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at Collision an Angle
Two cars, both of mass , collide and stick together. Prior to the collision, one car had been traveling north at speed , while the second was traveling at speed at an angle south of east (as indicated in the figure). After the collision, the two-car system travels at speed at
an angle Part A
east of north.
Find the speed of the joined cars after the collision. Hint A.1 Determine the conserved quantities Which of the following statements is true for the collision described? ANSWER: Momentum is conserved but kinetic energy is not conserved. Kinetic energy is conserved but momentum is not conserved. Both kinetic energy and momentum are conserved. Neither kinetic energy nor momentum is conserved. Correct Apply conservation of momentum: . Find both components (north and east) of the initial momentum in the problem introduction. The magnitude of momentum vector for the two-car system after the collision: Hint A.2 The component of the final velocity in the east-west direction Find the component of in the east-west direction. using the information .
is equal to the magnitude of the
Hint A.2.1 Find the east-west component of the initial momentum What is , the magnitude of the total momentum of the two cars in the east-west direction? (Take eastward to be positive, westward negative.) Express your answer in terms of ANSWER: = Correct , , and .
Now use the conservation of momentum equation to find Express your answer in terms of and . ANSWER: (east-west) = Correct
.
Hint A.3 Find the north-south component of the final momentum Find the component of in the north-south direction. Hint A.3.1 Find the north-south component of the initial momentum What is the magnitude of the total momentum of the two cars in the north-south direction? (Take northward to be positive, southward negative). Express your answer in terms of ANSWER: = Correct , , and .
Express your answer in terms of and . ANSWER: (north-south) = Correct Hint A.4 Math help Let be the east-west component of , and the north-south component. Then . You will also need to use the following trignometric identity when you evaluate the righthand side of the above equation in terms of and : . Express your answer in terms of and .
ANSWER: = Correct Part B What is the angle with respect to north made by the velocity vector of the two cars after the collision? Hint B.1 A formula for Let be the east-west component of , and the north-south component. Then
, since the angle asked for is the angle east of north. Express your answer in terms of function. ANSWER: = Correct . Your answer should contain an inverse trigonometric
A Ball Hits a Wall Elastically
A ball of mass moving with velocity strikes a vertical wall.
The angle between the ball's initial velocity vector and the wall is as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is , and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.
Part A What is the final angle that the ball's velocity vector makes with the negative y axis? Hint A.1 How to approach the problem Relate the vector components of the ball's initial and final velocities. This will allow you to determine in terms of . Hint A.2 Find the y component of the ball's final velocity What is , the component of the final velocity of the ball? Hint A.2.1 How to approach this part There is no force on the ball in the y direction. From the impulse-momentum theorem, this means that the change in the y component of the ball's momentum must be zero. Express your answer in terms of quantities given in the problem introduction and/or and , the and components of the ball's initial velocity. ANSWER: = Correct Hint A.3 Find the component of the ball's final velocity
What is , the component of the ball' final velocity? Hint A.3.1 How to approach this problem Since energy is conserved in this collision, the final speed of the ball must be equal to its initial speed. Express your answer in terms of quantities given in the problem introduction and/or and , the and components of the ball's initial velocity. ANSWER: = Correct The wall exerts a force on the ball in the direction. However, because energy is conserved in this collision, the final speed of the ball must be equal to its initial speed. Since there is no force on the ball in the y direction, the magnitude of the component of the ball's velocity is constant. Therefore, the magnitude of the component of the velocity must be constant as well. However, the sign of the velocity will change as the ball moves first toward, then away from, the wall. Hint A.4 Putting it together Once you find the vector components of the final velocity in terms of the initial velocity, use the geometry of similar triangles to determine in terms of . Express your answer in terms of quantities given in the problem introduction. ANSWER: = Correct Part B
What is the magnitude of the average force exerted on the ball by the wall? Hint B.1 What physical principle to use Use the impulse-momentum theorem, , along with the definition of impulse, . Putting everything
. In this case, only one force is acting, so together, .
Hint B.2 Change in momentum of the ball The fact that implies that the component of the ball's momentum does not change
during the collision. What is , the magnitude of the change in the ball's momentum? Express your answer in terms of quantities given in the problem introduction and/or . ANSWER: = Correct Express your answer in terms of variables given in the problem introduction and/or ANSWER: = Correct .
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 11.8 . Olaf's mass is 72.2 .
Part A If Olaf catches the ball, with what speed do Olaf and the ball move afterward?
Hint A.1 How to approach the problem Using conservation of momentum and the fact that Olaf's initial momentum is zero, set the initial momentum of the ball equal to the final momentum of Olaf and the ball, then solve for the final velocity. Hint A.2 Find the ball's initial momentum What is , the initial momentum of the ball? Express your answer numerically in kilogram centimeters per second. ANSWER: 472 = Correct Express your answer numerically in centimeters per second. ANSWER: 6.50 = Correct Part B If the ball hits Olaf and bounces off his chest horizontally at 7.00 direction, what is his speed after the collision? Hint B.1 How to approach the problem in the opposite
The initial momentum of the ball is the same as in Part A. Apply conservation of momentum, keeping in mind that both Olaf and the ball have a nonzero final momentum. Hint B.2 Find the ball's final momentum Taking the direction in which the ball was initially traveling to be positive, what is the ball's final momentum? Express your answer numerically in kilograms times centimeters per second. ANSWER: -280 = Correct Express your answer numerically in centimeters per second. ANSWER: 10.4 = Correct ,
Conservation of Momentum in Two Dimensions Ranking Task
Part A The figures below show bird's-eye views of six automobile crashes an instant before they occur. The automobiles have different masses and incoming velocities as shown. After impact, the automobiles remain joined together and skid to rest in the direction shown by . Rank these crashes according to the angle which the wreckage initially skids. , measured counterclockwise as shown, at
Hint A.1 Conservation of momentum in two dimensions Since momentum is a vector quantity, the x component of momentum and the y component of momentum must be individually conserved in any collision. Thus, the total x momentum before the collision must be equal to the total x momentum of the sliding wreckage after the collision. The same is true for the total y momentum. Hint A.2 Determining the angle Once the x and y momenta of the wreckage are determined, the exact angle through which the wreckage skids can be determined by trigonometry. Determining the exact angle of this final momentum vector is accomplished the same way you would find the angle of any vector, typically by finding the inverse tangent of the y component over the x component. (You can also determine the ranking without calculating the exact angle at which the wreckage skids.) Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:
View Correct
Surprising Exploding Firework
A mortar fires a shell of mass at speed . The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass and a larger piece of mass
. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance from the mortar. If there had been no explosion, the shell would have landed a distance from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.
Part A Find the distance from the mortar at which the larger piece of the shell lands. Hint A.1 Find the position of the center of mass in terms of The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance from the mortar to the center of mass of the exploded pieces? Hint A.1.1 Key idea The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so the center of mass will continue along the original trajectory of the shell. Express your answer in terms of . ANSWER: = Correct Hint A.2 Find the position of the center of mass in terms of The larger piece of the shell lands a distance from the mortar, and the smaller piece lands a distance zero from the mortar. What is , the final distance of the shell's center of mass from the mortar? Hint A.2.1 A helpful figure Here is a figure to help you visualize the situation.
Express your answer in terms of ANSWER: = Correct Express in terms of ANSWER: = Correct .
.
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: 1. The original mass of Chuck and his cart does not include the 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.
Part A Find the relative speed between Chuck and the ball after Chuck has thrown the ball. Hint A.1 How to approach the problem All this question is asking is: "How fast are Chuck and the ball moving away from each other?" If two objects are moving at the same speed (with respect to the ground) in the same direction, their relative speed is zero. If they are moving at the same speed, , in opposite directions, their relative speed is . In this problem, you are given variables for the speed of Chuck and the ball with respect to the ground, and you know that Chuck and the ball are moving directly away from each other. Express the speed in terms of and . ANSWER: = Correct Make sure you understand this result; the concept of "relative speed" is important. In general, if two objects are moving in opposite directions (either toward each other or away from each other), the relative speed between them is equal to the sum of their speeds with respect to the ground. If two objects are moving in the same direction, then the relative speed between them is the absolute value of the difference of the their two speeds with respect to the ground. Part B What is the speed of the ball (relative to the ground) while it is in the air? Hint B.1 How to approach the problem Apply conservation of momentum. Equate the initial (before the ball is thrown) and final (after the ball is thrown) momenta of the system consisting of Chuck, his cart, and the ball. Use the result from Part A to eliminate from this equation and solve for . Hint B.2 Initial momentum of Chuck, his cart, and the ball Before the ball is thrown, Chuck, his cart, and the ball are all at rest. Therefore, their total initial momentum is zero. Hint B.3 Find the final momentum of Chuck, his cart, and the thrown ball What is the total momentum of Chuck, his cart, and the ball after the ball is thrown? Express your answer in terms of , , , and . Remember that and are speeds, not velocities, and thus are positive scalars. ANSWER: = Correct Since equation for , you can use this equation to write in the above part to find a relation between in terms of and . . Then use the
Express your answer in terms of ANSWER: = Correct Part C
,
, and
.
What is Chuck's speed (relative to the ground) after he throws the ball? Hint C.1 How to approach the problem Use the answer to Part B to eliminate . Express your answer in terms of , ANSWER: = Correct Part D Find Jackie's speed (relative to the ground) after she catches the ball, in terms of Hint D.1 How to approach the problem . from the equation derived in Part A. Then solve for , and .
Apply conservation of momentum. Equate the initial (before Jackie catches the ball) and final (after the ball is caught) momenta of the system consisting of Jackie, her cart, and the ball, and solve for . Hint D.2 Initial momentum Just before Jackie catches the ball, the momentum of the system consisting of Jackie, her cart, and the ball is equal to the momentum of the ball as it flies through the air: . Hint D.3 Find the final momentum What is the final momentum of the system after Jackie catches the ball? Express your answer in terms of , , and . ANSWER: = Correct Express in terms of , ANSWER: = Correct Part E Find Jackie's speed (relative to the ground) after she catches the ball, in terms of Hint E.1 How to approach the problem . , and .
In Part B, you found an expression for in terms of . You can substitute this expression for into the equation you found in Part D, which will give you an expression for in terms of the desired quantities. Express in terms of , , and . ANSWER: = Correct
A Bullet Is Fired into a Wooden Block
A bullet of mass is fired horizontally with speed at a wooden block of mass resting on a frictionless table. The bullet hits the block and becomes completely embedded within it. After the bullet has come to rest within the block, the block, with the bullet in it, is traveling
at speed Part A
.
Which of the following best describes this collision? Hint A.1 Types of collisions An inelastic collision is a collision in which kinetic energy is not conserved. In a partially inelastic collision, kinetic energy is lost, but the objects colliding do not stick together. From this information, you can infer what completely inelastic and elastic collisions are. ANSWER: perfectly elastic partially inelastic perfectly inelastic Correct Part B Which of the following quantities, if any, are conserved during this collision?
Hint B.1 ANSWER:
When is kinetic energy conserved?
Kinetic energy is conserved only in perfectly elastic collisions. kinetic energy only momentum only kinetic energy and momentum neither momentum nor kinetic energy Correct Part C What is the speed of the block/bullet system after the collision? Hint C.1 Find the momentum after the collision What is the total momentum of the block/bullet system after the collision? Express your answer in terms of and other given quantities. ANSWER: = Correct Hint C.2 Use conservation of momentum The momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for , this time expressed as the total momentum of the system before the collision. Express your answer in terms of and other given quantities. ANSWER: = Correct Express your answer in terms of ANSWER: = Correct , , and .
Momentum in a Collision Graphing Question
Two asteroids, drifting at constant velocity, collide. The masses and velocities of the asteroids before the collision are indicated in the figure.
Part A Sketch graphs of the momenta of asteroids A and B before the collision. Hint A.1 Initial momentum The momentum of any object is determined by the product of the object's mass and velocity. Hint A.2 Shape of the momentum vs. time graph The asteroids are drifting at constant velocity before the collision, and have constant mass. ANSWER:
View Correct The total momentum of the system of two asteroids before the collision is equal to the sum of the individual momenta. After the collision, the two asteroids join together to form a single megaasteroid. When two objects stick together after a collision, the collision is called perfectly inelastic. Part B During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitude of the force of asteroid B on asteroid A? Hint B.1 Collision forces The forces specified in this question must obey Newton's third law. ANSWER:
equal
Correct Part C
Sketch a graph of the total momentum in the system of the two asteroids after the collision. Hint C.1 How to approach the problem The law of conservation of momentum states that the total momentum in an isolated system of objects must remain constant, regardless of the interactions (or collisions) between the objects. Thus, the total momentum of the two asteroids after the collision must be equal to the total momentum of the two asteroids before the collision. ANSWER:
View Correct Part D When the two asteroids collide, they stick together. Based on your graph in Part C, determine the velocity of the new megaasteroid. Hint D.1 Definition of momentum Momentum is equal to (total) mass times velocity. ANSWER: 16.7 Correct
Hanging Chandelier
A chandelier with mass is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectural decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier couldn't attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension and makes an angle of with the ceiling. Cable 2 has tension and makes an angle of
with the ceiling. Part A Find an expression for , the tension in cable 1, that does not depend on Hint A.1 Find the sum of forces in the x direction .
The chandelier is static; hence the vector forces on it sum to zero. Type in the sum of the x components of the forces acting on the chandelier, using the coordinate system shown.
Express your answer in terms of some or all of the variables ANSWER: Correct Hint A.2 Find the sum of forces in the y direction
,
,
,
, and
.
Now type the corresponding equation relating the y components of the forces acting on the chandelier, again using the coordinate system shown. Express your answer in terms of some or all of the variables as the magnitude of the acceleration due to gravity . ANSWER: Correct Hint A.3 Putting it all together There are two unknowns in this problem, and solve for . , , and , as well as the and . Each of the previous two hints leads from this pair of equations , , , , and , as well
you to an equation involving these two unknowns. Eliminate
Express your answer in terms of some or all of the variables magnitude of the acceleration due to gravity .
ANSWER: = Correct
The Window Washer
A window washer of mass is sitting on a platform suspended by a system of cables and
pulleys as shown
. He is pulling on the
cable with a force of magnitude . The cables and pulleys are ideal (massless and frictionless), and the platform has negligible mass. Part A Find the magnitude of the minimum force that allows the window washer to move upward. Hint A.1 Find a simple expression for the tension Find an expression for the tension in the cable on which the man is pulling. , , and .
Express your answer in terms of some or all of the variables ANSWER: = Correct Hint A.2 Upward forces on window washer What forces pull the window washer upward?
force equal to and a force equal to
A force equal to and a force equal to the tension in the string attached to the platform A force equal to the tension in the string attached to the platform and a
force equal to A force equal to and a force equal to the tension in the string attached to the platform and a force equal to Correct Hint A.3 All forces on window washer What objects exert forces on the window washer? ANSWER: Only the platform and the string being pulled Only the platform and the earth Only the earth and the cable supporting the platform Only the cable being pulled and the cable supporting the platform Only the platform; the earth; and the cable. Correct Hint A.4 What about the platform? The platform is given as having "negligible mass," so the net force on the platform must be zero. The upward force on the platform is provided by the cable connecting the left-hand pulley to the platform, and the downward force on the platform is the contact force between the window washer and the platform. This should allow a fairly simple determination of the upward force that the platform exerts on the window washer in terms Express of your answer in terms of the mass gravity . ANSWER: = Correct . and the magnitude of the acceleration due to
Block on an Incline Adjacent to a Wall
A wedge with an inclination of angle rests next to a wall. A block of mass is sliding down the plane, as shown. There is no friction between the wedge and the block or between the
wedge and the horizontal surface. Part A Find the magnitude, , of the sum of all forces acting on the block. Hint A.1 Direction of the net force on the block The net force on the block must be the force in the direction of motion, which is down the incline. Hint A.2 Determine the forces acting on the block What forces act on the block? Keep in mind that there is no friction between the block and the wedge. ANSWER: The weight of the block and friction The weight of the block and the normal (contact) force The weight of the block and the weight of the wedge The weight of the block and the force that the wall exerts on the wedge Correct Hint A.3 Find the magnitude of the force acting along the direction of motion Consider a coordinate system with the x direction pointing down the incline and the y direction perpendicular to the incline. In these coordinates, what is , the component of the block's weight in the x direction? Express in terms of , , and ANSWER: = Correct .
"Normal," in this context, is a synonym for "perpendicular." The normal force has no component in the direction of the block's motion (down the incline). Express in terms of and , along with any necessary constants.
ANSWER: = Correct Part B Find the magnitude, , of the force that the wall exerts on the wedge. Hint B.1 The force between the wall and the wedge There is no friction between the wedge and the horizontal surface, so for the wedge to remain stationary, the net horizontal force on the wedge must be zero. If the block exerts a force with a horizontal component on the wedge, some other horizontal force must act on the wedge so that the net force is zero. Hint B.2 Find the normal force between the block and the wedge What is the magnitude, , of the normal (contact) force between the block and the wedge? (You might have computed this already while answering Part A.) Express in terms of , , and ANSWER: = Correct .
Hint B.3 Find the horizontal component of the normal force In the previous hint you found the magnitude of the normal force between the block and the wedge. What is the magnitude, , of the horizontal component of this normal force? Express in terms of and . ANSWER: = Correct Express in terms of and ANSWER: = Correct , along with any necessary constants.
Your answer to Part B could be expressed as either
or
. In either
form, we see that as gets very small or as approaches 90 degrees ( radians), the contact force between the wall and the wedge goes to zero. This is what we should expect; in the first limit ( small), the block is accelerating very slowly, and all horizontal forces are small. In the second limit ( about 90 degrees), the block simply falls vertically and exerts no horizontal force on the wedge.
Two Blocks and Two Pulleys
A block of mass is attached to a massless, ideal string. This string wraps around a massless pulley and then wraps around a second pulley that is attached to a block of mass that is free to slide on a frictionless table. The string is firmly anchored to a wall and the whole
system is frictionless. Use the coordinate system indicated in the figure when solving this problem.
Part A Assuming that is the magnitude of the horizontal acceleration of the block of mass , what is , the tension in the string? Hint A.1 Which physical principle to use You should use Newton's 2nd law: ,
where , , ... are forces acting on the block of mass system is frictionless. Hint A.2 Force diagram for the block of mass
. Keep in mind that the whole
Which figure correctly illustrates the forces acting on the block of mass
?
The vectors , , , and denote the normal force, the gravitational force, the tension in the string, and the friction force, respectively. ANSWER: Figure 1 Figure 2 Figure 3 None of the above Correct Express the tension in terms of ANSWER: = Correct Part B Given , the tension in the string, calculate the block of mass . Hint B.1 Which physical principle to use Apply Newton's 2nd law: , , the magnitude of the vertical acceleration of and .
where
,
, ... are forces acting on the block of mass
. ?
Hint B.2 Force diagram for the block of mass Which figure correctly illustrates the forces acting on the block of mass
The vectors , , and denote the gravitational force, the tension in the string, and the inertial force, respectively. ANSWER: Figure 1 Figure 2 Figure 3 None of the above Correct Express the acceleration magnitude ANSWER: = Correct Part C Given the magnitude of the acceleration of the block of mass of the horizontal acceleration of the block of mass . Hint C.1 Method 1: String constraint (uses calculus) , find , the magnitude in terms of , , and .
Define and as the vertical coordinate of the block of mass coordinate of the block of mass , respectively.
and the horizontal
It is clear that is ,
, the length of the string,
where
is a constant that accounts for the wound portions of the string and the length of
string between the y axis and the wall. Do not worry about the value of , as it will vanish in the next step. By differentiating this equation twice with respect to time, you should obtain a relation between and . The variables and will vanish upon differentiation. Hint C.2 Method 2: Intuition (does not involve calculus) You should notice that, while the block of mass descends a height , the other will
move only half of . Hence, at each instant, , where and are the speeds of the blocks of masses and , respectively. The formula for versus should be obvious. Express in terms of . ANSWER: = Correct Part D Using the result of Part C in the formula for express as a function of . Express your answer in terms of ANSWER: = Correct and . that you previously obtained in Part A,
Part E Having solved the previous parts, you have all the pieces needed to calculate , the magnitude of the acceleration of the block of mass . Write an expression for . Hint E.1 How to approach this problem In Part B, using Newton's 2nd law, you derived a relation between string, . In Part D you found as a function of two linear equations and solve for . Express the acceleration magnitude in terms of ANSWER: = Correct . Now eliminate , , and . and the tension in the from this system of
A Modified Atwood Machine
Consider the situation depicted in this applet. The red block moves along a rough surface. The system begins at rest. You may assume that the pulley and rope are both massless and frictionless. Part A Which of the following forces act upon the red block? Check all that apply. ANSWER: tension weight friction normal force Correct Part B In what direction does the force of tension acting on the red block point? ANSWER: up down left right Correct Part C In what direction does the frictional force point?
ANSWER:
up down left right Correct
Part D Which of the following forces act upon the light gray block? Check all that apply. ANSWER: tension weight friction normal force Correct Part E In what direction does the force of tension acting on the light gray block point? ANSWER: up down left right Correct Part F Open the next applet. Now, you can see all of the forces. Which of the following shows the relation between the magnitude of the tension magnitude of the tension ANSWER: acting on the light gray block and the
acting on the red block?
Correct Part G Find an expression for the acceleration of the red block after it is released. Use for the mass of the red block, for the mass of the gray block, and for the coefficient of kinetic friction between the table and the red block. Hint G.1 How to approach the problem Hint not displayed Hint G.2 Find the net force on the red block in terms of Hint not displayed Hint G.3 Find the tension in the string Express your answer in terms of ANSWER: = Correct Part H Run the simulation a few times, and use your expression from the previous part to find a relation between the masses of the two blocks in this problem. Type an expression for the mass of the red block in terms of the mass of the gray block. Notice that the acceleration is given to you for any value of magnitude of the acceleration of gravity. Express your answer in terms of . ANSWER: = Correct Part I Now consider this applet, where the two blocks have different masses. Plug the starting values into your formula from Part G, using 5.0 for the mass of the red block and 10 for the magnitude of the acceleration of gravity. Then, run the applet. Hopefully, the results will look peculiar! Which of the following statements describes an error made by the applet in modeling this situation. You can change the values and run it again to help you determine just what the error is. Check all that apply. ANSWER: It allows the string to transmit compressive (pushing) forces, but a string can only transmit tensile (pulling) forces. It does not properly ensure that the friction force points in opposition to the direction of motion. that you choose. Use 10 for the Hint not displayed , , , and .
It allows gravity to act with an upward force, but gravity always acts with a downward force. It assumes that the normal force is , but that is actually the maximum possible normal force. It assumes the force of friction always equals , but this does not account for static friction. Correct The strange behavior of this applet comes from its use of a very simplified model of friction. When dealing with friction, you must always remember that it acts in opposition to motion. The speed of an object will never increase because of friction. When dealing with static friction, recall that its magnitude does not always take the value ; rather, it takes a value that cancels out the effect of any force with magnitude up to . If static friction always took the value , then whenever you set a book (or anything else) down on a table and did not apply a horizontal force, the book would begin to accelerate (presumably in some random direction)! Also, note that the coefficient of static friction is always greater than or equal to the coefficient of kinetic friction.
Velocity from Force Diagram Ranking Task
Below are birds-eye views of six identical toy cars moving to the right at 2 . Various forces act on the cars with magnitudes and directions indicated below. All forces act in the horizontal plane and are either parallel or at 45 or 90 degrees to the car's motion. Part A Rank these cars on the basis of their speed a short time after the forces are applied. Hint A.1 How to approach the problem First, added up the force vectors to find the net force acting on the car. Since you are asked about the speed a short time after the forces are applied, the speeds of all cars will be close to 2 . For each car, the small difference from 2 will be due to the acceleration caused by the net force acting on the car. A net force acting to the right will result in a speed greater than 2 2 law, , while a net force acting to the left will result in a speed less than
. Recall that the acceleration may be found from the net force by using Newton's 2nd .
Hint A.2 Summing force vectors Forces are vectors and sum in the same way that velocity or acceleration vectors sum. Be careful when resolving vectors into components, if you need to. Once the force vectors are summed into a single total, or net force , the acceleration of the car of mass can be
determined by Newtons 2nd law: . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER:
View Correct
Binary Star System
A binary star system consists of two stars of masses and . The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass has a centripetal acceleration of magnitude . Note that you do not need to understand universal gravitation to solve this problem. Part A Find , the magnitude of the centripetal acceleration of the star with mass Hint A.1 What causes acceleration? .
First consider what makes the stars orbit around their center of mass. Which of the following statements most correctly explains the physical origin of the centripetal acceleration of the star of mass ? 1. The gravitational force between the star of mass and the star of mass provides the centripetal acceleration necessary for circular motion. 2. The circular motion of the star of mass generates the centripetal acceleration. 3. The centripetal acceleration is the sum of the acceleration due to gravity and the acceleration due to the circular motion. 4. Once moving, the inertia of star 2 causes it to continue moving along its circular orbit, at constant speed, as dictated by Newton's First Law. ANSWER: 1 2 3 4 Correct
Hint A.2 Relationship between forces on each star What is the magnitude of the force acting on the star of mass Express your answer in terms of mass . ANSWER: = Correct ? , the magnitude of the force that acts on the star with
The forces and are equal in magnitude, but they act in opposite directions--in fact they pull both planets toward the center of mass (which lies on a line between them). Hint A.3 How to relate acceleration and force According to Newton's 2nd law, how can you express acts on the star with mass , in terms of and ? ANSWER: = Correct Hint A.4 Putting it together If you now apply Newton's 2nd law to the star of mass , you should be able to find its acceleration by combining the formulas from the previous hints. Express the acceleration in terms of quantities given in the problem introduction. ANSWER: = Correct To make sure you understand this result, consider the actual gravitational force acting on each star. The magnitude of the gravitational force on either star due to the other one is given by , the magnitude of the force that
, where is the separation between the stars. . The net external force is , so
Now, consider Newton's 2nd law for the star of mass .
Now consider Newton's 2nd law applied to the star of mass force acting on this star will be . You can see that the same force,
. Once again, the net external is
, so Newton's 2nd law for the star of mass
, appears in both the equation for the star of mass
and
that for the star of mass therefore write
. (Think about how this relates to Newton's 3rd law.) You can . Solving for the acceleration you find the equation
. Note that you did not need to know the exact form of the gravitational force, nor did you need to know or . Newton's 3rd law allows you to realize that is the same for the two stars, and Newton's 2nd law allows you to solve for in terms of , , and .
Pushing a Chair along the Floor
A chair of weight 150 lies atop a horizontal floor; the floor is not frictionless. You push on directed at an angle of 42.0 below the horizontal and the chair with a force of = 45.0 the chair slides along the floor. Part A
Using Newton's laws, calculate , the magnitude of the normal force that the floor exerts on the chair. Hint A.1 How to approach the problem To solve this problem you need to focus on the forces that have a vertical component. In fact, because the direction of motion is horizontal (the chair slides along the floor), both the acceleration of the chair and the net force acting on the chair are purely horizontal. This can be true only if the net vertical force acting on the chair is zero, that is, if the normal force exerted on the chair by the floor balances all the other vertical forces acting on the chair. Also, in any problem involving forces you should always draw a free-body diagram that shows all the forces acting on the system. To do that, choose a coordinate system and identify all the forces acting on the chair. Hint A.2 Choosing the correct free-body diagram Let be the friction force, be the force you exert on the chair, and be the weight of the chair. If you choose the x axis to be parallel to the floor, which free-body diagram would correctly represent the situation described in the introduction of this problem?
Hint A.2.1 Direction of the friction force Hint not displayed b c d Correct Hint A.3 Find the vertical net force Let the positive y axis point upward, and let be the magnitude of the normal force exerted on the chair by the floor, be the magnitude of the weight of the chair, and be the magnitude of the y component of the force you exert on the chair when you push it. What is the vertical net force, that is, the y component of the net force, acting on the chair? Hint A.3.1 How to approach the problem From your free-body diagram you can see that the normal force points upward, whereas the weight points downward. Moreover, the force that you exert on the chair has a vertical component, which is parallel to the weight. ANSWER:
Correct Recall that the vertical net force is zero because there is no vertical acceleration. To find the
normal force exerted on the chair by the floor, find the vertical component of the force you exert on the chair and solve for . Hint A.4 Find the vertical component of the force that you exert on the chair Find , the magnitude of the vertical component of the force that you exert on the chair when you push it. Hint A.4.1 Components of a vector Consider a vector components of that forms an angle are, respectively, and where is the magnitude of the vector. Express your answer in newtons. ANSWER: 30.1 = Correct Express your answer in newtons. ANSWER: 180 = Correct with the positive x axis. The x and the y
Tension Definition
This problem concerns the concept of tension in a rope. Consider a rope subjected to a pulling
force on its two ends as shown. The rope is stationary. An arbitrary point P divides the rope into a left-hand segment L and a right-hand segment R. Part A
For segment R and segment L to hold together, they must exert forces on each other. What is the direction of the force exerted on segment R by segment L? Hint A.1 Look at the forces The forces on segment R are 1. the force due to L and 2. the pull (to the right) at the right end of the rope. What must be true about these two forces if the rope is to remain stationary? ANSWER: left right Correct Part B Assume that segment R exerts a force of magnitude of the force exerted on segment R by segment L? Hint B.1 How to approach the problem Apply Newton's 3rd law. Give your answer in terms of and other constants such as ANSWER: = Correct . on segment L. What is the magnitude
The magnitude of the force exerted by one part of a rope on another at a certain point is called the tension at that point. Tension is often designated with the symbol . The tension is also the magnitude of the force with which the rope pulls on whatever is attached to its ends. Part C Now imagine two points, Q and P, that divide the rope into segments L, M ,and R.
The rope remains stationary. Assume that segment L exerts a force of magnitude on segment M. What is the magnitude force exerted by segment R on segment M? Hint C.1 How to approach the problem Draw a force diagram for the section M of the rope. Hint C.2 Relative magnitudes of the two forces The force exerted by segment L on segment M is ______ the force exerted by segment R on segment M? ANSWER: equal to Correct Give your answer in terms of ANSWER: = Correct and constants such as . of the
This answer implies that, in the absence of gravity, the tension in a stationary rope may be taken to be constant. Part D Now consider a rope that, unlike those usually studied in mechanics problems, actually has a significant mass . The tension at the right end of this rope is and that at the left end is
. the right.
The rope has an acceleration
to
Complete the following equation for the acceleration of the section of the rope of mass taking the positive direction to be to the right. Hint D.1 How to approach the problem Use Newton's 2nd law of motion. Give your answer in terms of , , and constants such as ANSWER: = Correct Part E Which of the following phrases, if they appear in a problem, allow you to assume that in a horizontally oriented rope? A. The rope is massless. B. The rope is moving at constant speed. C. The rope is stretched with negligible sag. only B only C only A or B A or C B or C A or B or C .
,
Correct The reason why a massless rope would cause . The force is given by . Since means that the force must be zero ( can be seen using Newton's 2nd law: then for any value of . . This ). Therefore,
Tension in a Hanging Massive Rope
Consider a rope with length and mass per unit length , hanging vertically as shown.
Let the bottom of the rope. Part A
refer to the height of a point P above
The force exerted on the rope by the ceiling is in the _____ direction. Hint A.1 Consider the weight of the rope To support the rope, the ceiling must exert a force opposite to the weight of the rope. ANSWER:
upw ard
Correct Part B Find , the magnitude of the force exerted on the rope by the ceiling. Hint B.1 Use Newton's 2nd law Take to be the weight of the rope. Apply Newton's 2nd law to find an expression for 0 = Correct
. ANSWER:
Hint B.2 Find the weight of the rope What is the weight of the rope, ? Hint B.2.1 Find the mass of the rope What is the mass ANSWER: of the rope? = Correct Express your answer in terms of ANSWER: = Correct , , and .
Express in terms of quantities given in the problem introduction. ANSWER: = Correct Part C What is the tension at point P in the rope? Hint C.1 How to approach this problem The tension at point P is the magnitude of the force exerted by the part of the rope above P on the part of the rope below P. Remember that point P is located a distance above the bottom of the rope. Hint C.2 Find the weight of the rope below point P What is the weight of the part of the rope below point P? Express your answer in terms of quantities given in the problem introduction. ANSWER: = Correct Now apply Newton's 2nd law to the point P on the rope. Express in terms of quantitites given in the problem introduction. ANSWER: = Correct

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UC Davis - ENG - 102

UC Davis - ENG - 102

UC Davis - ENG - 102

UC Davis - ENG - 102

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UC Davis - ENG - 102

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