\vec\{v\}_2 M 1 { u } 1 + m 2 { u } 2 = m 1 { v } 1 + m 2 { v } 2 Elastic and Inelastic Collisions Type Kinetic Energy Momentum Perfectly elastic Conserved Conserved Inelastic NOT conserved Conserved Perfectly inelastic Maximum loss Conserved (objects stick together)
A 2 k g 2\mathrm{ kg} 2 kg 5 m / s 5\mathrm{ m/s} 5 m/s 3 k g 3\mathrm{ kg} 3 kg
M 1 u 1 + m 2 u 2 = ( m 1 + m 2 ) v M_1 u_1 + m_2 u_2 = (m_1 + m_2)v M 1 u 1 + m 2 u 2 = ( m 1 + m 2 ) v 2 ( 5 ) + 3 ( 0 ) = 5 v 2(5) + 3(0) = 5v 2 ( 5 ) + 3 ( 0 ) = 5 v V = 2 { m / s } V = 2\mathrm\{ m/s\} V = 2 { m / s } KE before : 1 2 ( 2 ) ( 25 ) = 25 J \dfrac{1}{2}(2)(25) = 25\mathrm{ J} 2 1 ( 2 ) ( 25 ) = 25 J
KE after : 1 2 ( 5 ) ( 4 ) = 10 J \dfrac{1}{2}(5)(4) = 10\mathrm{ J} 2 1 ( 5 ) ( 4 ) = 10 J
15 J 15\mathrm{ J} 15 J
Work-Energy Theorem Work Done Work is done when a force causes displacement:
W = F d cos θ W = Fd\cos\theta W = F d cos θ Where θ \theta θ
Work is a scalar quantity (SI unit: joule, J \mathrm J J Work done by friction is always negative (opposes motion). If F F F d d d W = F d W = Fd W = F d If F ⊥ d F \perp d F ⊥ d W = 0 W = 0 W = 0 Work-Energy Theorem W { { n e t } } = Δ E k = { 1 } { 2 } m v 2 − { 1 } { 2 } m u 2 W_\{\mathrm\{net\}\} = \Delta E_k = \frac\{1\}\{2\}mv^2 - \frac\{1\}\{2\}mu^2 W { { n e t }} = Δ E k = 1 { } { 2 } m v 2 − 1 { } { 2 } m u 2 The net work done on an object equals its change in kinetic energy.
Kinetic Energy E k = { 1 } { 2 } m v 2 E_k = \frac\{1\}\{2\}mv^2 E k = 1 { } { 2 } m v 2 Power P = { W } { t } = { ⃗ F } ⋅ { ⃗ v } P = \frac\{W\}\{t\} = \vec\{F\} \cdot \vec\{v\} P = W { } { t } = { F } ⋅ { v } SI unit: watt (W \mathrm W W 1 W = 1 J / s 1\mathrm{ W} = 1\mathrm{ J/s} 1 W = 1 J/s
Efficiency E f f i c i e n c y = u s e f u l e n e r g y o u t p u t t o t a l e n e r g y i n p u t × 100 % \mathrm{Efficiency} = \frac{\mathrm{useful energy output}}{\mathrm{total energy input}} \times 100\% Efficiency = totalenergyinput usefulenergyoutput × 100% Alternatively:
E f f i c i e n c y = u s e f u l p o w e r o u t p u t t o t a l p o w e r i n p u t × 100 % \mathrm{Efficiency} = \frac{\mathrm{useful power output}}{\mathrm{total power input}} \times 100\% Efficiency = totalpowerinput usefulpoweroutput × 100%
A car of mass 1200 k g 1200\mathrm{ kg} 1200 kg 15 m / s 15\mathrm{ m/s} 15 m/s 25 m / s 25\mathrm{ m/s} 25 m/s 6 s 6\mathrm{ s} 6 s 3000 N 3000\mathrm{ N} 3000 N
Δ E k = { 1 } { 2 } ( 1200 ) ( 625 − 225 ) = { 1 } { 2 } ( 1200 ) ( 400 ) = 240000 { J } \Delta E_k = \frac\{1\}\{2\}(1200)(625 - 225) = \frac\{1\}\{2\}(1200)(400) = 240000\mathrm\{ J\} Δ E k = 1 { } { 2 } ( 1200 ) ( 625 − 225 ) = 1 { } { 2 } ( 1200 ) ( 400 ) = 240000 { J } W { { n e t } } = W { { e n g i n e } } + W { { f r i c t i o n } } = 3000 ( 6 × 20 ) + W f W_\{\mathrm\{net\}\} = W_\{\mathrm\{engine\}\} + W_\{\mathrm\{friction\}\} = 3000(6 \times 20) + W_f W { { n e t }} = W { { e n g in e }} + W { { f r i c t i o n }} = 3000 ( 6 × 20 ) + W f Wait, let me recalculate properly:
W { { n e t } } = ( F { { e n g i n e } } − f ) × d W_\{\mathrm\{net\}\} = (F_\{\mathrm\{engine\}\} - f) \times d W { { n e t }} = ( F { { e n g in e }} − f ) × d A = { 2 5 − 15 } { 6 } = { 5 } { 3 } { m / s } 2 A = \frac\{25-15\}\{6\} = \frac\{5\}\{3\}\mathrm\{ m/s\}^2 A = 2 { 5 − 15 } { 6 } = 5 { } { 3 } { m / s } 2 D = 15 ( 6 ) + { 1 } { 2 } ( { 5 } { 3 } ) ( 36 ) = 90 + 30 = 120 { m } D = 15(6) + \frac\{1\}\{2\}\left(\frac\{5\}\{3\}\right)(36) = 90 + 30 = 120\mathrm\{ m\} D = 15 ( 6 ) + 1 { } { 2 } ( 5 { } { 3 } ) ( 36 ) = 90 + 30 = 120 { m } 240000 = ( 3000 − f ) ( 120 ) 240000 = (3000 - f)(120) 240000 = ( 3000 − f ) ( 120 ) 3000 − f = 2000 ⟹ f = 1000 { N } 3000 - f = 2000 \implies f = 1000\mathrm\{ N\} 3000 − f = 2000 ⟹ f = 1000 { N } IB Exam-Style Questions Question 1 (Paper 1 style) A 10 k g 10\mathrm{ kg} 10 kg 60 N 60\mathrm{ N} 60 N 4 m / s 2 4\mathrm{ m/s}^2 4 m/s 2
F { { n e t } } = F − f k = m a F_\{\mathrm\{net\}\} = F - f_k = ma F { { n e t }} = F − f k = ma 60 − μ k × 10 ( 9.81 ) = 10 ( 4 ) 60 - \mu_k \times 10(9.81) = 10(4) 60 − μ k × 10 ( 9.81 ) = 10 ( 4 ) 60 − 98.1 μ k = 40 60 - 98.1\mu_k = 40 60 − 98.1 μ k = 40 μ k = { 2 0 } { 98.1 } ≈ 0.204 \mu_k = \frac\{20\}\{98.1\} \approx 0.204 μ k = 2 { 0 } { 98.1 } ≈ 0.204 Question 2 (Paper 2 style) Two ice skaters, one of mass 60 k g 60\mathrm{ kg} 60 kg 3 m / s 3\mathrm{ m/s} 3 m/s 80 k g 80\mathrm{ kg} 80 kg
(a) Find their common velocity after collision.
60 ( 3 ) + 80 ( 0 ) = ( 60 + 80 ) v 60(3) + 80(0) = (60 + 80)v 60 ( 3 ) + 80 ( 0 ) = ( 60 + 80 ) v 180 = 140 v ⟹ v = 1.29 { m / s } 180 = 140v \implies v = 1.29\mathrm\{ m/s\} 180 = 140 v ⟹ v = 1.29 { m / s } (b) Calculate the kinetic energy lost.
Before: 1 2 ( 60 ) ( 9 ) = 270 J \dfrac{1}{2}(60)(9) = 270\mathrm{ J} 2 1 ( 60 ) ( 9 ) = 270 J
After: 1 2 ( 140 ) ( 9 7 ) 2 = 1 2 ( 140 ) ( 81 49 ) = 115.7 J \dfrac{1}{2}(140)\left(\dfrac{9}{7}\right)^2 = \dfrac{1}{2}(140)\left(\dfrac{81}{49}\right) = 115.7\mathrm{ J} 2 1 ( 140 ) ( 7 9 ) 2 = 2 1 ( 140 ) ( 49 81 ) = 115.7 J
Lost: 270 − 115.7 = 154.3 J 270 - 115.7 = 154.3\mathrm{ J} 270 − 115.7 = 154.3 J
Question 3 (Paper 2 style) A block of mass 4 k g 4\mathrm{ kg} 4 kg 30 ° 30\degree 30° 0.5 0.5 0.5 0.3 0.3 0.3
(a) Determine whether the block slides.
M g sin 30 ° = 4 ( 9.81 ) ( 0.5 ) = 19.62 { N } Mg\sin 30\degree = 4(9.81)(0.5) = 19.62\mathrm\{ N\} M g sin 30° = 4 ( 9.81 ) ( 0.5 ) = 19.62 { N } μ s m g cos 30 ° = 0.5 ( 4 ) ( 9.81 ) ( 0.866 ) = 17.0 { N } \mu_s mg\cos 30\degree = 0.5(4)(9.81)(0.866) = 17.0\mathrm\{ N\} μ s m g cos 30° = 0.5 ( 4 ) ( 9.81 ) ( 0.866 ) = 17.0 { N } Since 19.62 > 17.0 19.62 \gt 17.0 19.62 > 17.0
(b) Find the acceleration down the plane.
A = g ( sin θ − μ k cos θ ) = 9.81 ( 0.5 − 0.3 × 0.866 ) = 9.81 ( 0.240 ) = 2.36 { m / s } 2 A = g(\sin\theta - \mu_k\cos\theta) = 9.81(0.5 - 0.3 \times 0.866) = 9.81(0.240) = 2.36\mathrm\{ m/s\}^2 A = g ( sin θ − μ k cos θ ) = 9.81 ( 0.5 − 0.3 × 0.866 ) = 9.81 ( 0.240 ) = 2.36 { m / s } 2 Question 4 (Paper 1 style) A motor lifts a 500 k g 500\mathrm{ kg} 500 kg 20 m 20\mathrm{ m} 20 m 10 s 10\mathrm{ s} 10 s
P = \frac`\{mgh}`\{t\} = \frac\{500 \times 9.81 \times 20\}\{10\} = 9810\mathrm\{ W\} = 9.81\mathrm\{ kW\} Intuition Dynamics is the answer to kinematics’ question of why things move the way they do. Newton’s three laws are like the rules of a cosmic game: objects stay still or keep moving unless pushed, the harder you push the faster it accelerates, and every push gets an equal push back. The third law is often misunderstood: when you stand on the floor, your weight pushing down and the floor pushing up are not an action-reaction pair because they act on the same object. Friction is nature’s way of converting kinetic energy into heat, which is why sliding on ice is easy but sliding on sandpaper is not.
Summary Law Statement Newton’s First An object continues at rest or constant velocity unless acted on by net force Newton’s Second F ⃗ n e t = m a ⃗ \vec{F}_{\mathrm{net}} = m\vec{a} F net = m a F ⃗ = d p ⃗ d t \vec{F} = \dfrac{d\vec{p}}{dt} F = d t d p Newton’s Third Every action has an equal and opposite reaction
Quantity Formula Unit Momentum p ⃗ = m v ⃗ \vec{p} = m\vec{v} p = m v k g ⋅ m / s \mathrm{kg} \cdot \mathrm{m/s} kg ⋅ m/s Impulse J ⃗ = F ⃗ Δ t = Δ p ⃗ \vec{J} = \vec{F}\Delta t = \Delta\vec{p} J = F Δ t = Δ p N ⋅ s \mathrm N \cdot \mathrm s N ⋅ s Work W = F d cos θ W = Fd\cos\theta W = F d cos θ J \mathrm J J Kinetic energy E k = 1 2 m v 2 E_k = \dfrac{1}{2}mv^2 E k = 2 1 m v 2 J \mathrm J J Power P = W t = F v P = \dfrac{W}{t} = Fv P = t W = F v W \mathrm W W
Always draw free-body diagrams. Resolve forces into components along the chosen axes. For momentum Questions, define the positive direction. For efficiency questions, remember that efficiency Is always less than 100% in real systems.
Advanced Applications Motion in Lifts (Elevators) Situation Equation Apparent Weight Stationary or constant velocity N = m g N = mg N = m g Normal (m g mg m g Accelerating upward N = m ( g + a ) N = m(g + a) N = m ( g + a ) Greater than normal Accelerating downward N = m ( g − a ) N = m(g - a) N = m ( g − a ) Less than normal Free fall (a = g a = g a = g N = 0 N = 0 N = 0 Weightless
A person of mass 70 k g 70\mathrm{ kg} 70 kg 2 m / s 2 2\mathrm{ m/s}^2 2 m/s 2
N = m ( g + a ) = 70 ( 9.81 + 2 ) = 70 × 11.81 = 826.7 { N } N = m(g + a) = 70(9.81 + 2) = 70 \times 11.81 = 826.7\mathrm\{ N\} N = m ( g + a ) = 70 ( 9.81 + 2 ) = 70 × 11.81 = 826.7 { N } The scale reads 826.7 N 826.7\mathrm{ N} 826.7 N 84.3 k g 84.3\mathrm{ kg} 84.3 kg
Connected Bodies on Inclined Planes For two masses connected by a string over a pulley on an inclined plane, draw separate free-body Diagrams and apply Newton’s second law to each body.
Mass m 1 = 5 k g m_1 = 5\mathrm{ kg} m 1 = 5 kg m 2 = 8 k g m_2 = 8\mathrm{ kg} m 2 = 8 kg 30 ° 30\degree 30° μ k = 0.2 \mu_k = 0.2 μ k = 0.2
For m 1 m_1 m 1 m 1 g − T = m 1 a m_1 g - T = m_1 a m 1 g − T = m 1 a
For m 2 m_2 m 2 T − m 2 g sin θ − μ k m 2 g cos θ = m 2 a T - m_2 g\sin\theta - \mu_k m_2 g\cos\theta = m_2 a T − m 2 g sin θ − μ k m 2 g cos θ = m 2 a
Adding: m 1 g − m 2 g sin θ − μ k m 2 g cos θ = ( m 1 + m 2 ) a m_1 g - m_2 g\sin\theta - \mu_k m_2 g\cos\theta = (m_1 + m_2)a m 1 g − m 2 g sin θ − μ k m 2 g cos θ = ( m 1 + m 2 ) a
A = { 5 ( 9.81 ) − 8 ( 9.81 ) ( 0.5 ) − 0.2 ( 8 ) ( 9.81 ) ( 0.866 ) } { 13 } A = \frac\{5(9.81) - 8(9.81)(0.5) - 0.2(8)(9.81)(0.866)\}\{13\} A = 5 { ( 9.81 ) − 8 ( 9.81 ) ( 0.5 ) − 0.2 ( 8 ) ( 9.81 ) ( 0.866 )} { 13 } = { 4 9.05 − 39.24 − 13.60 } { 13 } = { − 3.79 } { 13 } = \frac\{49.05 - 39.24 - 13.60\}\{13\} = \frac\{-3.79\}\{13\} = 4 { 9.05 − 39.24 − 13.60 } { 13 } = − { 3.79 } { 13 } The negative value means the system accelerates in the opposite direction to what was assumed (i.e., m 2 m_2 m 2 m 1 m_1 m 1
Friction: Extended Analysis Static Friction Graph As the applied force increases from zero:
The static friction matches the applied force (up to μ s N \mu_s N μ s N At the limiting friction point, the object begins to move. Once moving, kinetic friction (μ k N \mu_k N μ k N Kinetic friction is approximately constant regardless of speed. Rolling Friction Rolling friction is much smaller than sliding friction, which is why wheels are so effective. It Arises from deformation of the rolling object and the surface.
Drag Force At low speeds: F d ∝ v F_d \propto v F d ∝ v
At high speeds: F d ∝ v 2 F_d \propto v^2 F d ∝ v 2
Terminal velocity is reached when drag equals the driving force (e.g., weight for a falling object):
Mg = kv_\{\mathrm\{terminal\}\}^2 \implies v_\{\mathrm\{terminal\}\} = \sqrt\{\frac`\{mg}`\{k\}\} Momentum: Extended Applications Rockets and Thrust A rocket expels mass (exhaust gases) at high velocity. By conservation of momentum:
{ T h r u s t } = v e { Δ m } { Δ t } \mathrm\{Thrust\} = v_e \frac\{\Delta m\}\{\Delta t\} { T h r u s t } = v e Δ { m } { Δ t } Where v e v_e v e Δ m Δ t \dfrac{\Delta m}{\Delta t} Δ t Δ m
Rocket Equation (Tsiolkovsky) Δ v = v e ln ( { m i } { m f } ) \Delta v = v_e \ln\!\left(\frac\{m_i\}\{m_f\}\right) Δ v = v e ln ( m { i } { m f } ) Where m i m_i m i m f m_f m f
Impulse-Momentum in Two Dimensions Momentum is conserved separately in each direction:
∑ m { ⃗ v } { { i n i t i a l } } = ∑ m { ⃗ v } { { f i n a l } } \sum m\vec\{v\}_\{\mathrm\{initial\}\} = \sum m\vec\{v\}_\{\mathrm\{final\}\} ∑ m { v } { { ini t ia l }} = ∑ m { v } { { f ina l }} Resolve into x x x y y y
A 3 k g 3\mathrm{ kg} 3 kg 4 m / s 4\mathrm{ m/s} 4 m/s 2 k g 2\mathrm{ kg} 2 kg 3 k g 3\mathrm{ kg} 3 kg 2 m / s 2\mathrm{ m/s} 2 m/s 30 ° 30\degree 30° 2 k g 2\mathrm{ kg} 2 kg
x-direction : 3 ( 4 ) = 3 ( 2 ) cos 30 ° + 2 v x 3(4) = 3(2)\cos 30\degree + 2v_x 3 ( 4 ) = 3 ( 2 ) cos 30° + 2 v x
12 = 5.196 + 2 v x ⟹ v x = 3.402 { m / s } 12 = 5.196 + 2v_x \implies v_x = 3.402\mathrm\{ m/s\} 12 = 5.196 + 2 v x ⟹ v x = 3.402 { m / s } y-direction : 0 = 3 ( 2 ) sin 30 ° + 2 v y 0 = 3(2)\sin 30\degree + 2v_y 0 = 3 ( 2 ) sin 30° + 2 v y
0 = 3 + 2 v y ⟹ v y = − 1.5 { m / s } 0 = 3 + 2v_y \implies v_y = -1.5\mathrm\{ m/s\} 0 = 3 + 2 v y ⟹ v y = − 1.5 { m / s } V = { 3.402 2 + ( − 1.5 ) 2 } = { 11.57 + 2.25 } = { 13.82 } = 3.72 { m / s } V = \sqrt\{3.402^2 + (-1.5)^2\} = \sqrt\{11.57 + 2.25\} = \sqrt\{13.82\} = 3.72\mathrm\{ m/s\} V = { 3.40 2 2 + ( − 1.5 ) 2 } = { 11.57 + 2.25 } = { 13.82 } = 3.72 { m / s } Direction: θ = arctan ( − 1.5 3.402 ) = − 23.8 ° \theta = \arctan\!\left(\dfrac{-1.5}{3.402}\right) = -23.8\degree θ = arctan ( 3.402 − 1.5 ) = − 23.8°
Additional IB Exam-Style Questions Question 5 (Paper 2 style) A 60 k g 60\mathrm{ kg} 60 kg 30 ° 30\degree 30° 100 m 100\mathrm{ m} 100 m 0.1 0.1 0.1
(a) Find the acceleration down the slope.
A = g ( sin θ − μ k cos θ ) = 9.81 ( sin 30 ° − 0.1 cos 30 ° ) = 9.81 ( 0.5 − 0.0866 ) = 9.81 ( 0.4134 ) = 4.06 { m / s } 2 A = g(\sin\theta - \mu_k\cos\theta) = 9.81(\sin 30\degree - 0.1\cos 30\degree) = 9.81(0.5 - 0.0866) = 9.81(0.4134) = 4.06\mathrm\{ m/s\}^2 A = g ( sin θ − μ k cos θ ) = 9.81 ( sin 30° − 0.1 cos 30° ) = 9.81 ( 0.5 − 0.0866 ) = 9.81 ( 0.4134 ) = 4.06 { m / s } 2 (b) Find the speed at the bottom of the slope.
V 2 = u 2 + 2 a s = 0 + 2 ( 4.06 ) ( 100 ) = 812 V^2 = u^2 + 2as = 0 + 2(4.06)(100) = 812 V 2 = u 2 + 2 a s = 0 + 2 ( 4.06 ) ( 100 ) = 812 V = 28.5 { m / s } V = 28.5\mathrm\{ m/s\} V = 28.5 { m / s } (c) The skier then travels across level ground with the same coefficient of friction. How far do they slide before stopping?
μ k m g ⋅ d = { 1 } { 2 } m v 2 \mu_k mg \cdot d = \frac\{1\}\{2\}mv^2 μ k m g ⋅ d = 1 { } { 2 } m v 2 D = { v 2 } { 2 μ k g } = { 8 12 } { 2 ( 0.1 ) ( 9.81 ) } = { 8 12 } { 1.962 } = 414 { m } D = \frac\{v^2\}\{2\mu_k g\} = \frac\{812\}\{2(0.1)(9.81)\} = \frac\{812\}\{1.962\} = 414\mathrm\{ m\} D = v { 2 } { 2 μ k g } = 8 { 12 } { 2 ( 0.1 ) ( 9.81 )} = 8 { 12 } { 1.962 } = 414 { m } Question 6 (Paper 2 style) Two objects, one of mass m m m 3 m 3m 3 m 6 m / s 6\mathrm{ m/s} 6 m/s 2 m / s 2\mathrm{ m/s} 2 m/s 2 m / s 2\mathrm{ m/s} 2 m/s
(a) Find the velocity of the heavier object after the collision.
Taking the direction of m m m
M ( 6 ) + 3 m ( − 2 ) = m ( − 2 ) + 3 m v M(6) + 3m(-2) = m(-2) + 3m v M ( 6 ) + 3 m ( − 2 ) = m ( − 2 ) + 3 m v 6 m − 6 m = − 2 m + 3 m v 6m - 6m = -2m + 3mv 6 m − 6 m = − 2 m + 3 m v 0 = − 2 + 3 v ⟹ v = { 2 } { 3 } { m / s } { ( i n t h e p o s i t i v e d i r e c t i o n ) } 0 = -2 + 3v \implies v = \frac\{2\}\{3\}\mathrm\{ m/s\} \mathrm\{ (in the positive direction)\} 0 = − 2 + 3 v ⟹ v = 2 { } { 3 } { m / s } {( in t h e p os i t i v e d i r ec t i o n )} (b) Is the collision elastic?
KE before = 1 2 m ( 36 ) + 1 2 ( 3 m ) ( 4 ) = 18 m + 6 m = 24 m = \dfrac{1}{2}m(36) + \dfrac{1}{2}(3m)(4) = 18m + 6m = 24m = 2 1 m ( 36 ) + 2 1 ( 3 m ) ( 4 ) = 18 m + 6 m = 24 m
KE after = 1 2 m ( 4 ) + 1 2 ( 3 m ) ( 4 9 ) = 2 m + 2 m 3 = 8 m 3 = \dfrac{1}{2}m(4) + \dfrac{1}{2}(3m)\!\left(\dfrac{4}{9}\right) = 2m + \dfrac{2m}{3} = \dfrac{8m}{3} = 2 1 m ( 4 ) + 2 1 ( 3 m ) ( 9 4 ) = 2 m + 3 2 m = 3 8 m
Since 24 m ≠ 8 m 3 24m \neq \dfrac{8m}{3} 24 m = 3 8 m
Question 7 (Paper 1 style) A force of F = 3 t 2 N F = 3t^2\mathrm{ N} F = 3 t 2 N 2 k g 2\mathrm{ kg} 2 kg 4 s 4\mathrm{ s} 4 s
F = m a ⟹ a = { 3 t 2 } { 2 } = 1.5 t 2 F = ma \implies a = \frac\{3t^2\}\{2\} = 1.5t^2 F = ma ⟹ a = 3 { t 2 } { 2 } = 1.5 t 2 V = ∫ 0 4 1.5 t 2 d t = [ 0.5 t 3 ] 0 4 = 0.5 ( 64 ) = 32 { m / s } V = \int_0^4 1.5t^2\,dt = \left[0.5t^3\right]_0^4 = 0.5(64) = 32\mathrm\{ m/s\} V = ∫ 0 4 1.5 t 2 d t = [ 0.5 t 3 ] 0 4 = 0.5 ( 64 ) = 32 { m / s } Advanced Dynamics Topics Systems with Friction on Multiple Surfaces When a system involves multiple surfaces with different coefficients of friction, draw separate Free-body diagrams for each object and apply Newton’s second law individually.
Motion on Curved Paths For an object moving along a curved path (not necessarily circular), the normal force provides the Centripetal component of acceleration:
N − m g cos θ = { m v 2 } { r } N - mg\cos\theta = \frac\{mv^2\}\{r\} N − m g cos θ = m { v 2 } { r } Where θ \theta θ
Friction on a Banked Curve with Speed Different from Ideal On a banked curve designed for speed v 0 v_0 v 0 v ≠ v 0 v \neq v_0 v = v 0
v > v 0 v \gt v_0 v > v 0 v < v 0 v \lt v_0 v < v 0 Additional IB Exam-Style Questions Question 8 (Paper 2 style) A block of mass 3 k g 3\mathrm{ kg} 3 kg μ s = 0.4 \mu_s = 0.4 μ s = 0.4 μ k = 0.3 \mu_k = 0.3 μ k = 0.3 15 N 15\mathrm{ N} 15 N 25 ° 25\degree 25°
(a) Determine whether the block moves.
Vertical: N + 15 sin 25 ° = 3 ( 9.81 ) N + 15\sin 25\degree = 3(9.81) N + 15 sin 25° = 3 ( 9.81 )
N = 29.43 − 6.34 = 23.09 { N } N = 29.43 - 6.34 = 23.09\mathrm\{ N\} N = 29.43 − 6.34 = 23.09 { N } Maximum static friction: f s , max = 0.4 × 23.09 = 9.24 N f_{s,\max} = 0.4 \times 23.09 = 9.24\mathrm{ N} f s , m a x = 0.4 × 23.09 = 9.24 N
Horizontal applied force: 15 cos 25 ° = 13.59 N 15\cos 25\degree = 13.59\mathrm{ N} 15 cos 25° = 13.59 N
Since 13.59 > 9.24 13.59 \gt 9.24 13.59 > 9.24
(b) Find the acceleration.
A = { 1 5 cos 25 ° − 0.3 ( 23.09 ) } { 3 } = { 1 3.59 − 6.93 } { 3 } = { 6 .66 } { 3 } = 2.22 { m / s } 2 A = \frac\{15\cos 25\degree - 0.3(23.09)\}\{3\} = \frac\{13.59 - 6.93\}\{3\} = \frac\{6.66\}\{3\} = 2.22\mathrm\{ m/s\}^2 A = 1 { 5 cos 25° − 0.3 ( 23.09 )} { 3 } = 1 { 3.59 − 6.93 } { 3 } = 6 { .66 } { 3 } = 2.22 { m / s } 2 (c) If the applied force is removed, how far does the block slide before stopping?
μ k m g ⋅ d = { 1 } { 2 } m v 2 \mu_k mg \cdot d = \frac\{1\}\{2\}mv^2 μ k m g ⋅ d = 1 { } { 2 } m v 2 Need to find v v v
This depends on how long the force was applied. If the question implies the block has some velocity v 0 v_0 v 0
D = { v 0 2 } { 2 μ k g } = { v 0 2 } { 5.886 } D = \frac\{v_0^2\}\{2\mu_k g\} = \frac\{v_0^2\}\{5.886\} D = v { 0 2 } { 2 μ k g } = v { 0 2 } { 5.886 } Question 9 (Paper 1 style) A 5 k g 5\mathrm{ kg} 5 kg F = 6 t − 2 N F = 6t - 2\mathrm{ N} F = 6 t − 2 N 0 ≤ t ≤ 5 s 0 \le t \le 5\mathrm{ s} 0 ≤ t ≤ 5 s t = 5 s t = 5\mathrm{ s} t = 5 s
Δ p = ∫ 0 5 F d t = ∫ 0 5 ( 6 t − 2 ) d t = [ 3 t 2 − 2 t ] 0 5 = 75 − 10 = 65 { k g } ⋅ { m / s } \Delta p = \int_0^5 F\,dt = \int_0^5 (6t - 2)\,dt = [3t^2 - 2t]_0^5 = 75 - 10 = 65\mathrm\{ kg\}\cdot\mathrm\{m/s\} Δ p = ∫ 0 5 F d t = ∫ 0 5 ( 6 t − 2 ) d t = [ 3 t 2 − 2 t ] 0 5 = 75 − 10 = 65 { k g } ⋅ { m / s } Since p 0 = 0 p_0 = 0 p 0 = 0 p = 65 k g ⋅ m / s p = 65\mathrm{ kg}\cdot\mathrm{m/s} p = 65 kg ⋅ m/s
Question 10 (Paper 2 style) Two trolleys, A (2 k g 2\mathrm{ kg} 2 kg 3 k g 3\mathrm{ kg} 3 kg 4 m / s 4\mathrm{ m/s} 4 m/s
(a) Find the velocity of trolley B.
0 = m A v A + m B v B 0 = m_A v_A + m_B v_B 0 = m A v A + m B v B 0 = 2 ( − 4 ) + 3 v B ⟹ v B = { 8 } { 3 } = 2.67 { m / s } { ( t o t h e r i g h t ) } 0 = 2(-4) + 3v_B \implies v_B = \frac\{8\}\{3\} = 2.67\mathrm\{ m/s\} \mathrm\{ (to the right)\} 0 = 2 ( − 4 ) + 3 v B ⟹ v B = 8 { } { 3 } = 2.67 { m / s } {( t o t h er i g h t )} (b) Calculate the elastic potential energy stored in the spring.
E e = { 1 } { 2 } m A v A 2 + { 1 } { 2 } m B v B 2 = { 1 } { 2 } ( 2 ) ( 16 ) + { 1 } { 2 } ( 3 ) ( { 6 4 } { 9 } ) = 16 + 10.67 = 26.67 { J } E_e = \frac\{1\}\{2\}m_A v_A^2 + \frac\{1\}\{2\}m_B v_B^2 = \frac\{1\}\{2\}(2)(16) + \frac\{1\}\{2\}(3)\!\left(\frac\{64\}\{9\}\right) = 16 + 10.67 = 26.67\mathrm\{ J\} E e = 1 { } { 2 } m A v A 2 + 1 { } { 2 } m B v B 2 = 1 { } { 2 } ( 2 ) ( 16 ) + 1 { } { 2 } ( 3 ) ( 6 { 4 } { 9 } ) = 16 + 10.67 = 26.67 { J } (c) If the spring has spring constant k = 2000 N / m k = 2000\mathrm{ N/m} k = 2000 N/m
E e = { 1 } { 2 } k x 2 ⟹ 26.67 = 1000 x 2 ⟹ x = 0.163 { m } E_e = \frac\{1\}\{2\}kx^2 \implies 26.67 = 1000x^2 \implies x = 0.163\mathrm\{ m\} E e = 1 { } { 2 } k x 2 ⟹ 26.67 = 1000 x 2 ⟹ x = 0.163 { m } For the A-Level treatment of this topic, see Dynamics .
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Common Pitfalls Incorrectly applying F ⃗ = m a ⃗ \vec{F} = m\vec{a} F = m a
Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.
Rounding intermediate answers too early, which compounds errors in multi-step calculations.
Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.
Cross-References Topic Site Link [Forces and Dynamics] A-Level View [Forces and Dynamics] IB View [Forces and Dynamics] DSE View
Worked Examples See Also 