Exam (elaborations) TEST BANK FOR Elementary Mechanics and Thermodynamics By Prof. John W. Norbury (Solutions Manual) SOLUTIONS MANUAL for elementary
... [Show More] 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 [Show Less]