Exam (elaborations) TEST BANK FOR Elementary Mechanics and Thermodynamics By Prof. John W. Norbury (Solutions Manual)

Exam (elaborations) TEST BANK FOR Elementary Mechanics and Thermodynamics By Prof. John W. Norbury (Solutions Manual) 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 Contents 1 MOTION ALONG A STRAIGHT LINE 5 2 VECTORS 15 3 MOTION IN 2 & 3 DIMENSIONS 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 THERMODYNAMICS 93 16 KINETIC THEORY OF GASES 99 3 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 = Ct2 iv) x = Dcos !t v) x = E sin !t where A;B;C;D;E; ! are constants. A) What are the units for A;B;C;D;E; !? 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 msec¡1, C in msec¡2. !t is always an angle, µ is radius and cos µ and sin µ have no units. Thus ! 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 = ¡!D sin !t v) v = !E cos !t 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 = ¡!2Dcos !t v) a = ¡!2E sin !t 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 ¯gures below show position-time graphs. Sketch the corresponding velocity-time and acceleration-time graphs. t x t x t x SOLUTION The velocity-time and acceleration-time graphs are: t v t t v t a t a t a 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 v2 = v2 0 + 2a(y ¡ y0) In the vertical direction we have: v0 = 0, a = ¡g, y0 = H, y = 0. Thus v2 = 0¡ 2g(0 ¡ H) = 2gH ) v = p 2gH 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 ¯rst car, by stepping on the gas pedal and moving at acceleration a. Derive a formula for how long it takes to catch up. (The ¯rst car travels at constant speed v1 and does not accelerate.) SOLUTION Suppose the second car catches up in a time interval t. During that interval, the ¯rst car (which is not accelerating) has travelled a distance d = v1t. The second car also travels this distance d in time t, but the second car is accelerating at a and so it's distance is given by x ¡ x0 = d = v0t + 1 2at2 = v1t = v2t + 1 2at2 because v0 = v2 v1 = v2 + 1 2at ) t = 2(v1 ¡ v2) a 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 1sec = 1 60 £ 60hour v = v0 + at a = v ¡ v0 t = 30 mph ¡ 0 10 sec = 3mph per sec = 3mph 1 sec = 3 mph 1 ( 1 60 £ 1 60hour) = 3£ 60 £ 60 miles hour ¡2 = 10; 800 miles per hour2 13 6. If you throw a ball up vertically at speed V , with what speed does it return to the ground ? Prove your answer using the constant acceleration 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 v2 = v2 0 + 2a(x ¡ x0) and x0 = 0; x= 0; v0 = V ) v2 = V 2 or v = 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 =^i + 2^j and ~t =^j ¡ ^k. SOLUTION ~r:~t ´ j~rjj~tj cos µ = (^i + 2^j ):(^j ¡ ^k) = ^i:^j + 2^j:^j ¡^i:^k ¡ 2^j:^k = 0 + 2 ¡ 0 ¡ 0 = 2 j~rjj~tj cos µ = p 12 + 22 q 12 + (¡1)2 cos µ = p 5 p 2 cos µ = p 10 cos µ ) cos µ = p2 10 = 0:632 ) µ = 50:80 17 2. Evaluate (~r + 2~t ): ~ f where ~r =^i + 2^j and ~t =^j ¡ ^k and ~ f =^i ¡^j . SOLUTION ~r + 2~t = ^i + 2^j + 2(^j ¡ ^k) = ^i + 2^j + 2^j ¡ 2^k = ^i + 4^j ¡ 2^k (~r + 2~t ): ~ f = (^i + 4^j ¡ 2^k):(^i ¡^j) = ^i:^i + 4^j:^i ¡ 2^k:^i ¡^i:^j ¡ 4^j:^j + 2^k:^j = 1 + 0 ¡ 0 ¡ 0 ¡ 4 + 0 = ¡3 18 CHAPTER 2. VECTORS 3. Two vectors are de¯ned as ~u =^j + ^k and ~v =^i +^j. Evaluate: A) ~u +~v B) ~u ¡~v C) ~u:~v D) ~u £~v SOLUTION A) ~u +~v = ^j + ^k +^i +^j =^i + 2^j + ^k B) ~u ¡~v = ^j + ^k ¡^i ¡^j = ¡^i + ^k C) ~u:~v = (^j + ^k):(^i +^j ) = ^j:^i + ^k:^i+ ^j:^j + ^k:^j = 0 + 0 + 1 + 0 = 1 D) ~u £~v = (^j + ^k) £ (^i +^j ) = ^j £^i + ^k £^i +^j £^j + ^k £^j = ¡^k +^j + 0 ¡^i = ¡^i +^j ¡ ^k Chapter 3 MOTION IN 2 & 3 DIMENSIONS 19 20