* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Student Class ______ Date ______ MULTIPLE
Length contraction wikipedia , lookup
Roche limit wikipedia , lookup
Jerk (physics) wikipedia , lookup
Aristotelian physics wikipedia , lookup
Classical mechanics wikipedia , lookup
Internal energy wikipedia , lookup
Electromagnetic mass wikipedia , lookup
Casimir effect wikipedia , lookup
Density of states wikipedia , lookup
Mass versus weight wikipedia , lookup
Faster-than-light wikipedia , lookup
Weightlessness wikipedia , lookup
Negative mass wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Conservation of energy wikipedia , lookup
Lorentz force wikipedia , lookup
Potential energy wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Anti-gravity wikipedia , lookup
Speed of gravity wikipedia , lookup
Time in physics wikipedia , lookup
Electrostatics wikipedia , lookup
Student _________________ Class ___________ MULTIPLE-CHOICE QUESTIONS 1. Scalar is to vector as (1) speed is to velocity (2) displacement is to distance (3) displacement is to velocity (4) speed is to distance 2. If a car accelerates uniformly from rest to 15 meters per second over a distance of 100. meters, the magnitude of the car’s acceleration is (1) 0.15 m/s2 (2) 1.1 m/s2 (3) 2.3 m/s2 (4) 6.7 m/s2 3. An object accelerates uniformly from 3.0 meters per second east to 8.0 meters per second east in 2.0 seconds. What is the magnitude of the acceleration of the object? Date _____________ EXPLAIN WHY 1. Speed is a scalar, while velocity is a vector. Therefore the answer is (1). Also, distance is a scalar, while displacement is a vector. 2. Formula from the Reference Table is: vf2= vi2 + 2ad From rest it means vi=0 m/s; vf=15 m/s and d=100. m, therefore: (15m/s)2 = 2(a)(100.m) or a=1.125 m/s2 or a=1.1 m/s2 (2 sig figs) Therefore, the answer is (2). 3. Formula from the Reference Table is: a=(vf-vi)/t a = (8.0 m/s – 3.0 m/s)/(2.0s) =2.5 m/s2 (1) 2.5 m/s2 (2) 5.0 m/s2 (3) 5.5 m/s2 (4) 11 m/s2 4. A rock is dropped from a bridge. What happens to the magnitude of the acceleration and the speed of the rock as it falls? [Neglect friction.] (1) Both acceleration and speed increase. (2) Both acceleration and speed remain the same. (3) Acceleration increases and speed decreases. (4) Acceleration remains the same and speed increases. 5. A soccer ball kicked on a level field has an initial vertical velocity component of 15.0 meters per second. Assuming the ball lands at the same height from which it was kicked, what is the total time the ball is in the air? [Neglect friction.] (1) 0.654 s (3) 3.06 s (2) 1.53 s (4) 6.12 s Therefore, the answer is (1). 4. The acceleration due to gravity remains constant, ag = -9.8 m/s2 (- because it points downward). If a rock falls it speeds up, which means that the speed increases. Therefore, the answer is (4). 5. For the vertical component of the velocity, when the ball reaches the top: vfy=0 and the time is tup=t/2. Formula from the Reference Table is: vf = vi+at which for this case is written as: vfy = viy-9.81tup so: 0 = 15.0 m/s-(9.81m/s2)(tup) Time to go up to the top is: tup = 15/9.81 s = 1.53 s so total time is: t=2tup = 3.06 s Therefore, the answer is (3). Student _________________ Class ___________ 6. A student is standing in an elevator that is accelerating downward. The force that the student exerts on the floor of the elevator must be (1) less than the weight of the student when at rest (2) greater than the weight of the student when at rest (3) less than the force of the floor on the student (4) greater than the force of the floor on the student 7. The magnitude of the centripetal force acting on an object traveling in a horizontal, circular path will decrease if the Date _____________ 6. Since the student is accelerating downward that means the net force acting on the student has to point downward as well. Fnet=Fg-F For Fnet to have the same direction as Fg, that means F (the force that the students exerts on the floor of the elevator) has to be less than Fg. Therefore the answer is (1). 7. Formula for the centripetal force from the Reference Table is: Fc =m v2/r (1) radius of the path is increased (2) mass of the object is increased (3) direction of motion of the object is reversed (4) speed of the object is increased Fc and r are in inverse proportion, so if the radius r is increased then the centripetal force Fc is decreased. Therefore the answer is (1). 8. The centripetal force acting on the space shuttle as it orbits Earth is equal to the shuttle’s (1) inertia (3) velocity (2) momentum (4) weight 8. The Earth acts on the space shuttle with the force of gravity. This force of gravity (weight) serves as the centripetal force. Therefore, the answer is (4). 9. As a box is pushed 30. meters across a horizontal floor by a constant horizontal force of 25 newtons, the kinetic energy of the box increases by 300. joules. How much total internal energy is produced during this process? (1) 150 J (3) 450 J (2) 250 J (4) 750 J 9. The work done to push the box is: W=Fd = (30. m)(25N) = 750 J If 300. J of this work goes for the increase of kinetic energy, the rest: 750 J-300 J=450J is produced as internal energy. Therefore, the answer is (3). 10. What is the power output of an electric motor that lifts a 2.0-kilogram block 15 meters vertically in 6.0 seconds? (1) 5.0 J (2) 5.0 W (3) 49 J (4) 49 W 10. Formula for the power from the Reference Table is: P=Fv Since the motor is lifting the block up, F=Fg=mg P=mg(v)=2.0kg(9.81m/s2)(15m/6.0s)=49W Therefore, the answer is (4). Student _________________ Class ___________ 11. Four identical projectiles are launched with the same initial speed, v, but at various angles above the level ground. Which diagram represents the initial velocity of the projectile that will have the largest total horizontal displacement? [Neglect air resistance.] Date _____________ 11. The formula for the horizontal displacement of a projectile is: Dx=vixt. Using this formula, some people think that the largest vix means the largest Dx, therefore making choice (1) the answer. The trick is that total time t a projectile is in the air depends on the initial angle as well. As a result, this problem is a bit more complicated. However, to make the long story short, it is a known fact that 45o gives you the largest horizontal displacement, always that is. Simple as that! Therefore the answer is (2). 12. Two forces act concurrently on an object on a horizontal, frictionless surface, as shown in the diagram below. What additional force, when applied to the object, will establish equilibrium? (1) 16 N toward the right (2) 16 N toward the left (3) 4 N toward the right (4) 4 N toward the left 13. As shown in the diagram below, an open box and its contents have a combined mass of 5.0 kilograms. A horizontal force of 15 newtons is required to push the box at a constant speed of 1.5 meters per second across a level surface. 12. The resultant of the two given forces is: 10N-6N=4N to the right. To establish equilibrium (Fnet=0) we need 4N to the left, therefore the answer is (4). Student _________________ Class ___________ The inertia of the box and its contents increases if there is an increase in the (1) speed of the box (2) mass of the contents of the box (3) magnitude of the horizontal force applied (4) coefficient of kinetic friction between the box and the level surface 14. Which statement describes the kinetic energy and total mechanical energy of a block as it is pulled at constant speed up an incline? (1) Kinetic energy decreases and total mechanical energy increases. (2) Kinetic energy decreases and total mechanical energy remains the same. (3) Kinetic energy remains the same and total mechanical energy increases. (4) Kinetic energy remains the same and total mechanical energy remains the same. 15. Which diagram represents the electric field lines between two small electrically charged spheres? Date _____________ 13. This is a typical Regents question. You are given lots of information that is simply trying to distract you. Inertia is the ability of an object to resist the motion. More mass means more inertia, therefore the answer is (2). 14. This is a tricky question. Remember that the block is being pulled up the incline plane, not moving by itself. This means that we cannot say that the mechanical energy of the system is conserved. (If you like to use a formula, you could say that: W=ET, which means that work that you do, equals the increase in the total mechanical energy of the block). If the speed is constant that means the kinetic energy is constant. If height is increasing that means the potential energy is increasing as well. Therefore, the kinetic energy remains the same, while the total mechanical energy of the system increases (don’t forget you are doing work to maintain the constant kinetic energy), therefore the answer is (3). 15. For a system of two unlike charges the electric field lines always start from the positive charge and point toward the negative charge; in other words these electric fields lines seem to “like” each other, therefore they attract. For a system of two like charges (two positives or two negatives) the electric field lines seem to “dislike” each other, therefore they repel. For a single positive charge the electric field line go out of the positive charge (think of a fountain). For a single negative charge the electric field lines go toward the negative charge (think of a sink). Therefore, the answer is (2). Student _________________ Class ___________ 16. Two metal spheres, A and B, possess charges of 1.0 microcoulomb and 2.0 microcoulombs, respectively. In the diagram below, arrow F represents the electrostatic force exerted on sphere B by sphere A. Date _____________ 16. Both spheres appear to have positive charges. Since the like charges repel, sphere B will exert the same force F (action equals reaction in magnitude but in opposite direction) that points towards the left. Therefore the answer is (1). Which arrow represents the magnitude and direction of the electrostatic force exerted on sphere A by sphere B? 17. The diagram below represents a positively charged particle about to enter the electric field between two oppositely charged parallel plates. The electric field will deflect the particle (1) into the page (2) out of the page (3) toward the top of the page (4) toward the bottom of the page 18. What is the total amount of work required to move a proton through a potential difference of 100. volts? (1) 1.60 × 10–21 J (2) 1.60 × 10–17 J (3) 1.00 × 102 J (4) 6.25 × 1020 J 17. The electric field of a parallel-plates capacitor starts from the positive charge and goes toward the negative charge, so for the given capacitor the electric field is downward. A positive charge will move along the electric field, therefore the positive charge will be deflected toward the bottom of the page. Some people think of it this way: Positive will attract with the negative, therefore the answer is (4). 18. From the Reference Table formula for work is: W=qV. We substitute: for a proton q=e=1.6x10-19 C and V=100. V, and we get: W=(1.6x10-19 C)(100.V)= 1.6x10-17 J therefore the answer is (2). Student _________________ Class ___________ 19. A baseball player runs 27.4 meters from the batter’s box to first base, overruns first base by 3.0 meters, and then returns to first base. Compared to the total distance traveled by the player, the magnitude of the player’s total displacement from the batter’s box is (1) 3.0 m shorter (2) 6.0 m shorter (3) 3.0 m longer (4) 6.0 m longer 20. A motorboat, which has a speed of 5.0 meters per second in still water, is headed east as it crosses a river flowing south at 3.3 meters per second. What is the magnitude of the boat’s resultant velocity with respect to the starting point? (1) 3.3 m/s (3) 6.0 m/s (2) 5.0 m/s (4) 8.3 m/s 21. A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0 seconds. The total distance traveled by the car in this 12.0-second time interval is (1) 36.0 m (2) 180. m (3) 216 m (4) 252 m 22. A 0.149-kilogram baseball, initially moving at 15 meters per second, is brought to rest in 0.040 second by a baseball glove on a catcher’s hand. The magnitude of the average force exerted on the ball by the glove is (1) 2.2 N (3) 17 N (2) 2.9 N (4) 56 N 23. Which body is in equilibrium? (1) a satellite moving around Earth in a circular orbit (2) a cart rolling down a frictionless incline (3) an apple falling freely toward the surface of Earth (4) a block sliding at constant velocity across a tabletop Date _____________ 19. If the player overruns first base by 3.0 meters and then returns to first base, that means the displacement from the first base is zero (player is on first base). However, distance the player traveled (from and to first base) is 3.0m+3.0m=6.0 m. This means the magnitude of displacement is 27.4 m, while the distance is 27.4m +6.0m. Therefore, the answer is (2). 20. This question in about addition of two perpendicular velocities. Here is the picture for this: Vmotorboat = 5.o m/s east Vriver=3.3 m/s south Vresultant Using Pythagorean theorem we get: vresultant 2 =(5.0m/s)2+(3.3m/s)2 =36 m2/s2 So, vresultant =6.0 m/s Therefore, the answer is (3). 21. Formula from the Reference Table is: d v t , where v vi v f is the average 2 velocity in the accelerated motion (this formula is not on the Reference Table). The average velocity is: 15.0m / s 21.0m / s v 18.0m / s 2 So, the total distance is: d 18.0m / s 12.0s 216m Therefore the answer is (3). 22. Formula from the Reference Table is: Ft=p. In the meantime: p=mv=m(vfvi)= 0.149kg(0m/s-15 m/s)=-2.235 kgm/s. Solving for F we get: F=p/t = (-2.235 kgm/s)/( 0.040s) = -56N. The negative sign means this force stops the motion. Question is asking for the magnitude, therefore answer is (4). 23. Constant velocity means acceleration a=0, so Fnet=0 (equilibrium). Answer is (4). Student _________________ Class ___________ 24. As shown in the diagram below, a student standing on the roof of a 50.0-meterhigh building kicks a stone at a horizontal speed of 4.00 meters per second. Date _____________ 24. This is a question about the horizontal projectile. In the vertical direction (y-axis is positive going up) the initial vertical velocity of the stone is viy=0. The distance formula from the Reference Table is: 1 d vi t at 2 2 For the vertical component of the motion this formula can be written as: 1 ayt 2 2 We substitute: dy = – 50.0 m (– sign because of the downward displacement), viy=0 (initial vertical velocity is zero) and ay=– 9.81 m/s2 (acceleration due to gravity is pointing downward, hence the – sign). d y viy t 50.0m How much time is required for the stone to reach the level ground below? [Neglect friction.] (1) 3.19 s (2) 5.10 s (3) 10.2 s (4) 12.5 s 25. On the surface of Earth, a spacecraft has a mass of 2.00 × 104 kilograms. What is the mass of the spacecraft at a distance of one Earth radius above Earth’s surface? (1) 5.00 × 103 kg (2) 2.00 × 104 kg (3) 4.90 × 104 kg (4) 1.96 × 105 kg 26. A student pulls a 60.-newton sled with a force having a magnitude of 20. newtons. What is the magnitude of the force that the sled exerts on the student? (1) 20. N (2) 40. N (3) 60. N (4) 80. N 1 (9.81m / s 2 )t 2 2 Solving for t we get: t=3.19s, therefore the answer is (1). 25. Mass is an intrinsic quality of matter that never changes, regardless of the position of the object (Earth, moon, other planets etc.). People get confused because they think of the weight (Fg=mg), which depends on mass m but also depends on gravity. Weight does change if you go somewhere else. Therefore the answer is (2). 26. Based on Newton’s 3rd law, every action has an equal but opposite direction. If the student pulls the sled with 20. Newtons, the sled exerts exactly the same magnitude of force the other way. Therefore the answer is (1). Student _________________ Class ___________ 27. The data table below lists the mass and speed of four different objects. Date _____________ 27. Inertia is the ability of an object to resist the motion. More mass means more inertia, therefore the answer is (4). Some people get confused about the velocity. A physical quantity that includes both mass and velocity is momentum: p=mv. If the question was asking about the object with the greatest momentum (momentum is the amount of motion in an object), then the answer would be (2). Which object has the greatest inertia? (1) A (2) B (3) C (4) D 28. The diagram below shows a horizontal 12-newton force being applied to two blocks, A and B, initially at rest on a horizontal, frictionless surface. Block A has a mass of 1.0 kilogram and block B has a mass of 2.0 kilograms. 28. The mass of the system A+B is: m=1.0kg + 2.0 kg = 3.0kg The net force acting on the system A+B is F=12N. Using the Newton’s 2nd law we get: a=F/m = 12N/3.0kg = 4.0 m/s2 This is the acceleration for the whole system A+B, so for block B as well. Therefore the answer is (4). 29. Formula from the Reference Table is: The magnitude of the acceleration of block B is (1) 6.0 m/s2 (2) 2.0 m/s2 (3) 3.0 m/s2 (4) 4.0 m/s2 29. A ball is thrown vertically upward with an initial velocity of 29.4 meters per second. What is the maximum height reached by the ball? [Neglect friction.] (1) 14.7 m (2) 29.4 m (3) 44.1 m (4) 88.1 m vf2= vi2 + 2ad This formula written for the vertical direction (y-axis is positive in the upward direction) is: vfy2= viy2 + 2aydy Initial velocity viy=29.4 m/s; final velocity at the top vf=0m/s (when the ball reaches the top the height becomes maximum as well) and ay=-9.81 m/s2, so we get: (0m/s)2= (29.4 m/s) 2 + 2(-9.81 m/s2)dy Solving for dy we get: dy=44.1 m, therefore the answer is (3). Student _________________ Class ___________ 30. The diagram below represents a mass, m, being swung clockwise at constant speed in a horizontal circle. Date _____________ 30. Centripetal force Fc always is directed toward the center of the circle (acceleration in the circular motion ac also is always directed toward the center of the circle). Therefore the answer is (3). If the question was asking about the direction of velocity at the given point, the answer would be (D). That is because the velocity in the circular motion is always directed toward the tangent of the circle at the given point. At the instant shown, the centripetal force acting on mass m is directed toward point (1) A (2) B (3) C (4) D 31. A 3.1-kilogram gun initially at rest is free to move. When a 0.015-kilogram bullet leaves the gun with a speed of 500. meters per second, what is the speed of the gun? (1) 0.0 m/s (2) 2.4 m/s (3) 7.5 m/s (4) 500. m/s 32. A wound spring provides the energy to propel a toy car across a level floor. At time ti ,the car is moving at speed vi across the floor and the spring is unwinding, as shown below. At time tf, the spring has fully unwound and the car has coasted to a stop. 31. This question is about the law of conservation of momentum. Formula from the Reference Table is: pbefore = pafter Before is before the shot (gun at rest, bullet at rest) so pbefore =0. After is after the shot (bullet is moving let’s say to the right, while is gun is moving to the left, opposite of the bullet). 0=mbvb+mgvg 0=(0.015kg)(500. m/s) +(3.1kg)(vg) Solving for vg we get: vg= - 2.4 m/s. The question is asking for speed, therefore the answer is (2). The – sign in velocity means that the gun moves in the direction opposite of the bullet. Student _________________ Class ___________ Date _____________ 32. At ti the car has both kinetic energy (the car is moving with speed vi) and elastic potential energy of the spring (the spring is unwinding). At time tf the car has stopped, which suggests that the two form of energy the system had at ti (kinetic energy of the car and the elastic potential energy of the spring) have converted to heat (which is a form of internal energy). Therefore the answer is (3). Which statement best describes the transformation of energy that occurs between times ti and tf? (1) Gravitational potential energy at ti is converted to internal energy at tf. (2) Elastic potential energy at ti is converted to kinetic energy at tf. (3) Both elastic potential energy and kinetic energy at ti are converted to internal energy at tf. (4) Both kinetic energy and internal energy at ti are converted to elastic potential energy at tf. 33. What is the magnitude of the electrostatic force between two electrons separated by a distance of 1.00 × 10–8 meter? (1) 2.56 × 10–22 N (3) 2.30 × 10–12 N (2) 2.30 × 10–20 N (4) 1.44 × 10–1 N 33. The magnitude of the electrostatic force between two charges q1 and q2 placed at a distance r away from each other, is given by Coulomb’s law, which can be found on the Reference Table: kq q Fe 12 2 r For two electrons: q1=q2=e=1.6x10-19 C (constant on the Reference Table as well) that are separated by a distance of r=1.00x10-8 m we get: Fe (8.99 x10 9 Nm 2 / C 2 )(1.6 x10 19 C )(1.6 x10 19 C ) (1.00 x10 8 m) 2 Fe=2.30x10-12 N therefore the answer is (3). Student _________________ Class ___________ 34. The diagram below represents a 155newton box on a ramp. Applied force F causes the box to slide from point A to point B. What is the total amount of gravitational potential energy gained by the box? Date _____________ 34. The total amount of gravitational energy gained by the box is given by a formula in the Reference Table: PEg=mgh PEg=(155N)(1.80m) = 279 J Therefore the answer is (2). Again, a question where extra information is provided in order to throw you off. (1) 28.4 J (2) 279 J (3) 868 J (4) 2740 J 35. The diagram below represents the electric field surrounding two charged spheres, A and B. What is the sign of the charge of each sphere? (1) Sphere A is positive and sphere B is negative. (2) Sphere A is negative and sphere B is positive. (3) Both spheres are positive. (4) Both spheres are negative. 35. For a system of two unlike charges the electric field lines always start from the positive charge and point toward the negative charge; in other words these electric fields lines seem to “like” each other, therefore they attract. Therefore the answer is (2). Student _________________ Class ___________ 36. What is the approximate diameter of an inflated basketball? (1) 2 × 10–2 m (3) 2 × 100 m (2) 2 × 10–1 m (4) 2 × 101 m Date _____________ 36. The approximate diameter of an inflated basketball is 20 cm=0.2 m, therefore the answer is (2). If you know the answer in inches instead (approximately 8 inches), multiply this by 2.5 to get the answer in cm (1inch = 2.54 cm). 37. The graph below shows the relationship between the speed and elapsed time for an object falling freely from rest near the surface of a planet. 37. In a graph of speed vs. time the distance is given by the area under the graph. The area under the graph for the first 3.0 seconds is the area of the triangle: 1 1 A bxh (3.0s)(8.0m / s) 12m 2 2 Therefore the answer is (1). What is the total distance the object falls during the first 3.0 seconds? (1) 12 m (2) 24 m (3) 44 m (4) 72 m 38. A 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average force required to stop the player in 0.65 second? (1) 120 N (2) 290 N (3) 690 N (4) 920 N Even if you are not sure about the area under the graph being the distance, try this: Area is always: length width, so: (m/s)(s)=m, hence area is distance. Using the same logic the slope of speed vs. time would be: slope = (change of y)/(change of x), so: (m/s)/s=m/s2 , hence the slope is the acceleration. 38. Formula from the Reference Table is: Ft=p. In the meantime: p=mv=m(vf-vi)= 75kg(0m/s-6.0 m/s)= - 450kgm/s. Solving for F we get: F=p/t = (- 450kgm/s)/( 0.65s) = -692N The negative sign means this force stops the motion. Question is asking for the magnitude, therefore answer is (3). Student _________________ Class ___________ 39. A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22-newton force to the wagon handle, which is inclined at 35° to the sidewalk as shown below. Date _____________ 39. The constant velocity means acceleration a=0, so Fnet=0. If Fnet=0 that means the component of Fnet along the direction of motion (x-axis) will be zero as well: Fnetx=0 The forces along the x-direction are: Fnetx=0=Fappcos35o-Ffk Ffk=Fappcos35o=22Ncos35o=18N Therefore the answer is (3). What is the magnitude of the force of friction on the wagon? (1) 11 N (2) 13 N (3) 18 N (4) 22 N 40. The diagram below shows the arrangement of three small spheres, A, B, and C, having charges of 3q, q, and q, respectively. Spheres A and C are located distance r from sphere B. 40. The magnitude of the electrostatic force between two charges q1 and q2 placed at a distance r away from each other, is given by Coulomb’s law, which can be found on the Reference Table: kq q Fe 12 2 r The magnitude of electrostatic force between charges B and C will be: FBC Compared to the magnitude of the electrostatic force exerted by sphere B on sphere C, the magnitude of the electrostatic force exerted by sphere A on sphere C is (1) the same (2) twice as great (3) 3/4 as great (4) 3/2 as great kqq kq 2 2 r2 r The magnitude of electrostatic force between charges A and C will be: FAC k 3qq 3kq 2 3 FBC 4 ( 2r ) 2 4r 2 Therefore the answer is (3). Student _________________ Class ___________ 41. A space probe is launched into space from Earth’s surface. Which graph represents the relationship between the magnitude of the gravitational force exerted on Earth by the space probe and the distance between the space probe and the center of Earth? Date _____________ 41. The magnitude of the gravitational force between two objects with masses m1 and m2 placed at a distance r away from each other (in this problem the two objects interacting are the space probe and Earth), is given by Newton’s law of universal gravitation, which can be found on the Reference Table: Fg Gm1m2 r2 1 , which is an inverse r2 square relationship represented by a hyperbola. Therefore the answer is (2). As we can see Fg ~ 42. Which graph represents the relationship between the gravitational potential energy (GPE) of an object near the surface of Earth and its height above the surface of Earth? 42. The formula for the gravitational potential energy can be found on the Reference Table: PEg=mgh As we can see PEg~h, which is a direct relationship represented by a straight line. Therefore the answer is (2). Student _________________ Class ___________ 43. Two parallel metal plates are connected to a variable source of potential difference. When the potential difference of the source is increased, the magnitude of the electric field strength between the plates increases. The diagram below shows an electron located between the plates. Which graph represents the relationship between the magnitude of the electrostatic force on the electron and the magnitude of the electric field strength between the plates? Date _____________ 43. The formula of the electric force acting on charge placed in an electric field is found on the Reference Table: Fe=qE For an electron Fe=eE (this is not important for this question anyway). As we can see Fe~E, which is a direct relationship represented by a straight line. Therefore the answer is (4). Again, this question is designed to look complicated on purpose. You should be able to see beyond unnecessary information and get to the point directly. Student _________________ Class ___________ 44. The diagram below represents a circuit consisting of two resistors connected to a source of potential difference. What is the current through the 20.-ohm resistor? Date _____________ 44. This is a series connection of two resistors (one loop only), so from the Reference Table for the series connection we get: Req=R1+R2=10.+20. =30. To find the current we use Ohm’s law found on the Reference Table: I=I1=I2=V/Req = (120V)/( 30. ) = 4.0A Therefore the answer is (4). (1) 0.25 A (2) 6.0 A (3) 12 A (4) 4.0 A 45. A small electric motor is used to lift a 0.50-kilogram mass at constant speed. If the mass is lifted a vertical distance of 1.5 meters in 5.0 seconds, the average power developed by the motor is (1) 0.15 W (2) 1.5 W (3) 3.8 W (4) 7.5 W 46. A ball is dropped from the top of a cliff. Which graph best represents the relationship between the ball’s total energy and elapsed time as the ball falls to the ground? [Neglect friction.] 45. The power formula from the Reference Table is: P=Fv Substituting we get: P=mg(d/t) = (0.50kg9.81m/s2)(1.5m/5.0s) P= 1.5 W Therefore the answer is (2). 46. If we neglect friction that means the total mechanical energy of the ball remains constant. Therefore the answer is (4). Student _________________ Class ___________ 47. A child, starting from rest at the top of a playground slide, reaches a speed of 7.0 meters per second at the bottom of the slide. What is the vertical height of the slide? [Neglect friction.] (1) 0.71 m (2) 1.4 m (3) 2.5 m (4) 3.5 m 48. The graph below represents the relationship between the current in a metallic conductor and the potential difference across the conductor at constant temperature. Date _____________ 47. This question is about the law of conservation of energy. If we neglect friction that means the total mechanical energy of the child remains constant. The child starts from rest at the top (KEtop=0) and has zero potential energy at the bottom of the slide (PEgbottom=0). So we can write: MEtop=MEbottom PEgtop+KEtop=PEgbottom+KEbottom mgh+0=0+(1/2)mv2 m’s cancel out and we get: h=v2/(2g)=(7.0m/s)2/(29.81m/s2)=2.5m Therefore the answer is (3). 48. The Ohm’s law is a formla you can find on the Reference Table: R=V/I or, we can write: 1/R=I/V The given graph is of current I vs potential difference V, therefore: slope = I/V=1/R Slope = I/V = (0.5-0)A/(1.0-0)V = 0.5A/V R=1/slope = 1/(0.5) V/A = 2.0 The resistance of the conductor is (1) 1.0 (3) 0.50 (2) 2.0 (4) 4.0 Therefore the answer is (2). Student _________________ Class ___________ 49. A student throws a baseball vertically upward and then catches it. If vertically upward is considered to be the positive direction, which graph best represents the relationship between velocity and time for the baseball? [Neglect friction.] 50. A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the total work that must be done on the mass to increase its speed to 11.5 meters per second? (1) 120. J (2) 422 J (3) 570. J (4) 992 J Date _____________ 49. Velocity is a vector, which means it could be either positive or negative. If a ball is thrown upward, the ball goes from a positive value to zero (at the top when the ball stops for a split of a second). On the way back the velocity becomes negative and the baseball speeds up until it reaches the value it had when it was initially thrown up (speed is always positive, while the velocity on the way down is negative). Therefore the answer is (4). 50. This is horizontal motion, for which the gravitational potential energy does not change. To solve this question we will use the formula you have on the Reference Table: W=ET Since we do not consider PEg at all, the formula becomes: W=KE=KEf-KEi which we called the work-change of kinetic energy theorem during class (this formula is not on Reference Table as such). W=(1/2)mvf2-(1/2)mvi2 = =0.5(15.0kg)[(11.5m/s)2-(7.50m/s)2] = 570. J Therefore the answer is (3). Student _________________ Class ___________ FREE-RESPONSE QUESTIONS 1. A 0.50-kilogram frog is at rest on the bank surrounding a pond of water. As the frog leaps from the bank, the magnitude of the acceleration of the frog is 3.0 meters per second2. Calculate the magnitude of the net force exerted on the frog as it leaps. [Show all work, including the equation and substitution with units.] Base your answers to questions 2 and 3 on the information below. A student and the waxed skis he is wearing have a combined weight of 850 newtons. The skier travels down a snow-covered hill and then glides to the east across a snowcovered, horizontal surface. Date _____________ SHOW ALL WORK 1. 2. As the skier glides across the horizontal surface there is no motion along the vertical direction, therefore the net force in the vertical direction must be zero: Fnety=0=FN-Fg FN=Fg=850N 2. Determine the magnitude of the normal force exerted by the snow on the skis as the skier glides across the horizontal surface. 3. 3. Calculate the magnitude of the force of friction acting on the skis as the skier glides across the snow-covered, horizontal surface. [Show all work, including the equation and substitution with units.] 4. 4. Calculate the kinetic energy of a particle with a mass of 3.34 × 10–27 kilogram and a speed of 2.89 × 105 meters per second. [Show all work, including the equation and substitution with units.] Student _________________ Class ___________ 5. The graph below represents the relationship between the work done by a person and time. Date _____________ 5. From the Reference Table we get the formula: P=W/t The slope of the given graph is the rate at which work is being done, which is power. 6. Identify the physical quantity represented by the slope of the graph. 6. The heating element in an automobile window has a resistance of 1.2 ohms when operated at 12 volts. Calculate the power dissipated in the heating element. [Show all work, including the equation and substitution with units.] 7. Ohm’s law for an electric circuit can be found on the Reference Table as: Base your answers to questions 7 and 8 on the information below. R=V/I = (1.50V)/(24.0A) =0.0625 =6.25x10-2 A 3.50-meter length of wire with a cross sectional area of 3.14 × 10–6 meter2 is at 20° Celsius. The current in the wire is 24.0 amperes when connected to a 1.50-volt source of potential difference. 8. 7. Determine the resistance of the wire. 8. Calculate the resistivity of the wire. [Show all work, including the equation and substitution with units.] Student _________________ Class ___________ Date _____________ Base your answers to questions 9 and 10 on the information below. In an experiment, a 0.028-kilogram rubber stopper is attached to one end of a string. A student whirls the stopper overhead in a horizontal circle with a radius of 1.0 meter. The stopper completes 10. revolutions in 10. seconds. 9. The period of rotations is: T=10.s/10 = 1.0s The speed of rotations is: v=d/t = (2r)/T = (23.141.0m)/(1.0s) = 6.28m/s 6.3 m/s 9. Determine the speed of the whirling stopper. 10. Calculate the magnitude of the centripetal force on the whirling stopper. [Show all work, including the equation and substitution with units.] 10. Student _________________ Class ___________ Base your answers to questions 11 through 13 on the information below. 11. A kicked soccer ball has an initial velocity of 25 meters per second at an angle of 40.° above the horizontal, level ground. [Neglect friction.] 11. Calculate the magnitude of the vertical component of the ball’s initial velocity. [Show all work, including the equation and substitution with units.] 12. 12. Calculate the maximum height the ball reaches above its initial position. [Show all work, including the equation and substitution with units.] 13. On the diagram on the right, sketch the path of the ball’s flight from its initial position at point P until it returns to level ground. Base your answers to questions 65 through 67 on the information and diagram below. A 15-ohm resistor, R1, and a 30.-ohm resistor, R2, are to be connected in parallel between points A and B in a circuit containing a 90.-volt battery. 13. Date _____________ Student _________________ Class ___________ 14. Complete the diagram on your right to show the two resistors connected in parallel between points A and B. Date _____________ 14. 15. Determine the potential difference across resistor R1. 15. Since we have a parallel connection of two resistors, the potential difference across both resistors is the same as the voltage provided by the battery, which is 90. V. 16. Calculate the current in resistor R1. [Show all work, including the equation and substitution with units.] The spring in a dart launcher has a spring constant of 140 newtons per meter. The launcher has six power settings, 0 through 5, with each successive setting having a spring compression 0.020 meter beyond the previous setting. During testing, the launcher is aligned to the vertical, the spring is compressed, and a dart is fired upward. The maximum vertical displacement of the dart in each test trial is measured. The results of the testing are shown in the table below. 16. Student _________________ Class ___________ Date _____________ 17. Directions (17–20): Using the information in the data table, construct a graph on the grid on your right, following the directions below. 17. Plot the data points for the dart’s maximum vertical displacement versus spring compression. 18. Draw the line or curve of best fit on the grid on your right. Explain how you drew the line of best fit. 18. The curve seems to be the best fit in this case, not the line. We drew the curve of best fit by going through the middle of as many points as possible (which doesn’t necessarily mean connecting all the dots). The curve is a much better fit than a straight line 19. 19. Using information from your graph, calculate the energy provided by the compressed spring that causes the dart to achieve a maximum vertical displacement of 3.50 meters. [Show all work, including the equation and substitution with units.] 20. Determine the magnitude of the force, in newtons, needed to compress the spring 0.040 meter. 20. The formula for Hooke’s law can be found on the Reference Table: Fs=kx Fs=(140 N/s)(0.040m) = 5.6 N Student _________________ Class ___________ A pop-up toy has a mass of 0.020 kilogram and a spring constant of 150 newtons per meter. A force is applied to the toy to compress the spring 0.050 meter. 21. 21. Calculate the potential energy stored in the compressed spring. [Show all work, including the equation and substitution with units.] 22. The toy is activated and all the compressed spring’s potential energy is converted to gravitational potential energy. Calculate the maximum vertical height to which the toy is propelled. [Show all work, including the equation and substitution with units.] A horizontal force of 8.0 newtons is used to pull a 20.-newton wooden box moving toward the right along a horizontal, wood surface, as shown. 22. Date _____________ Student _________________ Class ___________ 23. Calculate the magnitude of the frictional force acting on the box. [Show all work, including the equation and substitution with units.] 24. Determine the magnitude of the net force acting on the box. Date _____________ 23. 24. The motion is along the horizontal direction, therefore the net force must be along that direction as well. Fnetx = F-Ff = 8.0N – 6.0 N = 2.0N to the right 25. 25. Calculate the magnitude of the acceleration of the box. [Show all work, including the equation and substitution with units.] Suggested time: 3 hours Important: I declare that all the work shown above is solely based on my individual effort. I have not consulted any tutors, any friends and/or other materials regarding this homework. Total credit: 40 points Due date: April 2, 2012 Student (please print) ___________________________ Class: Physics Honors Student (please sign) ___________________________ Teacher: B. Fejzo Student _________________ June 2011 (Part A) 1. 1 Class ___________ 23. 4 24. 1 2. 2 25. 2 3. 1 26. 1 4. 4 27. 4 5. 3 28. 4 6. 1 29. 3 7. 1 30. 3 8. 4 31. 2 9. 3 32. 3 10. 4 33. 3 11. 2 34. 2 12. 4 35. 2 13. 2 14. 3 June 2011 (Part B1) 15. 2 36. 2 16. 1 37. 1 17. 4 38. 3 18. 2 39. 3 40. 3 June 2010 (Part A) 41. 2 19. 2 42. 2 20. 3 43. 4 21. 3 44. 4 22. 4 Date _____________ Student _________________ Class ___________ June 2010 (Part B1) June 2008 (Part C) 45. 2 11. Date _____________ 46. 4 47. 3 48. 2 49. 4 12. 50. 3 June 2011 (Part B2) 1. Fnet = 1.5 N 2. FN = 850 N 3. Ff = 40 N 4. KE = 1.39x10-16 J 5. slope = power 13. (parabola starts at P, ends at the ground) 14. 6. P = 120 W June 2010 (Part B2) 7. R = 6.25 x10-2 8. = 5.61x10-8 m 15. 90. V 9. v = 6.3 m/s 16. 10. Fc = 1.1 N Student _________________ Class ___________ 17. error: +- .3 grid space 24. Fnet = 2.0 N 18. curve of best fit (parabola) 25. 19. 20. Fs = 5.6 N June 2007 (Part C) 21. 22. 23. Date _____________