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Physics 9 - Semester 2 Exam review Please notify me of errors! Practice problems (Chapters 2 - 37) 1. A car is at rest, then it accelerates at a constant rate of 4 m/s/s for 5 seconds. a. how fast is the car moving after exactly 5 seconds? a = ∆v/∆t ==> v = 20m/s 2. A ball is dropped (v = 0) from a height of 45 meters. a. how long does it take the ball to reach the ground? d = (1/2)at2 ==> t = 3 sec b. what is the speed of the ball the instant before it hits the ground? a = ∆v/∆t ==> v = 30 m/s 3. A ball is thrown horizontally at 15 m/s from a window 45 meters high. a. how long does it take the ball to reach the ground? d = (1/2)at2 ==> t = 3 sec b. how far from the building does the ball land? dx = (vx )t ==> dx = 45 m 4. A 2 kg object is on earth. a. what is the weight of the object? mg = 20N b. what is the mass of the object? m = 2 kg c. what is the force of gravity acting on the object? mg = 20N d. what is the acceleration when it falls (ignoring air resistance)? a = 10 m/s/s e. what is the value of “g” for this object? g = 10 m/s/s 5. A 1 kg ball and a 10 kg ball of the same size are dropped from 50 meters above the earth without air resistance. a. After falling exactly 2 seconds,which ball feels the greater force of gravity? the 10 kg ball b. After falling exactly 2 seconds,which ball experiences the greater acceleration? both the same c. After falling exactly 2 seconds,which ball experiences the greater net force? the 10 kg ball d. Which ball reaches the ground first? both the same e. Which ball has a greater speed the instant before it reaches the ground? both the same 6. A 1 kg ball and a 10 kg ball of the same size are dropped from the same height on earth with air resistance present. a. After falling exactly 2 seconds,which ball feels the greater force of gravity? the 10 kg ball b. After falling exactly 2 seconds,which ball experiences the greater acceleration? the 10 kg ball c. After falling exactly 2 seconds,which ball experiences the greater net force? the 10 kg ball d. Which ball reaches the ground first? the 10 kg ball e. Which ball has a greater speed the instant before it reaches the ground? the 10 kg ball 7. How much net force is needed to keep a 1,000 kg car moving at a constant velocity of 15 m/s? net Force = ma = (1000 kg)(0 m/s/s) = zero ... none! [note the correction - Thanks Erica!] 8. A car has a maximum acceleration of 4 m/s/s. a. what is the maximum acceleration if the engine is replaced by one that produces twice the force? 2 x F ==> 2 x a so: 8 m/s/s b. what is the maximum acceleration if, instead of replacing the engine, the car’s mass is doubled by loading it with cargo? 2 x m ==> (1/2) x a so: 2 m/s/s 9. A 50 kg rocket is being pushed upward by a thrust of 3,000 N, while at the same time being pulled downward by the force of gravity. a. Determine the weight of the rocket Wt. = mg = 500N b. Determine the net force acting on the rocket net F = the sum of the forces acting on the rocket net F = Thrust - Weight = 3,000 N - 500 N = 2,500 N c. Determine the acceleration of the rocket net F = ma 2,500 N = (50 kg)(a) so: a = 50 m/s/s 10. A girl exerts a force of 30 N to push a 20 kg box across the floor at a constant speed of 2 m/s. a. determine the acceleration of the box a = ∆v/∆t = 0/∆t ==> a = 0 b. determine the net force on the box net F = ma = 0 ==> net F = 0 c. determine the amount of friction present as the box slides push - friction = net F = 0 ==> friction = 30 N d. how hard must the girl push to accelerate the box at a constant 3 m/s/s? net F = ma = (20 kg)(3 m/s/s) = 60 N push - friction = 60 N push - 30 N = 60 N ==> push = 90 N 11. If a second crate was stacked on top of the crate in #10, how much friction would be present? double the wt ==> double the friction so: friction = 60 N 12. A 20 kg box is suspended by two vertical strings. How much tension is in each string? each tension = (1/2)(weight) = 100N 13. A 20 kg box is suspended by two strings angled at 45 degrees. a. What is the vertical component of tension in each string? 100 N b. What is the horizontal component of tension in each string? 100 N c. What is the actual tension in each string? 141 N (pyth. thm) 14. A 2 kg box is pulled by a horizontal rope and accelerated at 3 m/s/s on a frictionless surface. Determine the tension in the rope. net F = T = ma = 6 N 15. Two 2 kg boxes connected by a rope are pulled by a second rope and accelerated at 3 m/s/s on a frictionless surface. Determine the tension in each rope. pull Left Tension = ma = (2 kg)(3 m/s/s) = 6 N Right Tension = ma = (4 kg)(3 m/s/s) = 12 N 16. A 2 kg glider rests on a horizontal air track is attached by a string that runs over a pulley and a hanging 0.5 kg mass. Determine the accelerations of the glider and hanging mass. accel. of the system = (net force on the system)/(mass of the system) a = (5 N)/(2.5 kg) = 2 m/s/s [note the correction! - thanks Erica!] 17. Identify two “action/reaction pairs” involved when a hammer hits a tent stake into the ground. 1. hammer pushes stake down/stake pushes hammer up 2. stake pushes ground down/ground pushes stake up (also: Earth pulls stake down/stake pulls Earth up) 18. A 2 kg box sits at rest on a table. a. determine the force of gravity acting on the box mg = 20 N b. determine the support force acting on the box 20 N c. are these two forces an action/ reaction pair? No!! (A/R pairs act on differnet objects and therefore never cancel) 19. Imagine a Honda Odyssey loaded with bricks (5,000 kg) moving at 20 m/s collides head-on, bumper to bumper with an empty Honda Odyssey (2,500 kg) at rest. a. Which car experiences a greater force? Neither: the cars experience equal and opposite forces! b. Which car experiences a greater acceleration? the less massive car c. Which car experiences more damage Neither: the cars experience equal and opposite forces! 20. If action/reaction pairs are equal and opposite, why does a football accelerate when you kick it? action/reaction pairs don’t cancel (see #18 c) 21. A 2 kg ball (initially at rest) is accelerated to a velocity of +30 m/s (due north). a. What is the initial momentum? the final momentum? initial p = mv = 0 final p = mv = +60 kgm/s [north] b. What is the change in momentum? ∆p = m∆v = +60 kgm/s [north] c. What is the impulse experienced by the ball? I = ∆p = +60 kgm/s [north] 22. When a 2 kg ball of clay moving at + 20 m/s hits a wall and sticks, a. what is the initial momentum of the ball of clay? p = mv = +40 kgm/s b. what is the change in momentum of the ball when it strikes the wall? ∆p = final p - initial p = - 40 kgm/s (note the neg. sign!) c. what is the impulse experienced by the ball of clay when it hits the wall? impulse = ∆p = - 40 kgm/s d. what is the impulse experienced by the wall when the ball hits it? impulse = +40 kgm/s e which object experiences a greater force during the collision? equal and opposite forces! f. is this collision inelastic or elastic? inelastic 23. A 100 kg skater moving to the right at 6 m/s colides and hangs onto a 50 kg skater moving to the left at 6 m/s. a. What is the total momentum before the collision? ... after the collision? before: m1v1 + m2v2 = +600 kgm/s + (-300 kgm/s) = +300 kgm/s after: conservation of momentum = +300 kgm/s b. What is the final velocity of the pair of skaters? p = mv ==> +300 kgm/s = (150 kg)(v) ==> v = +2m/s c. How much impulse experienced by each skater? 100 kg skater: I = ∆p = final p - inital p = (100kg)(+2m/s) - (100kg)(+6m/s) I = - 400 kgm/s 50 kg skater: I = + 400 kgm/s d. If the actual collision lasts 0.2 seconds, how much force does each skater experience? 100 kg skater: I = Ft - 400 kgm/s = (F)(0.2 sec) F = - 2000 N 50 kg skater: F = + 2000 N 24. A box on the floor weighing 200N is lifted 2 meters into the air, and held there. a. How much work is done on the box? W = Fd = (200N)(2m) = 400 J b. What is the change in potential energy? PE = mgh = 400J or W = ∆PE = 400J c. If the box is dropped from this height, what will it’s kinetic energy be the instant before it hits the ground? initial PE = final KE ==> KE = 400J 25. A 2 kg ball is thrown horizontally at a speed of 20 m/s from a window 25m high. a. What is the total mechanichal energy of the ball the instant after it is thrown? TME = PE + KE = 900 J half way to the ground? TME = constant ==> 900J the instant before it hots the ground? TME = 900 J b. What is the speed of the ball the instant before it hits the ground? TME = KE = 900J ==>v = 30m/s c. Does the mass of the ball affect your answers to (a) or (b)? mass does effect the answer to (a) mass does not effect answer to (b) 26. A 1,000 kg car moving at 2 m/s skids to a stop in 4 meters. Find the force of friction present. W = ∆E (F)(d) = ∆KE = (1/2)mv2 (F)(4m) = (1/2)(1,000 kg)(2)2 ==> F = 500N 27. An inclined plane has dimensions: vertical ht = 2 m & diagonal length = 8 meters. a. Determine the theoretical mechanical advantage of this simple machine TMA = (input dist.)/(output dist.) = 8/2 ==> TMA = 4 b. How much force is needed to slide a block of ice weighing 200N up the 8 meter incline? (no friction) Force = (200N)/4 = 50 N c. Later, you notice it takes a 75 N force to slide a 200N wooden crate up the 8 meter incline. How much friction is present? 75 N - 50 N = 25 N d. What is the actual mechanical advantage in (c)? AMA = (output force)/(input force) = (200 N)/(75 N) = 8/3 = 2.67 e. What is the efficiency of the inclined plane in (b)? in (c)? in (b): 100% (anytime there is no friction!) in (c): eff. = AMA/TMA = 2.67/4 = 2/3 = 0.67 or 67% 28. A 60 ohm light bulb is connected to a 120 V outlet.in your house. a. how much current flows through the bulb? I = V/R = 2A b. is the current AC or DC? AC (in your house wiring) c. what is the power of the bulb? P = IV = (2A)(120V) = 240 W 29. How much energy (in kWh) do you use to keep a 100W bulb lit for 20 hours? P = E/t so: E = (P)(t) = (0.1kW)(10h) = 1kWh or: E = (P)(t) = (100J/s)(36000 s) = 3,600,000 J 30. A flashlight uses a single 6V battery connected to a single 60 ohm bulb. a. How much current flows thru the bulb? I = V/R = 0.1 A b. How many coulombs pass thru the bulb every 10 seconds? 1 coulomb c. How many joules of energy does the battery give each coulomb that passes thru it? 6 Joules 31. A 4 ohm and 8 ohm resistor are all connected in SERIES to a 6 volt battery. a. what is the total resistance of the circuit? 12 ohms b. How much current flows thru the battery and each resistor? I = V/R = 6V/12 ohms = 0.5 A c. How much does the voltage drop across each resistor? 8 ohm: V = IR = (0.5A)(8 ohms) = 4V 4 ohm: V = IR = (0.5A)(4 ohms) = 2V 32. A 3 ohm and 6 ohm resistor are all connected in PARALLEL to a 6 volt battery. a. How much does the voltage drop across each resistor? 6 V each b. How much current flows thru the battery and each resistor? 3 ohm: I = V/R = 6V/3 ohms = 2A 6 ohm: I = V/R = 6V/3]6 ohms = 1A c.. What is the total resistance of this circuit? total I = (total V)/R 3A = 6V/R ==> total R = 2 ohms 33. Draw magnetic the field lines around a bar magnet Look in your notes! (remember that the field lines point from N to S outside a magnet, S to N inside) 34. Draw the magnetic field lines around a wire carrying current out of the page. Look in your notes! (LH rule: clockwise concentric circles) 35. An electron moves between the poles of a horseshoe magnet. Determine the direction of force on the electron. N S LH rule#2: Force is out of the page (toward you!) 36. The primary coil of a transformer is plugged into the wall (120V AC), while the secondary coil is connected to a cell phone. There are 1000 turns in the primary coil and 100 turns in the secondary coil. a. What is the voltage in the secondary coil? 1/10 the turns ==> 1/10 the V so: V = 12V b. If the curent in the primary coil is 1 amp, what is the current in the secondary coil? 1/10 the turns ==> 10 times the I so: I = 10 A 37. Describe how a DC motor works, how a speaker works, how a generator works, how a microphone works, how a transformer works. Look in your notes! (remember that motors and speakers take electical energy to produce KE) (remember that generators and microphones take KE and turn it into electrical energy)