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SOLUTIONS MANUAL for elementary mechanics & thermodynamics Professor John W. Norbury Physics Department University of Wisconsin-Milwaukee P.O. Box 413 Milwaukee, WI 53201 November 20, 2000 2 Contents 1 MOTION ALONG A STRAIGHT LINE 5 2 VECTORS 15 3 MOTION IN2&3DIMENSIONS 19 4 FORCE & MOTION - I 35 5 FORCE & MOTION - II 37 6 KINETIC ENERGY & WORK 51 7 POTENTIAL ENERGY & CONSERVATION OF ENERGY 53 8 SYSTEMS OF PARTICLES 57 9 COLLISIONS 61 10 ROTATION 65 11 ROLLING, TORQUE & ANGULAR MOMENTUM 75 12 OSCILLATIONS 77 13 WAVES - I 85 14 WAVES - II 87 15 TEMPERATURE, HEAT & 1ST LAW OF THERMODY- NAMICS 93 16 KINETIC THEORY OF GASES 99 3 4 CONTENTS 17 Review of Calculus 103 Chapter 1 MOTION ALONG A STRAIGHT LINE 5 6 CHAPTER 1. MOTION ALONG A STRAIGHT LINE 1. The following functions give the position as a function of time: i) x = A ii) x = Bt iii) x = Ct 2 iv) x = D cos ωt v) x = E sinωt where A, B, C, D, E, ω are constants. A) What are the units for A, B, C, D, E, ω? B) Write down the velocity and acceleration equations as a function of time. Indicate for what functions the acceleration is constant. C) Sketch graphs of x, v, a as a function of time. SOLUTION A) X is always in m. Thus we must have A in m; B in m sec −1 , C in m sec −2 . ωt is always an angle, θ is radius and cos θ and sin θ have no units. Thus ω must be sec −1 or radians sec −1 . D and E must be m. B) v = dx dt and a = dv dt .Thus i) v = 0 ii) v = B iii) v = Ct iv) v = −ωD sin ωt v) v = ωE cos ωt and notice that the units we worked out in part A) are all consistent with v having units of m· sec −1 . Similarly i) a = 0 ii) a = 0 iii) a = C iv) a = −ω 2 D cos ωt v) a = −ω 2 E sin ωt 7 i) ii) iii) x t v a x x v v a a t t t t t t t t C) 8 CHAPTER 1. MOTION ALONG A STRAIGHT LINE iv) v) 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 x 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 x 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 v 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 v 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 a 0 1 2 3 4 5 6 t -1 -0.5 0 0.5 1 a 9 2. The figures below show position-time graphs. Sketch the correspond- ing velocity-time and acceleration-time graphs. t x t x t x SOLUTION The velocity-time and acceleration-time graphs are: t v t tt t v t t a t t a t t a t v 10 CHAPTER 1. MOTION ALONG A STRAIGHT LINE 3. If you drop an object from a height H above the ground, work out a formula for the speed with which the object hits the ground. SOLUTION v 2 = v 2 0 +2a(y − y 0 ) In the vertical direction we have: v 0 =0, a = −g, y 0 = H, y =0. Thus v 2 =0− 2g(0 −H) =2gH ⇒ v = 2gH [...]... i j) . (j ˆ = ˆ ˆ + 2ˆ ˆ − ˆ k − 2ˆ k i .j j .j i.ˆ j. ˆ = 0+2−0−0 = 2 |r||t| cos θ = 12 + 22 12 + (−1)2 cos θ √ √ = 5 2 cos θ √ 10 cos θ = 2 √ = 0.632 10 ⇒ θ = 50.80 ⇒ cos θ = 17 2 Evaluate (r + 2t ).f where r = ˆ + 2ˆ and t = ˆ − k and f = ˆ − ˆ i j j ˆ i j SOLUTION r + 2t = ˆ + 2ˆ + 2(ˆ − k) i j j ˆ ˆ = ˆ + 2ˆ + 2ˆ − 2k i j j ˆ = ˆ + 4ˆ − 2k i j (r + 2t ).f ˆ i j) = (ˆ + 4ˆ − 2k).(ˆ − ˆ i j ˆ i i .j j. .. + 2k.ˆ i.i j. i j. j = 1+0−0−0−4+0 = −3 18 CHAPTER 2 VECTORS 3 Two vectors are defined as u = ˆ + k and v = ˆ + ˆ Evaluate: j ˆ i j A) u + v B) u − v C) u.v D) u × v SOLUTION A) u + v = ˆ + k + ˆ + ˆ = ˆ + 2ˆ + k j ˆ i j i j ˆ B) u − v = ˆ + k − ˆ − ˆ = −ˆ + k j ˆ i j i ˆ C) u.v = (ˆ + k).(ˆ + ˆ j ˆ i j) = ˆ ˆ + k.ˆ + ˆ ˆ + k.ˆ j. i ˆ i j. j ˆ j = 0+0+1+0 = 1 D) u × v = (ˆ + k) × (ˆ + ˆ j ˆ i j) = ˆ ׈+... (ˆ + ˆ j ˆ i j) = ˆ ׈+ k ׈+ ˆ × ˆ + k × ˆ j i ˆ i j j ˆ j ˆ j = −k + ˆ + 0 − ˆ i = −ˆ + ˆ − k i j ˆ Chapter 3 MOTION IN 2 & 3 DIMENSIONS 19 20 CHAPTER 3 MOTION IN 2 & 3 DIMENSIONS 1 A) A projectile is fired with an initial speed vo at an angle θ with respect to the horizontal Neglect air resistance and derive a formula for the horizontal range R, of the projectile (Your formula should make no explicit... equations, and neglect air resistance SOLUTION We would guess that the ball returns to the ground at the same speed V , and we can actually prove this The equation of motion is 2 v 2 = v0 + 2a(x − x0 ) and ⇒ v or 2 = V v = V 2 x0 = 0, x = 0, v0 = V 14 CHAPTER 1 MOTION ALONG A STRAIGHT LINE Chapter 2 VECTORS 15 16 CHAPTER 2 VECTORS 1 Calculate the angle between the vectors r = ˆ + 2ˆ and t = ˆ − k i j j ˆ... DIMENSIONS 2 A projectile is fired with an initial speed vo at an angle θ with respect to the horizontal Neglect air resistance and derive a formula for the maximum height H, that the projectile reaches (Your formula should make no explicit reference to time, t) SOLUTION v0 height, H v0 y θ v0 x We wish to find the maximum height H At that point vy = 0 Also in the y direction we have ay = −g and H ≡ y − y0... distance d in time t, but the second car is accelerating at a and so it’s distance is given by 1 x − x0 = d = v0 t + at2 2 1 = v1 t = v2 t + at2 2 1 v1 = v2 + at 2 ⇒t = 2(v1 − v2 ) a because v0 = v2 12 CHAPTER 1 MOTION ALONG A STRAIGHT LINE 5 If you start your car from rest and accelerate to 30mph in 10 seconds, what is your acceleration in mph per sec and in miles per hour2 ? SOLUTION 1hour = 60 × 60sec 1...11 4 A car is travelling at constant speed v1 and passes a second car moving at speed v2 The instant it passes, the driver of the second car decides to try to catch up to the first car, by stepping on the gas pedal and moving at acceleration a Derive a formula for how long it takes to catch up (The first car travels at constant speed v1 and does not accelerate.) SOLUTION Suppose the second car... rifle in order to hit the bulls-eye Assume the bullet leaves the rifle with speed v0 B) How much bigger is L compared to the projectile height H ? Note: In this problem use previous results found for the range R and height H, namely R = 2 v0 sin 2θ g = 2 2v0 sin θ cos θ v 2 sin2 θ and H = 0 2g g SOLUTION L height, H θ range, R 24 CHAPTER 3 MOTION IN 2 & 3 DIMENSIONS A) From previous work we found the range... planet of mass M and radius R at an altitude of H Derive a formula for the additional speed that the satellite must acquire to completely escape from the planet Check that your answer has the correct units SOLUTION The gravitational potential energy is U = −G Mrm where m is the mass of the satellite and r = R + H Conservation of energy is Ui + Ki = Uf + Kf To escape to infinity then Uf = 0 and Kf = 0 (satellite... in circular motion of radius R and period T Due to the centrifugal force, the spring stretches by a certain amount x from its equilibrium position Derive a formula for x in terms of k, R and T Check that x has the correct units SOLUTION ΣF = ma mv 2 kx = r m( 2πR )2 mv 2 4π 2 mR T x = = = kR kR kT 2 Check units: The units of k are N m−1 (because F = −kx for a spring), and 2 N ≡ kg m Thus 4π mR has . 5 6 t -1 -0 .5 0 0.5 1 x 0 1 2 3 4 5 6 t -1 -0 .5 0 0.5 1 x 0 1 2 3 4 5 6 t -1 -0 .5 0 0.5 1 v 0 1 2 3 4 5 6 t -1 -0 .5 0 0.5 1 v 0 1 2 3 4 5 6 t -1 -0 .5 0 0.5 1 a 0. = ˆ i +2 ˆ j and t = ˆ j − ˆ k and f = ˆ i − ˆ j. SOLUTION r +2 t = ˆ i +2 ˆ j +2( ˆ j − ˆ k) = ˆ i +2 ˆ j +2 ˆ j − 2 ˆ k = ˆ i +4 ˆ j − 2 ˆ k (r