Total number of questions is 28 but score will be out of 27 (so one question is extra credit ). Questions are numbered

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1 Physics 1 Spring 007 Final Exam PHYSICS 1 Spring 007 FINAL EXAM: May 1st 007 4:0pm 6:0pm Name (printed): ID Number: Section Number: INSTRUCTIONS: Each question is of equal weight, answer all questions. All questions are multiple choice. Choose the best answer to each question. Before turning over this page, put away all materials except for pens, pencils, erasers, rulers, your calculator and aid sheet. An aid sheet is one two sided 8½ 11 page of notes prepared by the student. There is also a list of possibly useful equations at the end of the exam. "In general, any calculator, including calculators that perform graphing numerical analysis functions, is permitted. Electronic devices that can store large amounts of text, data or equations are NOT permitted." If you are unsure whether or not your calculator is allowed for the exam ask your TA. Examples of allowed calculators: Texas Instruments TI-0XII/8/8+/89, 9+ Casio FX115/50HCS/60/7400G/FX7400GPlus/FX9750 Sharp EL9900C. Examples of electronic devices that are not permitted: Any laptop, palmtop, pocket computer, PDA or e-book reader. In marking the multiple-choice bubble sheet use a number pencil. Do NOT use ink. If you did not bring a pencil, ask for one. Fill in your last name, middle initial, and first name. Your ID is the middle 9 digits on your ISU card. Special codes K to L are your recitation section; for the Honors sections please encode your section number as follows: H1 0; H 1 and H 5. If you need to change any entry, you must completely erase your previous entry. Also, circle your answers on this exam. Before handing in your exam, be sure that your answers on your bubble sheet are what you intend them to be. It is strongly suggested that you circle your choices on the question sheet. You may also copy down your answers on a piece of paper to take with you and compare with the posted answers. You may use the table at the end of the exam for this. When you are finished with the exam, place all exam materials, including the bubble sheet, and the exam itself, in your folder and return the folder to your recitation instructor. No cell phone calls allowed. Either turn off your cell phone or leave it at home. Anyone answering a cell phone must hand in their work; their exam is over. Total number of questions is 8 but score will be out of 7 (so one question is extra credit ). Questions are numbered Best of luck, David Atwood and Paula Herrera-Siklody

2 Physics 1 Spring 007 Final Exam Neglect air resistance in all problems. 56. The acceleration of a particle constrained to move along the x-axis 4 is ax = ( 4.0 m/s ) t. At t = 0, the particle is at x = 0 and its velocity is.0 m/s. Where is the particle at t =.0 s? (A) x = 18 m (B) x = 0 m (C) x = 96 m (D) x = 150 m (E) x = 0 m We must use integrals. First, v(t) (dropping units for simplicity): v = v(0) + adt 0 ( ) v= t v =.0 + ( 4.0) t 0 t t dt Then, x(t): x= x(0) + vdt x = + 4 t x = (.0) t + t 0 t 0.0+ ( 4.0) 0 t dt At t =.0 s: 4 (.0) x (.0 s) = (.0 )(.0) + = 18 m

3 Physics 1 Spring 007 Final Exam 57. You want to kick a ball into your neighbor s yard, over the.0-m high fence. The ball leaves the ground at 8.0 m/s and just barely clears the fence when it is at the top of its trajectory. At what angle was the ball moving relative to the horizontal right after the kick? (A) (B) 8 (C) 44 (D) 5 (E) 6 v 0 θ v y = 0 h Using conservation of energy (or the v projectile motion formula): v v = gh 0 But v = v, so x y 0 y 0 0x v v = gh 0 v sin θ = gh sinθ = θ gh v 0 gh (9.8)() 1 1 = sin = sin = 5 v 0 8

4 Physics 1 Spring 007 Final Exam 58. A coin is placed near the edge of a turntable with radius 0 cm. The turntable makes 0 turns in one minute. What is the minimum coefficient of static friction needed to ensure that the coin moves along with the turntable? (A) 0.1 (B) 0.4 (C) 0.51 (D) 0.70 (E) 0.8 The friction is the force that produces the required centripetal acceleration: f = mrω Since static friction cannot be larger than its maximum value, f μsn = μsmg μsmg mrω Rω μs g The angular speed needs to be in rad/s: turns π rad 1 minute π ω = 0 = rad/s minute 1 turn 60 seconds π ( 0.) Rω μs = = 0.1 g 9.8

5 Physics 1 Spring 007 Final Exam 59. Two blocks 1 and (m 1 = 4m ) on an incline are connected through an ideal, massless string that goes through an ideal, massless pulley as shown in the figure. The incline makes an angle of 0 with the horizontal. The blocks are released from rest and there is no friction between the blocks or between block 1 and the incline. T T 1 m g sinθ m 1 g sinθ 0 Find the acceleration of block 1 right after it is released. (A) 0 (B).9 m/s (C) 5.9 m/s (D) 7.4 m/s (E) 9.8 m/s In the free-body diagram shown above, only the forces along the incline have been drawn (the forces perpendicular to the incline normal forces and the rest of the weights- are irrelevant for this problem). Newton s second law for each block, taking positive down for 1 and up for (it s intuitively how the system will move): m1: m1gsinθ T = ma 1 m : T m gsinθ = m a Adding both equations: ( ) mgsinθ mgsinθ = m + m a 1 1 m a= m m + m 1 1 gsinθ Since m 1 = 4m, sin.9 m/s a= g θ = 5

6 Physics 1 Spring 007 Final Exam 60. A 100g ball is launched vertically using a spring as shown below. The spring is compressed by 10cm from its relaxed length so that the ball is at ground level. The ball rises to a maximum height of 1m. What is the force constant, k, of the spring? 1m 100g 10 cm (A) 49 N/m (B) 98N/m (C) 196 N/m (D) 9 N/m (E) 588 N/m Energy is conserved. Since the kinetic energy is zero both at the bottom and at the top, 1 kx = mgh mgh (0.1 kg)(9.8 m/s )(1.0 m) k = = = 196 N/m x (0.1 m)

7 Physics 1 Spring 007 Final Exam 61. Consider the two situations depicted below where a block is attached to an ideal string which winds around the given pulley without slipping. In both cases the mass of the pulley is m, but the blocks have different masses m and M (see figure). The blocks accelerate downwards at the same rate in both cases. What is the ratio M:m? (A) :1 (B) :1 (C) 1:1 (D) 1: (E) 1: Solid uniform disk of mass m T Hoop of mass m (all mass concentrated on rim) T T T m M mg Mg The free-body diagram is formally the same in both cases. The only differences between the two cases is the mass of the block and the moment of inertia of the pulley: I = γmr with γ = in case 1 γ = 1 in case 1 We ll save time if we solve the situation in the more general case where the mass of the block is not equal to the mass of the pulley: τ = Iα with I = γmr F net net = ma a = Rα Rotational part (torque): TR = γmr α a TR = γ mr R T = γ ma [1]

8 Translation of the block: Physics 1 Spring 007 Final Exam Mg T = Ma [] Putting [1] and [] together: Mg γ ma = Ma a g M = M + γ m For the solid disk, M = m and γ = ½, so a= g. For the hoop, γ =1, so a g M =. M + m Both accelerations should be the same: M M = = M + m m

9 Physics 1 Spring 007 Final Exam 6. Two particles of masses m 1 and m = 4m 1 head toward each other along the x axis with velocities v 1x = 0 m/s and v x = 10 m/s. The particles collide elastically. What is the speed of particle 1 right after the collision? (A) 0 (B) 6.0 m/s (C) 10 m/s (D) 5 m/s (E) 4 m/s This is a 1D elastic collision. Total linear momentum is conserved, and total kinetic energy is also conserved. Instead of using conservation of kinetic energy, though, we can use the relation between relative velocities before and after the collision, which will make the algebra easier: mv + mv = mv + mv 1 1i i 1 1f f ( ) v v = v v 1i i 1f f m(0) + 4 m( 10) = mv + 4mv f 1 f ( v1f v f ) 0 ( 10) = 10 = v + 4v 1f f 40 = v + v 1f f Solving this system yields: v 1 f v f = 4 m/s = 6.0 m/s

10 Physics 1 Spring 007 Final Exam 6. John wants to ride a jet ski across a river that is 1.0-km wide and flows from north to south at a rate of 5.0 km/h. Starting at point A on the west bank he would like to drive his jet ski in a straight line to point B due east of point A. If the jet ski can move at a speed of 10 km/h with respect to the water, how long does it take to cross the river? (A) 0.01 hours (B) hours (C) 0.10 hours (D) 0.1 hours (E) 0.0 hours A 5.0 km/h John's trajectory B Let us draw a diagram of the velocities: v ski relative to shore v water relative to shore (5 km/h) v ski relative to water (10 km/h) From the diagram, using Pythagoras s theorem: v = v v v ski, shore ski, water water, shore ( ) ( ), = 10 km/h 5 km/h = 8.7 km/h ski shore Thus the time to cross the river is: d 1 km t = = = 0.1 h v 8.7 km/h ski, shore

11 Physics 1 Spring 007 Final Exam 64. A uniform sphere of mass 5M and radius R is glued to a uniform rod of mass M and length R. The rod is perpendicular to the surface of the sphere. The system rotates about axis P, which is perpendicular to the page through a point in the surface of the sphere (see figure). What is the moment of inertia of this rigid body for this axis? (A) MR (B) 6MR (C) 8MR (D) 11MR (E) 5MR P R R 5M M We need the moment of inertia about an axis through the center of mass for each object (sphere and rod) and the corresponding parallel axis theorem correction for each. Isphere = (5 M ) R + ( 5 M) R = 7 MR 5 1 Irod = ( M ) ( R) + M ( R) = 8 MR 1 I total = 5MR

12 Physics 1 Spring 007 Final Exam Net Force 65. The graph below shows a position-dependent net force acting on a 1.0-kg block moving along the x-axis. Initially the block is at x = 0 m, moving in the +x direction. When it reaches 8.0 m, it has a kinetic energy of 0 J. What was the initial kinetic energy of the block? (A) 1 J (B) 18 J (C) 4 J (D) 6 J (E) 48 J N N 1N 1m m m 4m 5m 6m 7m 8m x Work-kinetic-energy theorem: W=K f K i The work is the area under the F-x curve between x = 0 and x = 8 m.. Breaking the figure into two triangles and one rectangle: 1 1 W = ( m)( N) + ( m)( N) + ( m)( N) = 1 J The initial kinetic energy is thus: K = K W = 0 J 1 J = 18 J i f

13 Physics 1 Spring 007 Final Exam 66. A uniform plank of length 4.0 m and mass 4.0 kg is placed on the edge of a table with 75% of its length on the table. We want to place a 6.0 kg cube of side 1.0 m on the free end of the plank without tipping over the system. What is the maximum allowed value of d? 1.0 m 4.0 kg x = 0 d? 6.0 kg CM CM x 4.0 m 1.0 m (A) 0.05 m (B) 0.17 (C) 0. (D) 0.67 (E) The system tips over for any (positive) value of d. In order to have equilibrium, the center of mass of the whole system (plank + block) must be on the table. If we take +x to the right and x=0 at the edge of the table (see figure), the condition is x CM 0. In this reference frame, the centers of mass of the plank and block are respectively x = 1.0 m x = d m plank The CM of the entire system: x CM x m + x m = m + m plank plank block block plank block block x 0 x m + x m 0 CM plank plank block block ( 1.0 m)(4.0 kg) + ( d m)(6.0 kg) 0 d 0.17 m

14 Physics 1 Spring 007 Final Exam 67. A 1.0-kg particle attached to a spring with k = 400 N/m oscillates with simple harmonic motion along the x axis. As it passes the equilibrium position, the velocity of the particle is v x =.0 m/s. What is the position of the particle.0 s after? (A) x = 6.7 cm (B) x =. cm (C) x = +1. cm (D) x = +7.5 cm (E) x = +10 cm As a function of time, the velocity of the particle is: v= Aω ωt+ ϕ sin ( ) We know Aω because this is the speed of the particle at the equilibrium position (and the maximum possible speed), vmax = Aω =.0 m/s k (400 N/m) 1 We can find ω = = = 0 s m 1.0 kg vmax.0 m/s Therefore, the amplitude is A = = = 0.10 m = 10 cm 1 ω 0 s But we need the phase angle φ. For this, we use the initial conditions on position and velocity. At t=0, ( ϕ) ( ) 0 = Acos 0 + cosϕ = 0.0 = Aωsin 0 + ϕ sinϕ < 0 The solution to this system is π ϕ = We can then go back to the position equation: ( ω ϕ) x= Acos t+ 1 π x= ( 10 cm) cos ( 0 s ) t x(.0 s) = 7.5 cm

15 Physics 1 Spring 007 Final Exam The solar system of the Ugly Star (mass 10 0 kg) has two planets, 1 and, that have circular orbits of radii r 1 = m and r = r 1. A 1000-kg spacecraft is to be sent from planet 1 to planet. The spacecraft leaves planet 1 when this is in the position shown in the figure, and reaches planet on the other side of the star, traveling along the trajectory shown in the figure below. Note that the velocity vector of the spacecraft is parallel to the velocity vector of planet 1 at launch and parallel to the velocity vector of planet on arrival. Planet 1 (position upon launching) Ugly Star Planet (position upon spacecraft arrival) spacecraft 68. How long will the spacecraft trip take? (A) 700 hours (B) 650 hours (C) 600 hours (D) 1100 hours (E) 900 hours The path of the satellite is half of an elliptical orbit about the Ugly Star with semimajor axis a = = m. Thus the time needed for the trip is half of r1+ r 11 the period of the entire orbit, t trip T = π = a GM Ugly Star 11 ( m) 11 0 ( Nm /kg )( 10 kg) = π = 600 h

16 Physics 1 Spring 007 Final Exam The situation below refers to the next two questions: Two charges are fixed at the positions shown below. y Q 1 =.0μC Q =.0μC Q 1 Q x x =.0 cm 69. Find the force on Q 1 by Q. 7 (A).0 10 i N 7 (B).0 10 i N (C) 90 i N (D) 90 i N (E) None of the above The interaction is attractive, so the force on Q 1 must point to the right (ie, direction +x). So (B) and (D) cannot be the right answer. The magnitude of the force is given by Coulomb s law: F = k e QQ 1 r 9 ( Nm /C ) = = 90 N Thus the answer is (C). 6 6 (.0 10 C)(.0 10 C) (.0 10 m)

17 Physics 1 Spring 007 Final Exam 70. Determine the y-component of the total electric field at point (x, y) = (0,.0 cm). (A) E y = 0 (B) E y = N/C (C) E y = N/C (D) E y = N/C (E) E y = N/C We need to add the E fields by Q 1 and by Q : y E 1 a E a Q 1 =.0μC Q =.0μC Q 1 Q x a =.0 cm The magnitude of each of these electric fields is: E E Q k a = e = = Q = ke = = ( a ) (.0 10 C) ( Nm /C ).0 10 N/C (.0 10 m) We only need the y components: (.0 10 C) ( Nm /C ) N/C (.0 10 m) E = E E = y 7 1 cos N/C

18 Physics 1 Spring 007 Final Exam 71. Shown below are four conducting spheres, each with a total charge Q. Point P is in each case at the same distance from the center of the sphere. In cases 1 and, point P is outside the spheres. In cases and 4, it is inside the spheres. (Each sphere is to be considered an isolated system, very far from any other charged object) 1 4 P P P P Rank the magnitude of the electric field at point P in each case. (A) E 1 = E = E = E 4 (B) E 1 < E = E = E 4 (C) E 1 < E < E < E 4 (D) E 1 = E > E 4 > E (E) E 1 = E > E = E 4 E = E 4 = 0 because these points are inside a conductor in equilibrium. For 1 and, we can use the spherical shell theorem (the electric field produced by a spherically symmetric charge distribution, at points outside the charged object, is the same as if all the charge was concentrated at the center of the system). Thus for both 1 and, the electric field is the same (same total charge Q, same distance from the center of the sphere to point P), but not zero (unless Q = 0).

19 Physics 1 Spring 007 Final Exam 7. A sphere of radius R has a charge Q uniformly distributed throughout its volume. Consider an imaginary sphere of radius R/ with center located at distance R/ from the real sphere (see figure). What is the electric flux through the imaginary sphere? (A) 0 Q (B) (C) (D) (E) ε 0 Q ε 0 Q 4ε 0 Q 8ε 0 R R/ R/ This is a straightforward application of Gauss s law. The charge enclosed by the little, imaginary sphere is q enclosed = (charge density)(volume sphere) Q 4 R = π 4 π R Q = 8 qenclosed Q Thus Φ= = ε 8ε 0 0

20 Physics 1 Spring 007 Final Exam 7. Two very thin surfaces with equal uniform charge density σ = nc/m are placed parallel to the yz axes, at x 1 = 0 and at x =.0 cm, respectively. If we take the electric potential to be zero at x = 0, what is the electric potential at x = 6.0 cm? (A) 0 V (B) 60 V (C) 60 V (D) 40 V (E) 40 V σ σ x =.0 cm x = 6.0 cm x The electric field is the superposition of the contribution from each sheet ( appropriate direction). Overall: σ ε 0 in the E x σ x < 0 ε0 = 0 0 < x< cm σ x > cm ε 0 Since E = 0 between the plates, the potential is constant in that region. Thus the potential difference between 0 and 6 cm is the same as the potential difference between cm and 6 cm. For x > cm, the electric field is uniform, so σ Δ V = ExΔ x= Δ x= = ε Nm /C C/m 1 (6 ) 10 m 40 V This is just the magnitude of the change in potential. Since the electric field points in the positive x direction, the potential must be decreasing in that direction, so V(6 cm) < V( cm). Since V( cm ) = V(0) = 0, we conclude V(6 cm) = -40 V.

21 Physics 1 Spring 007 Final Exam 74. The figure below shows the electric field lines produced by an electric dipole. Compare the electric potential at points A, B and C. B A C (A) V B < V A < V C (B) V A = V B < V C (C) V B < V C < V A (D) V A > V B = V C (E) V A < V B = V C All the points on the dotted line have the same potential because the electric field is perpendicular to the line (no electric field component along the line means no changes in potential). So V B = V C. An electric field line flows from B to A. Thus for a trajectory along this line, the electric potential is constantly decreasing (the electric field always points in the direction of decreasing potential). Thus V B > V A.

22 Physics 1 Spring 007 Final Exam 75. Three capacitors are connected to an ideal battery as shown below. When the capacitors are all fully charged, a dielectric with κ > 1 is inserted between the plates of capacitor C. Due to this, and compared with the values before the dielectric was inserted, the charge in C 1 and the charge in C. C V 0 C 1 C (A) decreases, decreases (B) decreases, remains the same (C) increases, remains the same (D) remains the same, increases (E) remains the same, decreases No matter what we do with C, the potential difference between the plates of C 1 is fixed by the battery, and thus remains the same. Therefore the charge on C 1 remains the same. The charge on C must be the same as the charge on C, and the same as that of the combined equivalent capacitor C - (series connection). Since the capacitance of C increases due to the dielectric, C - also increases. But again, the potential difference across C - is fixed by the battery. So the charge must increase, Q - = C - V 0.

23 Physics 1 Spring 007 Final Exam 76. An electric dipole made of two charges ± C separated a distance m is placed between the plates of a parallel plate capacitor that is connected to a 9.0-V battery. The plates of the capacitor have an area of 400 m and are separated by an air-filled gap 0.10 mm wide. The dipole moment makes an angle of 60 with the plates of the capacitor. Determine the magnitude of the torque on the dipole. (A) 0 (B) Nm (C) Nm (D) Nm (E) Nm 60 p E d The dipole moment is ( )( ) p= Qd between charges =.0 10 C m =.0 10 Cm The electric field between the plates is dictated by the potential difference between its plates and the distance between them: V 9.0 V E = = = d m N/C τ The torque on the dipole is = Epsin ( angle between E and p) 4 4 ( )( ) = N/C.0 10 C sin N =

24 Physics 1 Spring 007 Final Exam The situation below refers to the next two questions: An ideal battery and four resistors are connected as shown below. All wires are ideal. The ideal ammeter reads 0.40 A. R 1 R R 1 = 10 Ω R = 0 Ω R = 0 Ω R 4 = 40 Ω R A R 4 V Find the equivalent resistance of the circuit. (A) 6.7 Ω (B) 17 Ω (C) 6 Ω (D) 55 Ω (E) 100 Ω Resistors 1 and are in series: R1 = R1+ R= 0 Ω Resistor R 1- is in parallel to R : = + R1 R1 R R1 = 15 Ω Resistor R 1-- is in series with R 4 : R = R + R = Ω all

25 Physics 1 Spring 007 Final Exam 78. Determine the power dissipated in resistor R. (A) 1. W (B).1 W (C) 4.8 W (D) 8.8 W (E) 16 W Since both R 1- and R are 0 Ω each, the current splits evenly between the two branches. The current through R is thus 0.40/ = 0.0 A. The power dissipated in R is thus: P ( ) ( ) = IR = 0.0 A 0 Ω = 1. W

26 Physics 1 Spring 007 Final Exam 79. In the circuit below, all resistors are 10 Ω and all batteries are 10 V. The wires and the battery are ideal. Determine the current through the central wire. I 1 I I (A) 0. A (B) 0.67 A (C) 1.0 A (D) 1. A (E).0 A Due to the left-right symmetry of the circuit, I 1 = I, so in fact we only have two unknowns. We need to equations using Kirchhoff s laws. Junction equation: I1+ I = I I1 = I [1] Loop equation (see loop in the figure): ( ) ( ) ( ) 10 V 10 Ω I + 10 V 10 Ω I 10 Ω I = 0 0 0I 10I = I I = 0 [] 1 1 Using [1] and [], I = 0 I = 1.0 A

27 Physics 1 Spring 007 Final Exam Laboratory Final 80. Consider the motion of a cart like the one you used in lab, and which is adjusted to have significant frictional drag. Assume that the friction is well described by the "laws" of (dry) friction that you studied this semester. Also assume that the cart is given an initial velocity toward the left on the horizontal track and then released. (Assume the sensor gives positions relative to the X axis illustrated in the figure). 0 X Which of the following best illustrates the acceleration, a X, versus time of the cart during the same time interval? a x a x a x a x a x 0 t t t t t A B C D E If the cart is moving to the left, the frictional force points to the right (i.e., is positive). Thus the acceleration will also be positive. Since the magnitude of the frictional force is constant, so is the acceleration.

28 Physics 1 Spring 007 Final Exam 81. Consider a hollow metal sphere mounted on a thin insulating rod. Using standard apparatus (e.g., an electrophorus), the largest possible electric charge is placed on the sphere. Where is the electric field largest, and for a sphere of a given size, what factor determines the magnitude of the maximum charge? location where electric field is largest factor that determines the magnitude of the maximum charge A center of sphere dielectric strength of air B surface of sphere type of metal C surface of sphere dielectric strength of air D center of sphere type of metal E none of the above The electric field is strongest on the surface. It decreases as you move away from the sphere, and it is zero anywhere inside the sphere because it is a conductor. The maximum charge is dictated by the maximum electric field that the insulator around the sphere (air) can withstand without becoming a conductor (dielectric breakdown). This maximum electric field is what is called the dielectric strength of air.

29 Physics 1 Spring 007 Final Exam 8. Using the rotating wheel apparatus such as you used in lab, a disk (which is not spinning) is dropped concentrically upon the wheel as it is rotating freely. Assume that the disk has a moment of inertia of 0.5I 0, where I 0 is the moment of inertia of the rotating wheel. Which of the following graphs best represents the angular velocity, ω, as a function of time before, during, and after the disk is dropped? ω A ω B C D ω ω ω E t t t t t When the disk is dropped on the wheel, no external torque is exerted, so angular momentum is conserved: ( ) I ω = I + I ω wheel 0 wheel disk f I I ω = ω = ω = ω f wheel Iwheel + Idisk I I0 So the answer is D. Note that B is qualitatively correct, too, but the angular speed drops to about 1/ of its initial value instead of /. The value of ω is decreasing all the time (slight slope in the lines) because of friction and air resistance.

30 Physics 1 Spring 007 Final Exam 8. Consider two air pucks (made of some unknown material) which can slide upon a smooth horizontal surface with little friction, leaving trails of spark marks, with marks produced at a fixed frequency. The pucks are of equal mass, and are pushed (and released) toward one another, and then collide. Assume that the pucks rotate little before and after the collision. For the record shown above, select the comment listed below which is most appropriate. (A) The data looks O.K., although the collision clearly is not elastic. (B) The data must be invalid since clearly momentum is not conserved. (C) The data must be invalid since the collision clearly is not elastic. (D) The data looks O.K.; both momentum and mechanical energy appear to be conserved. (E) One must know the time between sparks to make a definitive statement. NOTE: By the data being invalid is meant that, for the situation described, such data is impossible. To obtain such data, then some large extraneous factor must be at work (such as hidden magnets, angels, etc.) Both pucks seem to be moving at approximately equal speed (the spacing of the dots is the same for both) before and after the collision. From the trajectories before the collision, the net linear momentum should be zero. However, after the collision, the total linear momentum seems to be nonzero vector pointing down and to the left. Thus something went wrong in our experiment, because linear momentum appears to not be conserved.