1. (2 marks total) a. Multiply 109 x 309 b. Divide 1988 by 16 exactly 2. (4 marks total) a. Write 2/11 as a decimal to 2 decimal places b. Calculate 35% of 62 c. Add 103/4 to 92/3 d. Subtract 79.04 from 115.225 giving your answer correct to 2 decimal places 3. Circle the fractions in the list which are equivalent to 0.80 (2 marks) 2/7 32/40 8/10 8/20 8/25 9/24 36/45 40/50 4. Write the numerical value of: 3-3 (2 marks total) 5. Simplify z + 67 = 3z + 33 (1 mark total) 6. Solve to 1 decimal place 3y – 34 = 2y + 89 (1 mark total) 7. Solve the following equations to 2 decimal places (3 marks total) a. 37x + 1 = 35 b. 27 – a = 7.45 c. 3(y + 2) = 14 8. A 7-sided polygon is called a Heptagon. (3 marks total) a. What is the total of a Heptagon’s interior angles? b. If the Heptagon is regular (all angles the same), calculate the size of each interior angle to 2 decimal places. 9. Calculate the size of angle a and angle b. (2 mark total) 10. How many centilitres are there in 1.25 litres? (1 mark total) 11. The diagram below shows a stone carving with a hole on it; determine its volume (not including hole), if its thickness is 8 cm. Give your answer in cm3 to 2 decimal points. Assume π = 3.14 (6 marks total) 12. The diagram below shows a piece of alloy plate with a hole in it made from aluminium, copper and magnesium with a mass ratio of 35:3:2. Calculate the following to 2 decimal places. All measurements are in cm. (7 marks total) a. Using the formula A = 1/2(a+b)h calculate the height of the shape below. b. The volume of the solid part (not including the hole) of the shape below to 3 decimal places if it was 0.25cm thick. c. The mass of each material if the total mass of the plate is 62 kg. 10 cm Hole dia = 3 cm Cross sectional area of solid (not including hole) = 28.935 cm2 8 cm 13. A 66kg boy is running at 3 m/s. Calculate his Kinetic Energy using the formula KE = 1/2mv2 (2 marks total) 14. A rocket has a mass of 2,000 kg. What is its acceleration if the forces of its engines are 50kN? Show working out to receive full marks. (1 marks total) 250,000,000 m/s² 25 m/s² 25,000 m/s² 15. In the diagram below a force of 125N (F1) is applied to a lever 20cm (D1) away from the fulcrum, (4 marks total) Fulcrum (a) How far away in metres would a force of 5N (F2) need to be to balance the load? (b) How much force (F2) would need to be applied 0.7m away to balance the same load (F1)? 16. For the circuit shown in the diagram below, calculate: (3 mark total) a. The total circuit resistance. b. The value of the current I. c. Calculate the voltage of the battery cell if the current was 3amp and the resistors stayed the same. 17. In the diagram of a hydraulic system, the area of piston A is 8cm2 and the area of piston B is 48cm2. (2 mark total) If the Force IN is 16 N, calculate the force OUT. 18. Plot the graph 2y = x3 – 4 using a value range for x from 0 to 3 (3 marks total) 14 12 10 8 6 4 2 0 -2 Choosing appropriate scale (1 mark) Accurately plotting y values (1 mark) X 0 1 2 3 Y Accurately plotting line of best fit. (1 mark) SPARE PAPER

1. (2 marks total) a. Multiply 109 x 309 b. Divide 1988 by 16 exactly 2. (4 marks total) a. Write 2/11 as a decimal to 2 decimal places b. Calculate 35% of 62 c. Add 103/4 to 92/3 d. Subtract 79.04 from 115.225 giving your answer correct to 2 decimal places 3. Circle the fractions in the list which are equivalent to 0.80 (2 marks) 2/7 32/40 8/10 8/20 8/25 9/24 36/45 40/50 4. Write the numerical value of: 3-3 (2 marks total) 5. Simplify z + 67 = 3z + 33 (1 mark total) 6. Solve to 1 decimal place 3y – 34 = 2y + 89 (1 mark total) 7. Solve the following equations to 2 decimal places (3 marks total) a. 37x + 1 = 35 b. 27 – a = 7.45 c. 3(y + 2) = 14 8. A 7-sided polygon is called a Heptagon. (3 marks total) a. What is the total of a Heptagon’s interior angles? b. If the Heptagon is regular (all angles the same), calculate the size of each interior angle to 2 decimal places. 9. Calculate the size of angle a and angle b. (2 mark total) 10. How many centilitres are there in 1.25 litres? (1 mark total) 11. The diagram below shows a stone carving with a hole on it; determine its volume (not including hole), if its thickness is 8 cm. Give your answer in cm3 to 2 decimal points. Assume π = 3.14 (6 marks total) 12. The diagram below shows a piece of alloy plate with a hole in it made from aluminium, copper and magnesium with a mass ratio of 35:3:2. Calculate the following to 2 decimal places. All measurements are in cm. (7 marks total) a. Using the formula A = 1/2(a+b)h calculate the height of the shape below. b. The volume of the solid part (not including the hole) of the shape below to 3 decimal places if it was 0.25cm thick. c. The mass of each material if the total mass of the plate is 62 kg. 10 cm Hole dia = 3 cm Cross sectional area of solid (not including hole) = 28.935 cm2 8 cm 13. A 66kg boy is running at 3 m/s. Calculate his Kinetic Energy using the formula KE = 1/2mv2 (2 marks total) 14. A rocket has a mass of 2,000 kg. What is its acceleration if the forces of its engines are 50kN? Show working out to receive full marks. (1 marks total) 250,000,000 m/s² 25 m/s² 25,000 m/s² 15. In the diagram below a force of 125N (F1) is applied to a lever 20cm (D1) away from the fulcrum, (4 marks total) Fulcrum (a) How far away in metres would a force of 5N (F2) need to be to balance the load? (b) How much force (F2) would need to be applied 0.7m away to balance the same load (F1)? 16. For the circuit shown in the diagram below, calculate: (3 mark total) a. The total circuit resistance. b. The value of the current I. c. Calculate the voltage of the battery cell if the current was 3amp and the resistors stayed the same. 17. In the diagram of a hydraulic system, the area of piston A is 8cm2 and the area of piston B is 48cm2. (2 mark total) If the Force IN is 16 N, calculate the force OUT. 18. Plot the graph 2y = x3 – 4 using a value range for x from 0 to 3 (3 marks total) 14 12 10 8 6 4 2 0 -2 Choosing appropriate scale (1 mark) Accurately plotting y values (1 mark) X 0 1 2 3 Y Accurately plotting line of best fit. (1 mark) SPARE PAPER

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Derive an expression for the capillary-height change h for a fluid of surface tension Y and contact angle between two paraller plates W apart Evaluate h for water at 20c if W=0.5mm

Derive an expression for the capillary-height change h for a fluid of surface tension Y and contact angle between two paraller plates W apart Evaluate h for water at 20c if W=0.5mm

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

ELEC153 Circuit Theory II M2A3 Lab: AC Series Circuits Introduction Previously you worked with two simple AC series circuits, R-C and R-L circuits. We continue that work in this experiment. Procedure 1. Setup the following circuit in MultiSim.The voltage source is 10 volts peak at 1000 Hz. Figure 1: Circuit for analysis using MultiSim 2. Change R1 to 1 k and C1 to 0.1 uF. Connect the oscilloscope to measure both the source voltage and the voltage across the resistor.You should have the following arrangement. Figure 2: Circuit of figure 1 connected to oscilloscope To color the wires, right click the desired wire and select “Color Segment…” and follow the instructions. Start the simulation and open the oscilloscope. You should get the following plot: Figure 3: Source voltage (red) and the voltage (blue) across the resistor The red signal is the voltage of the source and the blue is the voltage across the resistor. The colors correspond to the colors of the wires from the oscilloscope. 3. From the resulting analysis plotdetermine the peak current. To determine the peak current measure the peak voltage across the resistor and divide by the value of the resistor (1000 Ohms). Record it here. Measured Peak Current 4. Determine the peak current by calculation. Record it here. Does it match the measured peak current? Explain. Calculated Peak Current 5 Determine the phase shift between the current in the circuit and the source voltage. We look at the time between zero crossings to determine the phase shift between two waveforms. In our plot, the blue waveform (representing the circuit current or the voltage across the resistor) crosses zero before the red waveform (the circuit voltage). So, current is leading voltage in this circuit. This is exactly what should happen when we have a capacitive circuit. 6. To determine the phase shift, we first have to measure the time between zero crossings on the red and blue waveforms. This is done by moving the oscillator probes to the two zero crossing as is shown in the following figure Figure 4: Determining the phase shift between the two voltage waveforms We can see from the figure that the zero crossing difference (T2 – T1) is approximately 134 us. The ratio of the zero-crossing time difference to the period of the waveform determines the phase shift, as follows: Using our time values, we have: How do we know if this phase shift is correct? In step 4 when you did your manual calculations to find the peak current, you had to find the total impedance of the circuit, which was: Now, the current will be: Here, the positive angle on the current indicates it is leading the circuit voltage. 7. Change the frequency of the voltage source to 5000 Hz. Estimulate and perform a Transient Analysis to find the new circuit current and phase angle. Measure them and record them here: Measured Current Measured Phase Shift 8. Perform the manual calculations needed to find the circuit current and phase shift. Record the calculated values here. Do they match the measured values within reason? What has happened to the circuit with an increase in frequency? Calculated Current Calculated Phase Shift Writeup and Submission In general, for each lab you do, you will be asked to setup certain circuits, simulate them, record the results, verify the results are correct by hand, and then discuss the solution. Your lab write-up should contain a one page, single spaced discussion of the lab experiment, what went right for you, what you had difficulty with, what you learned from the experiment, how it applies to our coursework, and any other comment you can think of. In addition, you should include screen shots from the MultiSim software and any other figure, table, or diagram as necessary.

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Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $p$ p1p2 −−−− ” p1 +p2 $p$ p1+p2 2 p1 + p2 $p$ |p1 − p2 | p1 + p2 $p$ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

Chapter 9 Practice Problems (Practice – no credit) Due: 11:59pm on Friday, April 18, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Momentum and Internal Forces Learning Goal: To understand the concept of total momentum for a system of objects and the effect of the internal forces on the total momentum. We begin by introducing the following terms: System: Any collection of objects, either pointlike or extended. In many momentum-related problems, you have a certain freedom in choosing the objects to be considered as your system. Making a wise choice is often a crucial step in solving the problem. Internal force: Any force interaction between two objects belonging to the chosen system. Let us stress that both interacting objects must belong to the system. External force: Any force interaction between objects at least one of which does not belong to the chosen system; in other words, at least one of the objects is external to the system. Closed system: a system that is not subject to any external forces. Total momentum: The vector sum of the individual momenta of all objects constituting the system. In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses and . To simplify the analysis, we will make several assumptions: The blocks can move in only one dimension, namely, 1. along the x axis. 2. The masses of the blocks remain constant. 3. The system is closed. At time , the x components of the velocity and the acceleration of block 1 are denoted by and . Similarly, the x components of the velocity and acceleration of block 2 are denoted by and . In this problem, you will show that the total momentum of the system is not changed by the presence of internal forces. m1 m2 t v1(t) a1 (t) v2 (t) a2 (t) Part A Find , the x component of the total momentum of the system at time . Express your answer in terms of , , , and . ANSWER: Part B Find the time derivative of the x component of the system’s total momentum. Express your answer in terms of , , , and . You did not open hints for this part. ANSWER: Why did we bother with all this math? The expression for the derivative of momentum that we just obtained will be useful in reaching our desired conclusion, if only for this very special case. Part C The quantity (mass times acceleration) is dimensionally equivalent to which of the following? ANSWER: p(t) t m1 m2 v1 (t) v2 (t) p(t) = dp(t)/dt a1 (t) a2 (t) m1 m2 dp(t)/dt = ma Part D Acceleration is due to which of the following physical quantities? ANSWER: Part E Since we have assumed that the system composed of blocks 1 and 2 is closed, what could be the reason for the acceleration of block 1? You did not open hints for this part. ANSWER: momentum energy force acceleration inertia velocity speed energy momentum force Part F This question will be shown after you complete previous question(s). Part G Let us denote the x component of the force exerted by block 1 on block 2 by , and the x component of the force exerted by block 2 on block 1 by . Which of the following pairs equalities is a direct consequence of Newton’s second law? ANSWER: Part H Let us recall that we have denoted the force exerted by block 1 on block 2 by , and the force exerted by block 2 on block 1 by . If we suppose that is greater than , which of the following statements about forces is true? You did not open hints for this part. the large mass of block 1 air resistance Earth’s gravitational attraction a force exerted by block 2 on block 1 a force exerted by block 1 on block 2 F12 F21 and and and and F12 = m2a2 F21 = m1a1 F12 = m1a1 F21 = m2a2 F12 = m1a2 F21 = m2a1 F12 = m2a1 F21 = m1a2 F12 F21 m1 m2 ANSWER: Part I Now recall the expression for the time derivative of the x component of the system’s total momentum: . Considering the information that you now have, choose the best alternative for an equivalent expression to . You did not open hints for this part. ANSWER: Impulse and Momentum Ranking Task Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest. Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. ANSWER: Both forces have equal magnitudes. |F12 | > |F21| |F21 | > |F12| dpx(t)/dt = Fx dpx(t)/dt 0 nonzero constant kt kt2 Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. If the ranking cannot be determined, check the box below. You did not open hints for this part. ANSWER: A Game of Frictionless Catch Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is . Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie’s speed relative to the ground after she catches the ball is . When answering the questions in this problem, keep the following in mind: The original mass of Chuck and his cart does not include the 1. mass of the ball. 2. The speed of an object is the magnitude of its velocity. An object’s speed will always be a nonnegative quantity. mcart mball vc vb vj mcart Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Express the speed in terms of and . You did not open hints for this part. ANSWER: Part B What is the speed of the ball (relative to the ground) while it is in the air? Express your answer in terms of , , and . You did not open hints for this part. ANSWER: Part C What is Chuck’s speed (relative to the ground) after he throws the ball? Express your answer in terms of , , and . u vc vb u = vb mball mcart u vb = vc mball mcart u You did not open hints for this part. ANSWER: Part D Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: Part E Find Jackie’s speed (relative to the ground) after she catches the ball, in terms of . Express in terms of , , and . You did not open hints for this part. ANSWER: vc = vj vb vj mball mcart vb vj = vj u vj mball mcart u Momentum in an Explosion A giant “egg” explodes as part of a fireworks display. The egg is at rest before the explosion, and after the explosion, it breaks into two pieces, with the masses indicated in the diagram, traveling in opposite directions. Part A What is the momentum of piece A before the explosion? Express your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: vj = pA,i Part B During the explosion, is the force of piece A on piece B greater than, less than, or equal to the force of piece B on piece A? You did not open hints for this part. ANSWER: Part C The momentum of piece B is measured to be 500 after the explosion. Find the momentum of piece A after the explosion. Enter your answer numerically in kilogram meters per second. You did not open hints for this part. ANSWER: pA,i = kg  m/s greater than less than equal to cannot be determined kg  m/s pA,f pA,f = kg  m/s ± PSS 9.1 Conservation of Momentum Learning Goal: To practice Problem-Solving Strategy 9.1 for conservation of momentum problems. An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be moving backward just after releasing the ball? PROBLEM-SOLVING STRATEGY 9.1 Conservation of momentum MODEL: Clearly define the system. If possible, choose a system that is isolated ( ) or within which the interactions are sufficiently short and intense that you can ignore external forces for the duration of the interaction (the impulse approximation). Momentum is conserved. If it is not possible to choose an isolated system, try to divide the problem into parts such that momentum is conserved during one segment of the motion. Other segments of the motion can be analyzed using Newton’s laws or, as you will learn later, conservation of energy. VISUALIZE: Draw a before-and-after pictorial representation. Define symbols that will be used in the problem, list known values, and identify what you are trying to find. SOLVE: The mathematical representation is based on the law of conservation of momentum: . In component form, this is ASSESS: Check that your result has the correct units, is reasonable, and answers the question. Model The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied. Part A Sort the following objects as part of the system or not. Drag the appropriate objects to their respective bins. ANSWER: kg kg m/s F = net 0 P = f P  i (pfx + ( + ( += ( + ( + ( + )1 pfx)2 pfx)3 pix)1 pix)2 pix)3 (pfy + ( + ( += ( + ( + ( + )1 pfy)2 pfy)3 piy)1 piy)2 piy)3 Part B This question will be shown after you complete previous question(s). Visualize Solve Part C This question will be shown after you complete previous question(s). Assess Part D This question will be shown after you complete previous question(s). Conservation of Momentum in Inelastic Collisions Learning Goal: To understand the vector nature of momentum in the case in which two objects collide and stick together. In this problem we will consider a collision of two moving objects such that after the collision, the objects stick together and travel off as a single unit. The collision is therefore completely inelastic. You have probably learned that “momentum is conserved” in an inelastic collision. But how does this fact help you to solve collision problems? The following questions should help you to clarify the meaning and implications of the statement “momentum is conserved.” Part A What physical quantities are conserved in this collision? ANSWER: Part B Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are and . What is the speed of the two-car system after the collision? the magnitude of the momentum only the net momentum (considered as a vector) only the momentum of each object considered individually v1 v2 You did not open hints for this part. ANSWER: Part C Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, what is the magnitude of their combined momentum? You did not open hints for this part. ANSWER: The answer depends on the directions in which the cars were moving before the collision. v1 + v2 v1 − v2 v2 − v1 v1v2 −−−− ” v1+v2 2 v1 + 2 v2 2 −−−−−−−  p1 p2 Part D Two cars collide inelastically and stick together after the collision. Before the collision, their momenta are and . After the collision, their combined momentum is . Of what can one be certain? You did not open hints for this part. ANSWER: Part E Two cars collide inelastically and stick together after the collision. Before the collision, the magnitudes of their momenta are and . After the collision, the magnitude of their combined momentum is . Of what can one be certain? The answer depends on the directions in which the cars were moving before the collision. p1 + p2 p1 − p2 p2 − p1 p1p2 −−−− ” p1+p2 2 p1 + 2 p2 2 −−−−−−−  p 1 p 2 p p = p1 + # p2 # p = p1 − # p2 # p = p2 − # p1 # p1 p2 p You did not open hints for this part. ANSWER: Colliding Cars In this problem we will consider the collision of two cars initially moving at right angles. We assume that after the collision the cars stick together and travel off as a single unit. The collision is therefore completely inelastic. Two cars of masses and collide at an intersection. Before the collision, car 1 was traveling eastward at a speed of , and car 2 was traveling northward at a speed of . After the collision, the two cars stick together and travel off in the direction shown. Part A p1 + p2 $p$ p1p2 −−−− ” p1 +p2 $p$ p1+p2 2 p1 + p2 $p$ |p1 − p2 | p1 + p2 $p$ p1 + 2 p2 2 −−−−−−−  m1 m2 v1 v2 First, find the magnitude of , that is, the speed of the two-car unit after the collision. Express in terms of , , and the cars’ initial speeds and . You did not open hints for this part. ANSWER: Part B Find the tangent of the angle . Express your answer in terms of the momenta of the two cars, and . ANSWER: Part C Suppose that after the collision, ; in other words, is . This means that before the collision: ANSWER: v v v m1 m2 v1 v2 v = p1 p2 tan( ) = tan = 1 45′ The magnitudes of the momenta of the cars were equal. The masses of the cars were equal. The velocities of the cars were equal. ± Catching a Ball on Ice Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 that is traveling horizontally at 11.2 . Olaf’s mass is 67.1 . Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: Part B kg m/s kg vf vf = m/s If the ball hits Olaf and bounces off his chest horizontally at 8.00 in the opposite direction, what is his speed after the collision? Express your answer numerically in meters per second. You did not open hints for this part. ANSWER: A One-Dimensional Inelastic Collision Block 1, of mass = 2.90 , moves along a frictionless air track with speed = 25.0 . It collides with block 2, of mass = 17.0 , which was initially at rest. The blocks stick together after the collision. Part A Find the magnitude of the total initial momentum of the two-block system. Express your answer numerically. m/s vf vf = m/s m1 kg v1 m/s m2 kg pi You did not open hints for this part. ANSWER: Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically. You did not open hints for this part. ANSWER: Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 0 points. pi = kg  m/s vf vf = m/s

1) Can two different forces, acting through the same point, produce the same torque on an object? Answer: Yes, as long as the component of the force perpendicular to the line joining the axis to the force is the same for both forces. 2) If you stand with your back towards a wall and your heels touching the wall, you cannot lean over to touch your toes. Why? Answer: As you bend over your center of gravity moves forward and eventually is beyond the area of the floor in touch with your feet. This does not happen when you do it away from the wall because part of your body moves back and the center of mass remains over your feet. 3) Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? A) the first applied perpendicular to the door B) the second applied at an angle C) both exert equal non-zero torques D) both exert zero torques E) Additional information is needed. 4) A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? A) It is impossible to say without knowing the masses. B) It is impossible to say without knowing the distances. C) The side the boy is sitting on will tilt downward. D) Nothing, the seesaw will still be balanced. E) The side the girl is sitting on will tilt downward. 5) A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her angular speed increases dramatically. The speed increase is a demonstration of A) conservation of energy: her moment of inertia is decreased, and so her angular speed must increase to conserve energy. B) conservation of angular momentum: her moment of inertia is decreased, and so her angular speed must increase to conserve angular momentum. C) Newton’s second law for rotational motion: she exerts a torque and so her angular speed increases. D) This has nothing to do with mechanics, it is simply a result of her natural ability to perform. 6) A girl weighing 450. N sits on one end of a seesaw that is 3.0 m long and is pivoted 1.3 m from the girl. If the seesaw is just balanced when a boy sits at the opposite end, what is his weight? Neglect the weight of the seesaw. 7) An 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward? FIGURE 11-4 8) The mobile shown in Figure 11-4 is perfectly balanced. What must be the masses of m1, m2, and m3?

1) Can two different forces, acting through the same point, produce the same torque on an object? Answer: Yes, as long as the component of the force perpendicular to the line joining the axis to the force is the same for both forces. 2) If you stand with your back towards a wall and your heels touching the wall, you cannot lean over to touch your toes. Why? Answer: As you bend over your center of gravity moves forward and eventually is beyond the area of the floor in touch with your feet. This does not happen when you do it away from the wall because part of your body moves back and the center of mass remains over your feet. 3) Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30° to the plane of the door. Which force exerts the greater torque? A) the first applied perpendicular to the door B) the second applied at an angle C) both exert equal non-zero torques D) both exert zero torques E) Additional information is needed. 4) A heavy boy and a lightweight girl are balanced on a massless seesaw. If they both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw? A) It is impossible to say without knowing the masses. B) It is impossible to say without knowing the distances. C) The side the boy is sitting on will tilt downward. D) Nothing, the seesaw will still be balanced. E) The side the girl is sitting on will tilt downward. 5) A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her angular speed increases dramatically. The speed increase is a demonstration of A) conservation of energy: her moment of inertia is decreased, and so her angular speed must increase to conserve energy. B) conservation of angular momentum: her moment of inertia is decreased, and so her angular speed must increase to conserve angular momentum. C) Newton’s second law for rotational motion: she exerts a torque and so her angular speed increases. D) This has nothing to do with mechanics, it is simply a result of her natural ability to perform. 6) A girl weighing 450. N sits on one end of a seesaw that is 3.0 m long and is pivoted 1.3 m from the girl. If the seesaw is just balanced when a boy sits at the opposite end, what is his weight? Neglect the weight of the seesaw. 7) An 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward? FIGURE 11-4 8) The mobile shown in Figure 11-4 is perfectly balanced. What must be the masses of m1, m2, and m3?

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/Two cables are tied togeather at B and are loaded , a. draw the free body diagram ((FBD) of the supporting book at B , This FBD must be a new figure , b) write the cable force the vector T in cartesian vector form, where the magnitude of the force vector is left as variable , c) write the cable force vector Tac in cartesian vector form, where the magnitude of the force vector is left as a variable, d) Determine the unit vector along the line from point A to C , e) Using the vector dot product calculate the angle between the two cable , f) write the two equations needed to calculate the tension in cable Tba and tbc

/Two cables are tied togeather at B and are loaded , a. draw the free body diagram ((FBD) of the supporting book at B , This FBD must be a new figure , b) write the cable force the vector T in cartesian vector form, where the magnitude of the force vector is left as variable , c) write the cable force vector Tac in cartesian vector form, where the magnitude of the force vector is left as a variable, d) Determine the unit vector along the line from point A to C , e) Using the vector dot product calculate the angle between the two cable , f) write the two equations needed to calculate the tension in cable Tba and tbc

Two cables are tied togeather at B and are loaded … Read More...
The diagram 3 forces F1, F2, F3 act on the body. The magnitude and direction of forces F1 and F2 are known and indicated . Also the resultant force R= F1+F2+F3 is know and indicated. Using force decomposition procedure in the Cartesian coordinate system find the magnitude of the force F3 and the angle between line of action of this force and the x direction.