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Browse subjects › Physics Form 4 › Chapter 2: Force and Motion I
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Form 4 · Physics · iSchool.myChapter 2: Force andMotion I2.1 Linear Motion–Distance andDisplacement–DistanceDisplacementPath dependenceSpeed, Velocity andAcceleration–Speed — v = d/t–v = d/tSymbols:v = speed (m s⁻¹)d = distance (m)t = time (s)Worked example: average speedDistance = 150 m, time = 20 sSpeed = distance/time = 150/20Answer: 7.5 m s⁻¹Velocity — v = s/t–v = s/tSymbols:v = velocity (m s⁻¹)s = displacement (m)t = time (s)Worked example: velocity with directionDisplacement = 80 m north, time = 10 sv = 80/10Answer: 8.0 m s⁻¹ northAcceleration — a = (v −u)/t–a = (v − u)/tSymbols:a = acceleration (m s⁻²)u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)t = time (s)Worked example: decelerationu = 20 m s⁻¹, v = 8 m s⁻¹, t = 4 sa = (8 − 20)/4Answer: a = −3.0 m s⁻²; deceleration = 3.0m s⁻²Uniform velocityUniform accelerationEquations of UniformAcceleration–First equation — v = u +at–v = u + atSymbols:u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)a = acceleration (m s⁻²)t = time (s)Worked example: use v = u + atu = 5 m s⁻¹, a = 3 m s⁻², t = 4 sv = 5 + 3(4)Answer: v = 17 m s⁻¹Second equation — s = 1/2(u + v)t–s = 1/2 (u + v)tSymbols:s = displacement (m)u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)t = time (s)Worked example: use s = ½(u + v)tu = 4 m s⁻¹, v = 16 m s⁻¹, t = 5 ss = ½(4 + 16)(5)Answer: s = 50 mThird equation — s = ut +1/2 at²–s = ut + 1/2 at²Symbols:s = displacement (m)u = initial velocity (m s⁻¹)a = acceleration (m s⁻²)t = time (s)Worked example: use s = ut + ½at²u = 2 m s⁻¹, a = 4 m s⁻², t = 3 ss = 2(3) + ½(4)(3²)Answer: s = 24 mFourth equation — v² = u²+ 2as–v² = u² + 2asSymbols:u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)a = acceleration (m s⁻²)s = displacement (m)Worked example: use v² = u² + 2asu = 0, a = 5 m s⁻², s = 20 mv² = 0 + 2(5)(20) = 200Answer: v = √200 = 14.1 m s⁻¹Choosing equationDerivation: SUVATequation links–From a = (v − u)/t → v = u+ atAverage velocity = (u +v)/2 → s = ½(u + v)tSubstitute v = u + at intos = ½(u + v)t → s = ut + ½at²Eliminate t between v = u+ at and s = ½(u + v)t → v² = u² + 2asTicker Tape Motion–Equal spacingIncreasing spacingDecreasing spacing2.2 Linear MotionGraphs–Displacement-TimeGraph–Horizontal linePositive gradientNegative gradientCurved graphWorked example: velocity from an s–tgraphDisplacement increases from 10 m to 70 min 12 sv = gradient = (70 − 10)/12Answer: v = 5.0 m s⁻¹Velocity-Time Graph–Horizontal linePositive gradientNegative gradientArea ruleDistance vs displacementWorked example: acceleration anddisplacement from a v–t graphVelocity rises uniformly from 0 to 20 ms⁻¹ in 5 sAcceleration = gradient = 20/5 = 4.0 m s⁻²Displacement = area = ½(5)(20) = 50 mAcceleration-TimeGraph–Zero accelerationPositive accelerationNegative accelerationWorked example: change in velocity from ana–t graphAcceleration = 3.0 m s⁻² for 4.0 sΔv = area = at = 3.0(4.0)Answer: Δv = 12 m s⁻¹Converting BetweenGraphs–s-t to v-tv-t to s-tv-t to a-tExam strategy2.3 Free Fall Motion–DefinitionGravitationalaccelerationVacuum resultFree Fall Equations–Velocity — v = u + gt–v = u + gtSymbols:u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)g = gravitational acceleration (m s⁻²)t = time (s)Worked example: speed of a dropped objectafter 2.0 sTake downward as positive: u = 0, g = 9.81m s⁻², t = 2.0 sv = u + gt = 0 + 9.81(2.0)Answer: v = 19.6 m s⁻¹ downwardDisplacement — s = ut +1/2 gt²–s = ut + 1/2 gt²Symbols:s = displacement (m)u = initial velocity (m s⁻¹)g = gravitational acceleration (m s⁻²)t = time (s)Worked example: height fallen in 3.0 su = 0, g = 9.81 m s⁻², t = 3.0 ss = ½gt² = ½(9.81)(3.0²)Answer: s = 44.1 mNo time — v² = u² + 2gs–v² = u² + 2gsSymbols:u = initial velocity (m s⁻¹)v = final velocity (m s⁻¹)g = gravitational acceleration (m s⁻²)s = displacement (m)Worked example: maximum height of anupward throwTake upward as positive: u = 20 m s⁻¹, v =0, g = −9.81 m s⁻²0 = 20² + 2(−9.81)sAnswer: s = 20.4 mSign Convention–Thrown upwardDropped objectMaximum height2.4 Inertia–Newton's First LawMass and inertiaDaily Examples–Table cloth trickCar stops suddenlyRoller coasterSafety Applications–Seat beltHeadrestAirbag2.5 Momentum–Momentum — p = mv–p = mvSymbols:p = momentum (kg m s⁻¹)m = mass (kg)v = velocity (m s⁻¹)Worked example: momentum of a movingobjectm = 0.50 kg, v = 12 m s⁻¹ eastp = mv = 0.50(12)Answer: p = 6.0 kg m s⁻¹ eastConservation ofMomentum–Two-body collision —m1u1 + m2u2 = m1v1 + m2v2–m1u1 + m2u2 = m1v1 + m2v2Symbols:m₁, m₂ = masses (kg)u₁, u₂ = initial velocities (m s⁻¹)v₁, v₂ = final velocities (m s⁻¹)Worked example: perfectly inelasticcollisionm₁ = 2 kg, u₁ = 6 m s⁻¹; m₂ = 1 kg, u₂ = 0Before = after: 2(6) + 1(0) = (2 + 1)vAnswer: v = 4.0 m s⁻¹ in the originaldirectionClosed systemTotal before = totalafterCollision Types–Elastic ideaInelastic ideaExplosion/recoilApplications2.6 Force–Newton's Second Law— F = ma–F = maSymbols:F = resultant force (N)m = mass (kg)a = acceleration (m s⁻²)Worked example: force from accelerationm = 800 kg, a = 2.5 m s⁻²F = ma = 800(2.5)Answer: F = 2.0 × 10³ NResultant forceBalanced forcesUnbalanced forcesNewton's Third Law–WalkingRowingRocket2.7 Impulse andImpulsive Force–Impulse — J = Ft = mv −mu–J = Ft = mv − muSymbols:J = impulse (N s)F = average force (N)t = contact time (s)m = mass (kg)u, v = initial and final velocities (m s⁻¹)Worked example: impulse from change inmomentumm = 0.20 kg, u = −15 m s⁻¹, v = 25 m s⁻¹J = Δp = m(v − u) = 0.20[25 − (−15)]Answer: J = 8.0 N s in the positivedirectionImpulsive force — F =(mv − mu)/t–F = (mv − mu)/tSymbols:F = average impulsive force (N)m = mass (kg)u, v = initial and final velocities (m s⁻¹)t = contact time (s)Worked example: average impulsive forceImpulse J = 8.0 N s, contact time Δt = 0.040sF = J/Δt = 8.0/0.040Answer: F = 200 NReducing ImpulsiveForce–Car crumple zoneAirbag and cushionBending kneesIncreasing ImpulsiveForce–Hammering nailGolf or tennis hit2.8 Weight–Weight — W = mg–W = mgSymbols:W = weight (N)m = mass (kg)g = gravitational field strength (N kg⁻¹)Worked example: weight on Earthm = 65 kg, g = 9.81 m s⁻²W = mg = 65(9.81)Answer: W = 638 N downwardMassGravitational fieldstrengthMoon comparison