Physics I – Final Exam – Spring 2003 Answer Key Pages 2-3 – Part A –Multiple Choice 1. A 2. A 3. B 4. E Give +2 points for choice F (sign error). 5. A Give +2 points for choice B (sign error). 6. B Give +2 points for choice A (sign error). 7. A Give +2 points for choice E (sign error). Page 4 – Part B – B-1, B-2, B-3 B-1 v1 0 a (2) v 2 30 v1 x 1 0 0 (2) 12 a (2) 2 10 a (8) 6 a a 5 m / s2 2 x 2 x 1 12 (v1 v 2 )(8) 10 4 (10 30) 170 m B-2 Total time is twice the time to drop from maximum height to Jack’s hand (or vice versa). 2h 2 19.6 2 t1 2 Total Time = 4 seconds. h 12 g t 1 g 9 .8 B-3 The answer is B. The time depends only on maximum height, not horizontal velocity. Page 5 – Part B – B-4 & B-5 B-4 p i m v i (2) (5 î ) 10 î p p f p i J 18 î 6 ĵ p f p i J 10 î 18 î 6 ĵ 8 î 6 ĵ v f p f / m (4 î 3 ĵ ) m / s B-5 Constant force: W F d Fx d x Fy d y (8 3) (6 4) 24 24 0 J 1 Page 6 – Part B – B-6 & B-7 B-6 There are no external forces so momentum is conserved. We will call the initial direction of Puck A the +X direction. Pafter Pbefore m v 0 12 m î Total mass of the system is M = 3 m. v cm Pafter / M 12 m î / (3 m) 4 î The answer is the speed or magnitude, which is 4 m/s. B-7 f 0 0 10 2 0.5 rev / s 2 2 2 100 Accept either + or – answer. (Asked for magnitude.) If student gives answer in rad/s2: = – rad/s2. 2 f 0 2 2 2 2 Page 7 – Part B – B-8 & B-9 B-8 I total I1 I 2 I 3 121 M1L2 M 2 r 2 M 3 r 2 1 12 0.6 0.5 2 0.5 0.25 2 0.5 0.25 2 0.075 kg m 2 B-9 Step 1: Use Conservation of Angular Momentum to find new rotation speed: I1 1 1.25 12 15 I 2 2 0.75 2 2 15 / 0.75 20 Step 2: Use Work-Kinetic Energy Theorem to find work done: 2 2 W K 2 K1 12 I 2 2 12 I11 0.5 0.75 20 2 0.5 1.25 12 2 150 90 60 J Page 8 – Part B – B-10 & B-11 B-10 We are only interested in magnitude. F q v B sin( ) 1.6 10 19 5.6 10 6 2.5 10 4 sin( 120) 1.94 10 16 N B-11 mv r qB m v q B r 1.6 10 19 1.0 0.5 8.0 10 20 kg m / s 2 Page 9 – Part B – B-12 – Matching Problem _12_ The Impulse-Momentum Theorem _22_ The Work-Kinetic Energy Theorem _26_ Change in Mechanical Energy from Non-Conservative Forces _25_ Potential Energy of an Ideal Spring __6_ Newton’s 2nd Law of Motion _46_ Newton’s Law of Universal Gravitation _47_ Coulomb’s Law _54_ The Magnetic or Lorentz Force on a Moving Charged Particle _18_ Finding the Scalar or Dot Product of Two Vectors _35_ Finding the Vector or Cross Product (Magnitude) of Two Vectors _44_ Collision in Two Dimensions with Conservation of Momentum _45_ Elastic Collision in One Dimension _39_ Definition of Torque _41_ Definition of Angular Momentum for a Particle Page 10 – Part C – C-1 K U (K f K i ) ( U f U i ) Wnoncons K f K i U f U i Wnoncons K f 0 0 12 k x 2 Ffriction d 0.5 *125 * (0.02) 2 0.0475 * 0.10 0.02025 J 2 Kf 2 * 0.02025 9m/s m 0.0005 2h 2 1 .0 t 0.45 s The time to drop is given by h 12 g t 2 g 9 .8 The horizontal distance = 9 t = 4.066 m. T Page 11 – Part C – C-2 1 2 m v2 Kf v Q E T sin( ) 0 T cos() M g 0 T sin( ) Q E tan( ) T cos() M g T cos() QE T sin() 3 QE 735 1 2 10 tan 1 tan tan Mg 0.3 9.8 1 26.565 1 2 Mg 3 Page 12 – Part C – C-3 & C-4 C-3 Since the two particles are equal mass and the CM is (0,0), the positron location is negative the electron position: X = –4.0 x 10–12 m, Y = –3.0 x 10–12 m. C-4 Since the two particles are equal mass and the velocity of the CM is 0, the positron velocity is negative the electron velocity: Vx = –6.0 x 10+6 m/s, Vy = 0.0 m/s. Page 13 – Part C – C-5 Angular momentum: l r p r (m v) m r v Direction by right-hand rule is into the page (–Z). Angle between r and v is = tan-1(3/4). Magnitude: l m r v sin( ) 9.1 10 31 5.0 10 12 6.0 10 6 sin(tan 1 3 4 ( )) 1.64 10 35 kg m 2 / s X and Y components are zero. Z component is –1.64 x 10-35 kg m2/s Page 14 – Part C – C-6 The problem is solved with conservation of energy at two positions: y = d and y = h. Since sphere A is not moving at the initial and final positions, we have no kinetic energy and two forms of potential energy: electrostatic and gravitational. q2 q2 k Wd k Wh where W = m g and k is the electrostatic constant. d h Note that h = d is a trivial solution and not what we want, but useful to know. Turn this into a quadratic equation for unknown h, multiplying by h and d: W d h 2 (k q 2 W d 2 ) h k q 2 d 0 There are two ways to solve this, the pedestrian way and the elegant way. The pedestrian way: Use the quadratic formula with numbers: (0.036) h 2 (0.03744) h 0.00144 0 0.03744 0.03744 2 4 0.036 0.00144 h 1 m 2 0.036 The elegant way is to realize that we can factor using (h – d) and get (h d) ( W d h k q 2 ) 0 The non-trivial solution is the second root or k q 2 0.036 h 1 m W d 0.036 4