Mastering Physics Ch 04 HW College Physics I Brian Uzpen LCCC
February 5, 2017 | Author: Samuel | Category: N/A
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Mastering Physics...
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5/6/2016
Ch 04 HW
Ch 04 HW Due: 9:30am on Monday, February 22, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy
Prelecture Video: Newton’s First Law Click Play to watch the video. Answer the ungraded questions in the video and the graded followup questions at right.
Part A Recall the portion of the video in which the girl pushes her brother on the sled at constant velocity. The pushing force she exerts on the sled is _____ the frictional force the ground exerts on the sled. ANSWER: greater than equal to less than
Correct Because the sled’s velocity is constant, the net force must be zero and the two horizontal forces must balance out.
Part B According to Newton’s first law, when the net force acting on an object is zero, the object must _____. ANSWER:
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have zero velocity have constant velocity be slowing down be speeding up
Correct When the net force is zero, the acceleration is zero, so the velocity must be constant. Note that zero velocity is simply a special case of constant velocity.
Part C An elevator moves straight upward at a constant speed. Newton’s first law predicts that the net force acting on the elevator will _____. ANSWER: be zero point upward be increasing
Correct An object moving in a straight line at constant speed has zero acceleration because the net force acting on it is also zero.
Part D Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? Ignore air resistance. Check all that apply. ANSWER: at the instant the falling ball hits your hands when you hold the ball still in your hands after catching it just after the ball first leaves your hands at the instant the ball reaches its highest point when you hold the ball still in your hands before it is thrown
Correct When an object’s velocity is changing, the net force on it is not zero, even if it stops for an instant. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Prelecture Video: Newton’s Second Law Click Play to watch the video. Answer the ungraded questions in the video and the graded followup questions at right.
Part A In the video, forces acting on the car that are parallel to the direction of motion are analyzed. How are these forces related? ANSWER: The forces are equal in size, act in the same direction, and produce a net force in the direction of motion. The forces are unequal in size, act in the same direction, and produce a net force in the direction of motion. The forces are equal in size, act in opposite directions, and produce a net force in the direction of motion. The forces are unequal in size, act in opposite directions, and produce a net force in the direction of motion. The forces are equal in size, act in opposite directions, and produce a zero net force.
Correct The car is subject to two forces acting along the direction of motion. The propelling force acts in the direction of motion, while the force of drag acts opposite the direction of motion. Because the two forces are equal in size, they cancel, producing a zero net force. As a result, the car travels at constant speed.
Part B The mass of a rocket decreases as it burns through its fuel. If the rocket engine produces constant force (thrust), how does the acceleration of the rocket change over time? ANSWER:
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It does not change. It increases. It decreases.
Correct The constantforce rocket’s acceleration is inversely related to its mass. Because its mass decreases over time, its acceleration increases.
Part C Newton’s second law relates an object’s acceleration to its mass and the net force acting on it. Does Newton’s second law apply to a situation in which there is no net force? Select the best explanation. ANSWER: No. The law applies only to situations in which a net force acts. Yes. The law applies and it tells us that the object has a nonzero acceleration. Yes. The law applies and it tells us that the object has constant velocity. No. The law does not apply because a zero net force would produce an infinite acceleration.
Correct Newton’s second law applies to all situations, whether there is a net force or not. In the case of zero net force, any finite mass object must have zero acceleration, which means that its velocity is constant.
Part D An experiment measures the acceleration of an object of known mass subject to a known net force. The experiment is then repeated while varying the mass and the net force. The table below summarizes the conditions used in each trial. Trial
Mass
Net force
1
doubled
doubled
2
halved
unchanged
3
doubled
unchanged
4
halved
doubled
Rank the accelerations measured in each trial from largest to smallest. ANSWER:
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Correct The accelerations from largest to smallest are as follows: Trial 4 (quadrupled), Trial 2 (doubled), Trial 1 (unchanged) and Trial 3, (halved).
An Object Accelerating on a Ramp Learning Goal: Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the xy plane), the definition of acceleration is a⃗ (t) =
⃗ ⃗ v (t+Δt)− v (t) Δt
in the limit Δt→0.
In picturing this vector derivative you can think of the derivative of a vector as an instantaneous quantity by thinking of the velocity of the tip of the arrow as the vector changes in time. Alternatively, you can (for small Δt) approximate the acceleration as a⃗ (t) ≈
⃗ ⃗ v (t+Δt)− v (t) Δt
.
⃗ + Δt) and v (t) ⃗ Obviously the difference between v (t is another vector that can lie in any direction. If it is longer but in ⃗ ⃗ + Δt) has the same magnitude as v (t) ⃗ the same direction, a⃗ (t) will be parallel to v (t) . On the other hand, if v (t but ⃗ + Δt) can differ from v (t) ⃗ is in a slightly different direction, then a⃗ (t) will be perpendicular to v ⃗ . In general, v (t in ⃗ both magnitude and direction, hence a⃗ (t) can have any direction relative to v (t) .
This problem contains several examples of this.Consider an object sliding on a frictionless ramp as depicted here. The object is already moving along the ramp toward position 2 when it is at position 1. The following questions concern the direction of the object's acceleration vector, a⃗ . In this problem, you should find the direction of the acceleration vector by drawing the velocity vector at two points near to the position you are asked about. Note that since the object moves https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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along the track, its velocity vector at a point will be tangent to the track at that point. The acceleration vector will point in the same direction as the vector difference of the two velocities. (This is a result of the equation ⃗ + Δt) − v (t))/Δt ⃗ given above.) a⃗ (t) ≈ (v (t
Part A Which direction best approximates the direction of a⃗ when the object is at position 1?
Hint 1. Consider the change in velocity At this point, the object's velocity vector is not changing direction; rather, it is increasing in magnitude. Therefore, the object's acceleration is nearly parallel to its velocity. ANSWER: straight up downward to the left downward to the right straight down
Correct
Part B Which direction best approximates the direction of a⃗ when the object is at position 2?
Hint 1. Consider the change in velocity At this point, the speed has a local maximum; thus the magnitude of v ⃗ is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of v ⃗ is changing there is an acceleration. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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straight up upward to the right straight down downward to the left
Correct Even though the acceleration is directed straight up, this does not mean that the object is moving straight up.
Part C Which direction best approximates the direction of a⃗ when the object is at position 3?
Hint 1. Consider the change in velocity At this point, the speed has a local minimum; thus the magnitude of v ⃗ is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of v ⃗ is changing there is an acceleration. ANSWER: upward to the right to the right straight down downward to the right
Correct
PhET Tutorial: Projectile Motion Learning Goal: To understand how the trajectory of an object depends on its initial velocity, and to understand how air resistance affects the trajectory. For this problem, use the PhET simulation Projectile Motion. This simulation allows you to fire an object from a cannon, see its trajectory, and measure its range and hang time (the amount of time in the air).
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Start the simulation. Press Fire to launch an object. You can choose the object by clicking on one of the objects in the scrolldown menu at top right (a cannonball is not among the choices). To adjust the cannon barrel’s angle, click and drag on it or type in a numerical value (in degrees). You can also adjust the speed, mass, and diameter of the object by typing in values. Clicking Air Resistance displays settings for (1) the drag coefficient and (2) the altitude (which controls the air density). For this tutorial, we will use an altitude of zero (sea level) and let the drag coefficient be automatically set when the object is chosen. Play around with the simulation. When you are done, click Erase and select a baseball prior to beginning Part A. Leave Air Resistance unchecked.
Part A First, you will investigate purely vertical motion. The kinematics equation for vertical motion (ignoring air resistance) is given by 2 y(t) = y + v 0 t − (1/2)gt , 0 where y0 = 0 is the initial position (which is 1.2 m above the ground due to the wheels of the cannon), v 0 is the initial speed, and g is the acceleration due to gravity. Shoot the baseball straight upward (at an angle of 90∘ ) with an initial speed of 20 m/s . How long does it take for the baseball to hit the ground? Express your answer with the appropriate units. ANSWER: 4.1 s
Correct Notice that this value could be determined from the kinematics equation: Approximating the final height to be y(t f ) = 0 (keep in mind that the location where y = 0 is inside the cannon, not the ground), the kinematics equation becomes (v 0 − 0.5gt) = 0 , or t = 2v 0 /g = 4.1 s. This calculation is interesting because it shows that, for vertical motion, the time the ball is in the air is proportional to its initial speed.
Part B When the baseball is shot straight upward with an initial speed of 20 m/s , what is the maximum height above its8/43 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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When the baseball is shot straight upward with an initial speed of 20 m/s , what is the maximum height above its initial location? (Note that the ball’s initial height is denoted by the horizontal white line. It is initially 1.2 m above the ground. The yellow box that is below the target on the grass is measuring tape that should be used for this part.) Express your answer with appropriate units.
Hint 1. How to approach the problem Use the measuring tape to determine the height. Align the plus sign at the beginning of the spool with the horizontal white line, and drag the end of the tape to the maximum height of the ball’s trajectory. You can zoom in or out using the magnification buttons above the Fire and Erase buttons. ANSWER: 20 m
Correct Notice that this value could be determined from the kinematics equation: Since you found it takes 4.1 s for the ball to reach the ground, it must take roughly 2.1 s to reach the maximum height (using the unrounded 2 answer), which is given by y(t = 2.1 s) = (20 m/s)(2.1 s) − (1/2)(9.8 m/s 2 )(2.1 s) = 20.4 m or 20 m using two significant figures. (Since 20 m is considerably longer than the initial height above the ground, it is a good approximation to pretend that the initial position of the baseball was the ground.)
Part C If the initial speed of the ball is doubled, how does the maximum height change?
Hint 1. How to approach the problem You can use the simulation to help with this question. Set the initial speed to 40 m/s , and again measure the maximum height with the measuring tape. You will likely have to zoom out to see the top of the ball’s trajectory. ANSWER: The maximum height increases by a factor of four. The maximum height increases by a factor of 1.4 (square root of 2). The maximum height increases by a factor of two.
Correct Since the amount of time it takes to reach the maximum height doubles, and since its average velocity in going upward also doubles (the average velocity is equal to half the initial velocity), the height it reaches before stopping increases by a factor of four (distance is equal to the average velocity multiplied by the time duration). https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Part D Erase all the trajectories, and fire the ball vertically again with an initial speed of 20 m/s . As you found earlier, the maximum height is roughly 20 m. If the ball isn’t fired vertically, but at an angle less than 90∘ , it can reach the same maximum height if its initial speed is faster. Set the initial speed to 25 m/s , and find the angle such that the maximum height is roughly 20 m. Experiment by firing the ball with many different angles. You can use the measuring tape to determine the maximum height of the trajectory and compare it to 20 m. What is this angle? ANSWER: ∘
30
∘
75
∘
53
∘
63
∘
45
Correct Notice that the initial speed in the vertical direction is given by (25 m/s)sin(53∘ ) = 20 m/s. The ball launched at this angle reaches the same height as the vertically launched ball because they have the same initial speeds in the vertical direction.
Part E In the previous part, you found that a ball fired with an initial speed of 25 m/s and an angle of 53∘ reaches the same height as a ball fired vertically with an initial speed of 20 m/s . Which ball takes longer to land? ANSWER: The ball fired vertically stays in the air longer. Both balls are in the air the same amount of time. The ball fired at an angle of 53∘ stays in the air longer.
Correct The vertical component of the velocity determines how long the ball will be in the air (and its maximum height). The horizontal component of the ball's velocity does not affect this hang time.
Part F The figure shows two trajectories, made by two balls launched with different angles and possibly different initial speeds.
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Based on the figure, for which trajectory was the ball in the air for the greatest amount of time?
Hint 1. How to approach the problem Think about the result of Part E (think about the relationship between the maximum height of something thrown upward and the amount of time it is in the air). Does the time in the air depend on the range of the trajectory? ANSWER: It’s impossible to tell solely based on the figure. Trajectory A The balls are in the air for the same amount of time. Trajectory B
Correct All that matters is the vertical height of the trajectory, which is based on the component of the initial velocity in the vertical direction (v 0 sinθ ). The higher the trajectory, the more time the ball will be in the air, regardless of the ball’s range or horizontal velocity.
Part G The range is the distance from the cannon when the ball hits the ground. This distance is given by the horizontal velocity (which is constant) times the amount of time the ball is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 20 m/s , and fire the ball several times while varying the angle between the cannon and the horizontal. Notice that the digital display near the top gives the range of the ball. For which angle is the range a maximum (with the initial speed held constant)? ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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0
∘
60
∘
30
∘
90
∘
45
Correct When the ball is launched near a level ground, 45∘ is the optimum angle. If launched with a greater angle, it stays in the air longer, but its horizontal speed is slower, and it won’t go as far. If launched with a smaller angle, its horizontal speed is faster, but it won’t stay in the air as long and it won’t go as far. The product between the horizontal speed and the amount of time in the air is largest when the angle is 45∘ .
Part H How does the range of the object change if its initial velocity is doubled (keeping the angle fixed and less than ∘ 90 )? ANSWER: The ball’s range is eight times as far. The ball’s range is twice as far. The ball’s range is four times as far.
Correct Since the vertical component of the velocity is twice as large, it takes twice as long to hit the ground. The horizontal component of the velocity is also twice as large, and since the range is equal to the horizontal velocity times the amount of time the ball is in the air, the range increases by a factor of four. The results of this question and the previous question can be summarized by the range equation, which is 2 R = 2v 0 sin(θ)cos(θ)/g.
Part I Now, let’s see what happens when the cannon is high above the ground. Click on the wheel of the cannon, and drag it upward as far as it goes (about 21 m above the ground). Set the initial velocity to 20 m/s , and fire several balls while varying the angle. For what angle is the range the greatest? ANSWER:
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30
∘
45
∘
35
∘
40
∘
50
Correct Since the cannon is very high off the ground, the ball will be in the air for an appreciable amount of time even if the ball is launched nearly horizontally. Thus, the amount of time the ball is in the air isn’t proportional to the vertical component of the initial velocity (as it was when the cannon was on the ground). This means that the initial horizontal velocity is more important, resulting in an optimal angle less than 45∘ . You should realize that the range equation given in Part H, sin(θ)cos(θ)/g, is not valid when the initial height is not zero. You can also verify that, if you change the initial velocity, the optimal angle also changes! R = 2v 0
2
Part J So far in this tutorial, you have been launching a baseball. Let’s see what happens to the trajectory if you launch something bigger and heavier, like a Buick car. Compare the trajectory and range of the baseball to that of the Buick car, using the same initial speed and angle (e.g., 45∘ ). (Be sure that air resistance is still turned off.) Which statement is true? ANSWER: The trajectories and thus the range of the Buick and the baseball are identical. The trajectories differ; the range of the Buick is shorter than that of the baseball. The trajectories differ; the range of the Buick is longer than that of the baseball.
Correct Since we are ignoring air resistance, the trajectory of the object does not depend on its mass or size. In the next part, you will turn on air resistance and discover what changes.
Part K In the previous part, you discovered that the trajectory of an object does not depend on the object’s size or mass. But if you have ever seen a parachutist or a feather falling, you know this isn’t really true. That is because we have been neglecting air resistance, and we will now study its effects here. Select Air Resistance for the simulation. Fire a baseball with an initial speed of roughly 20 m/s and an angle of ∘ 45 . Compare the trajectory to the case without air resistance. How do the trajectories differ? ANSWER:
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The trajectory with air resistance has a shorter range. The trajectory with air resistance has a longer range. The trajectories are identical.
Correct Air resistance is a force due to the object ramming through the air molecules, and is always in the opposite direction to the object’s velocity. This means the air resistance force will slow the object down, resulting in a shorter range (the simulation assumes the air is still; there is no strong tailwind).
Part L Notice that you can adjust the diameter (and mass) of any object (e.g., you can make a really big baseball). What happens to the trajectory (with air resistance on) when you increase the diameter while keeping the mass constant? ANSWER: Increasing the size makes the range of the trajectory decrease. The size of the object doesn’t affect the trajectory. Increasing the size makes the range of the trajectory increase.
Correct Since the surface area increases if the diameter increases, the object is sweeping through more air, causing more collisions, and a greater force of air drag (in fact, if the diameter is doubled, for a given speed, the force of air drag is increased by a factor of four). This greater force of air drag causes the object to slow down more quickly, resulting in a slower average speed and a shorter range.
Part M You might think that it is never a good approximation to ignore air resistance. However, often it is. Fire the baseball without air resistance, and then fire it with air resistance (same angle and initial speed). Then, adjust the mass of the baseball (increase it and decrease it) and see what happens to the trajectory. Don’t change the diameter. When does the range with air resistance approach the range without air resistance? ANSWER: The range with air resistance approaches the range without air resistance as the mass of the baseball is decreased. It never does. Regardless of the mass, the range with air resistance is always shorter than the range without. The range with air resistance approaches the range without air resistance as the mass of the baseball is increased.
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Correct As the mass is increased, the force of gravity on the baseball becomes larger. The force due to air drag just depends on the speed and the size of the object, so it doesn’t change if the mass changes. As the mass gets large enough, the force of gravity becomes much larger than the air drag force, and so the air drag force becomes negligible. This results in a trajectory nearly the same as when air resistance is turned off. Thus, for small, dense objects (like rocks and bowling balls), air resistance is typically unimportant, but for objects with a low density (like feathers) or a very large surface area (like parachutists), air resistance is very important.
PhET Interactive Simulations University of Colorado http://phet.colorado.edu
Video Tutor: Ball Fired Upward from Moving Cart First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.
Part A The crew of a cargo plane wishes to drop a crate of supplies on a target below. To hit the target, when should the crew drop the crate? Ignore air resistance.
Hint 1. How to approach the problem While the crate is on the plane, it shares the plane's velocity. What is the crate's velocity immediately after it is released?
Hint 2. What affects the motion of the crate? Gravity will accelerate the crate downward. What, if anything, affects the crate's horizontal motion? (Keep in mind that we are told to ignore air resistance, even though that's not very realistic in this situation.) ANSWER:
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After the plane has flown over the target When the plane is directly over the target Before the plane is directly over the target
Correct At the moment it is released, the crate shares the plane's horizontal velocity. In the absence of air resistance, the crate would remain directly below the plane as it fell.
Video Tutor: Dropped and Thrown Balls First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.
Part A Which ball (if either) has the greatest speed at the moment of impact?
Hint 1. How to approach this problem The balls strike the ground at the same time because the horizontal and vertical components of their velocities are independent, meaning that the two balls fall downward at the same rate even though one ball is also moving horizontally. However, the speed of a ball at any point is the magnitude of the ball's resultant (total) velocity. How does the ball's resultant velocity relate to the vertical and horizontal velocity components? If you need to, draw diagrams showing the velocities of the two balls at the moment of impact. For each ball, show the vertical and horizontal components of velocity (the latter being zero for the dropped ball), and show the resultant velocity. Which ball is moving faster? ANSWER:
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The ball thrown horizontally Both balls have the same speed. The dropped ball
Correct The two balls have the same vertical velocity when they land, but the thrown ball has an additional horizontal velocity component. Since speed is defined as the magnitude of the resultant velocity vector, the thrown ball is moving faster when it lands.
± Arrow Hits Apple ∘
An arrow is shot at an angle of θ = 45 above the horizontal. The arrow hits a tree a horizontal distance D = 220 m away, at the same height above the ground as it was shot. Use g = 9.8 m/s 2 for the magnitude of the acceleration due to gravity.
Part A Find t a , the time that the arrow spends in the air. Answer numerically in seconds, to two significant figures.
Hint 1. Find the initial upward component of velocity in terms of D. Introduce the (unknown) variables v y0 and v x0 for the initial components of velocity. Then use kinematics to relate them and solve for t a . What is the vertical component v y0 of the initial velocity? Express your answer symbolically in terms of t a and D.
Hint 1. Find v x0 Find the horizontal component v x0 of the initial velocity. Express your answer symbolically in terms of t a and given symbolic quantities. ANSWER: v x0
=
D ta
Hint 2. Find v y0 What is the vertical component v y0 of the initial velocity? Express your answer symbolically in terms of v x0 . ANSWER:
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=
v x0
ANSWER: v y0
=
D ta
Correct Hint 2. Find the time of flight in terms of the initial vertical component of velocity. From the change in the vertical component of velocity, you should be able to find t a in terms of v y0 and g . Give your answer in terms of v y0 and g .
Hint 1. Find v y (t a ) When applied to the ycomponent of velocity, in this problem the formula for v(t) with constant acceleration −g is v y (t) = v y0 − g t
What is v y (t a ) , the vertical component of velocity when the arrow hits the tree? Answer symbolically in terms of v y0 only. ANSWER: v y (t a )
=
−v y0
ANSWER:
ta
=
2v y0 g
All attempts used; correct answer displayed Hint 3. Put the algebra together to find t a symbolically. If you have an expression for the initial vertical velocity component in terms in terms of D and t a , and another in terms of g and t a , you should be able to eliminate this initial component to find an expression for t
2 a
Express your answer symbolically in terms of given variables. ANSWER: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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ta
=
2D g
All attempts used; correct answer displayed
ANSWER: ta
= 6.7 s
Correct
Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.
Part B How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree? Express your answer numerically in seconds, to two significant figures.
Hint 1. When should the apple be dropped The apple should be dropped at the time equal to the total time it takes the arrow to reach the tree minus the time it takes the apple to fall 6.0 meters.
Hint 2. Find the time it takes for the apple to fall 6.0 meters How long does it take an apple to fall 6.0 meters? Express your answer numerically in seconds, to two significant figures. ANSWER: tf
= 1.1 s
ANSWER: td
= 5.6 s
Correct
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Battleship Shells A battleship simultaneously fires two shells toward two identical enemy ships. One shell hits ship A, which is close by, and the other hits ship B, which is farther away. The two shells are fired at the same speed. Assume that air resistance is negligible and that the magnitude of the acceleration due to gravity is g . Note that after Part B the question setup changes slightly.
Part A What shape is the trajectory (graph of y vs. x) of the shells? ANSWER: straight line parabola hyperbola The shape cannot be determined.
Correct
Part B For two shells fired at the same speed which statement about the horizontal distance traveled is correct?
Hint 1. Two things to consider The distance traveled is the product of the x component of the velocity and the time in the air. How does the y component of the velocity affect the "air time"? What angle would give the longest range? ANSWER: The shell fired at a larger angle with respect to the horizontal lands farther away. The shell fired at an angle closest to 45 degrees lands farther away. The shell fired at a smaller angle with respect to the horizontal lands farther away. The lighter shell lands farther away.
Correct
Now, consider for the remaining parts of the question below that both shells are fired at an angle greater than 45 degrees with respect to the horizontal. Remember that enemy ship A is closer than enemy ship B.
Part C https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Which shell is fired at the larger angle?
Hint 1. Consider the limiting case Consider the case in which a shell is fired at 90 degrees above the horizontal (i.e., straight up). What distance x will the shell travel? Now lower the angle at which the shell is fired. What happens to the distance x ? ANSWER: A B Both shells are fired at the same angle.
Correct
Part D Which shell is launched with a greater vertical velocity, v y ? ANSWER: A B Both shells are launched with the same vertical velocity.
Correct
Part E Which shell is launched with a greater horizontal velocity, v x ? ANSWER: A B Both shells are launched with the same horizontal velocity.
Correct
Part F https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Which shell reaches the greater maximum height?
Hint 1. What determines maximum height? What determines the maximum height reached by the shell? ANSWER: horizontal velocity vertical velocity mass of the shell
ANSWER: A B Both shells reach the same maximum height.
Correct
Part G Which shell has the longest travel time (time elapsed between being fired and hitting the enemy ship)?
Hint 1. Consider the limiting case If a shell is fired exactly horizontally (0 degrees) the shell hits the ground right away. As the angle above the horizontal increases, what happens to the time of travel? Does this change as the angle becomes greater than 45 degrees? ANSWER: A B Both shells have the same travel time.
Correct
PSS 4.1 Projectile Motion Problems Learning Goal: https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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To practice ProblemSolving Strategy 4.1 for projectile motion problems. A rock thrown with speed 11.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 17.0 m before hitting the ground. From what height was the rock thrown? Use the value g = 9.810 m/s 2 for the freefall acceleration. PROBLEMSOLVING STRATEGY 4.1 Projectile motion problems MODEL: Make simplifying assumptions, such as treating the object as a particle. Is it reasonable to ignore air
resistance? VISUALIZE: Use a pictorial representation. Establish a coordinate system with the x axis horizontal and the y axis
vertical. Show important points in the motion on a sketch. Define symbols, and identify what you are trying to find. SOLVE: The acceleration is known: a x
and a y
= 0
= −g
. Thus, the problem becomes one of twodimensional
kinematics. The kinematic equations are x f = x i + v ix Δt,
y f = y i + v iy Δt −
1 2
g(Δt)
2
,
. Δt is the same for the horizontal and vertical components of the motion. Find Δt from one component, and then use that value for the other component. v f x = v ix = constant, and
v f y = v iy − gΔt
ASSESS: Check that your result has the correct units, is reasonable, and answers the question.
Model Start by making simplifying assumptions: Model the rock as a particle in free fall. You can ignore air resistance because the rock is a relatively heavy object moving relatively slowly.
Visualize Part A Which diagram represents an accurate sketch of the rock's trajectory?
Hint 1. The launch angle In a projectile's motion, the angle of the initial velocity v i⃗ above the horizontal is called the launch angle. ANSWER:
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Correct
Part B As stated in the strategy, choose a coordinate system where the x axis is horizontal and the y axis is vertical. Note that in the strategy, the y component of the projectile's acceleration, a y , is taken to be negative. This implies that the positive y axis is upward. Use the same convention for your y axis, and take the positive x axis to be to the right. Where you choose your origin doesn't change the answer to the question, but choosing an origin can make a problem easier to solve (even if only a bit). Usually it is nice if the majority of the quantities you are given and the quantity you are trying to solve for take positive values relative to your chosen origin. Given this goal, what location for the origin of the coordinate system would make this problem easiest? https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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ANSWER: At the point where the rock is released At ground level below the point where the rock is launched At the point where the rock strikes the ground At ground level below the peak of the trajectory At the peak of the trajectory
Correct It's best to place the origin of the coordinate system at ground level below the launching point because in this way all the points of interest (the launching point and the landing point) will have positive coordinates. (Based on your experience, you know that it's generally easier to work with positive coordinates.) Keep in mind, however, that this is an arbitrary choice. The correct solution of the problem will not depend on the location of the origin of your coordinate system. Now, define symbols representing initial and final position, velocity, and time. Your target variable is yi , the initial y coordinate of the rock. Your pictorial representation should be complete now, and similar to the picture below:
Solve Part C Find the height yi from which the rock was launched. Express your answer in meters to three significant figures.
Hint 1. How to approach the problem The time Δt needed to move horizontally to the final position x f = d = 17.0 m is the same time needed for the rock to rise from the initial position yi to the peak of its trajectory and then fall to the ground. Use the information you have about motion in the horizontal direction to solve for Δt. Knowing this time will allow https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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you to use the equations of motion for the vertical direction to solve for yi .
Hint 2. Find the time spent in the air How long (Δt) is the rock in the air? Express your answer in seconds to three significant figures.
Hint 1. Determine which equation to use Which of the equations given in the strategy and shown below is the most appropriate to calculate the time Δt the rock spent in the air? ANSWER: xf
= x i + v ix Δt
y
= y + v iy Δt −
f
i
1 2
g(Δt)
2
v f y = v iy − gΔt
Hint 2. Find the x component of the initial velocity What is the x component of the rock's initial velocity? Express your answer in meters per second to three significant figures. ANSWER: v ix
= 9.53 m/s
ANSWER: Δt
= 1.78 s
All attempts used; correct answer displayed If you use the answer to this hint in a subsequent part or hint, please use your unrounded answer and only round as a final step before submitting your answer. Now, use the equation for the other component of motion (vertical component) and find the initial height y . To do that, first you will need to calculate the y component of the rock's initial velocity. i
Hint 3. Find the y component of the initial velocity What is the y component of the rock's initial velocity? Express your answer in meters per second to three significant figures. ANSWER:
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= 5.50 m/s
Correct If you use the answer to this hint in a subsequent part or hint, please use your unrounded answer and only round as a final step before submitting your answer.
ANSWER: yi
= 5.81 m
All attempts used; correct answer displayed
Assess Part D A second rock is thrown straight upward with a speed 5.500 m/s . If this rock takes 1.785 s to fall to the ground, from what height H was it released? Express your answer in meters to three significant figures.
Hint 1. Identify the known variables What are the values of yf , v iy , Δt, and a for the second rock? Take the positive y axis to be upward and the origin to be located on the ground where the rock lands. Express your answers to four significant figures in the units shown to the right, separated by commas. ANSWER: y
f
, v iy , Δt, a = 0,5.500,1.785,9.810 m, m/s , s , m/s 2
Answer Requested Hint 2. Determine which equation to use to find the height Which equation should you use to find H ? Keep in mind that if the positive y axis is upward and the origin is located on the ground, yi = H . ANSWER:
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y f = y i + v iy Δt −
1 2
g(Δt)
2
v f y = v iy − gΔt v
2 fy
= v
2 iy
− 2g(y f
− yi )
Correct
ANSWER: H
= 5.81 m
Answer Requested Projectile motion is made up of two independent motions: uniform motion at constant velocity in the horizontal direction and freefall motion in the vertical direction. Because both rocks were thrown with the same initial vertical velocity, v iy = 5.500 m/s , and fell the same vertical distance of 5.81 m , they were in the air for the same amount of time. This result was expected and helps to confirm that you did the calculation in Part C correctly.
± Delivering a Package by Air A relief airplane is delivering a food package to a group of people stranded on a very small island. The island is too small for the plane to land on, and the only way to deliver the package is by dropping it. The airplane flies horizontally with constant speed of 452 km/hour at an altitude of 775 m . The positive x and y directions are defined in the figure. For all parts, assume that the "island" refers to the point at a distance D from the point at which the package is released, as shown in the figure. Ignore the height of this point above sea level. Assume that the acceleration due to gravity is g = 9.80 m/s 2 .
Part A After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? Assume that the package, like the plane, has an initial velocity of 452 km/hour in the horizontal direction. Express your answer numerically in seconds. Neglect air resistance.
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Hint 1. Knowns and unknowns: what are the initial conditions? Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. In this coordinate system, what are the values of x 0 , y0 , v 0x , v 0y of the package? Express your answers numerically and enter them, separated by commas, in the order x 0 , y0 , v 0x , v 0y . Use units of meters and km/hour for distances and speeds, respectively.
Hint 1. Initial velocity in the y direction Because the package is dropped horizontally, the vertical component of its initial velocity is zero. ANSWER: x0
, y0 , v 0x , v 0y = 0,775,452,0 m, m, km/hour, km/hour
Hint 2. What are the knowns and unknowns when the package hits the ground? Take the origin of the coordinate system to be at the point on the surface of the water directly below the point at which the package is released. The directions of the axes are shown in the figure in the problem introduction. Let t g be the time when the package hits the ground. In this coordinate system, v x (t g ) = 452 km/hour since there is no acceleration in the x direction. Which of the following values is/are known? Check all that apply. ANSWER: x(t g ) y(t g ) v y (t g )
Hint 3. Find the best equation to use Which of the equations below could you use to find the time t g when the packet hits the ground?
Hint 1. How to determine which equation to use Only one of the quantities x(t g ) , y(t g ) , v y (t g ) is known. Which one? You have to use the equation that contains this variable. ANSWER:
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x = x 0 + v 0x t v y = v 0y − gt
y = y 0 + v 0y t −
1 2
gt
2
ANSWER: t
= 12.6 s
Correct
Part B If the package is to land right on the island, at what horizontal distance D from the plane to the island should the package be released? Express the distance numerically in meters.
Hint 1. How to approach the problem You are asked to find D, which is also the change in the x coordinate of the package over the time spent in the air. You should have calculated this time interval in Part A. Use it to find D.
Hint 2. The equation for x(t) Since there is no acceleration in the horizontal direction, the equation for x(t) is x(t) = x 0 + v 0x t
.
ANSWER: D
= 1580 m
Correct
Part C What is the speed v f of the package when it hits the ground? Express your answer numerically in meters per second.
Hint 1. How to approach the problem −−−−−−
The speed is the magnitude of the velocity, i.e., √ v 2x
2
+ vy
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
. You already know that v x
452 km/hour ,
=
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so you need to find v y . Remember to be consistent in your usage of units when calculating your answer.
Hint 2. The equation for the velocity in the y direction The equation for the velocity in the y direction v y (t) is . If you have completed the earlier parts, you should know all the quantities on the right side. v y (t) = v 0y + a y t
ANSWER: vf
= 176 m/s
Answer Requested
Part D The speed at which the package hits the ground is really fast! If a package hits the ground at such a speed, it can be crushed and also cause some serious damage on the ground. Which of the following would help decrease the speed with which the package hits the ground? ANSWER: Increase the plane's speed and height Decrease the plane's speed and height
Correct This is why it would be nice for rescue teams to have hybrid airplanehelicopters. Of course, then they can just airlift the stranded group.
Video Tutor: Range of a Gun at Two Firing Angles First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the questions at right. You can watch the video again at any point.
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Part A Which projectile spends more time in the air, the one fired from 30∘ or the one fired from 60∘ ?
Hint 1. How to approach this problem Which component of the initial velocity vector affects the time the projectile spends in the air? ANSWER: The one fired from 60∘ They both spend the same amount of time in the air. The one fired from 30∘
Correct The projectile fired from 60∘ has a greater vertical velocity than the one fired from 30∘ , so it spends more time in the air.
± Crossing a River A swimmer wants to cross a river, from point A to point B, as shown in the figure. The distance d 1 (from A to C) is 200 m, the distance d 2 (from C to B) is 150 m, and the speed v r of the current in the river is 5 km/hour . Suppose that the swimmer's velocity relative to the water makes an angle of θ = 45 degrees with the line from A to C, as indicated in the figure.
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Part A To swim directly from A to B, what speed u s , relative to the water, should the swimmer have? Express the swimmer's speed numerically, to three significant figures, in kilometers per hour.
Hint 1. Use the motion in the y direction Suppose it takes the swimmer time t to cross the river and arrive at B. Find an expression for t by considering only v y (the y component of the swimmer's motion with respect to the shore). Answer in terms of d 1 , u s , and θ .
Hint 1. The y component of the velocity relative to the shore The y component (see the figure in the problem introduction) of the swimmer's velocity relative to the shore is .
v y = u s cos(θ)
ANSWER: d1 us
d1
u s cos(θ) d1 u s sin(θ)
d1 vr
Correct https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Hint 2. Use the motion in the x direction Now find an expression for t by considering only the x component of the swimmer's motion. Express your answer in terms of d 2 , u s , v r , and θ .
Hint 1. The x component of the velocity relative to the shore The x component (see the figure in the problem introduction) of the swimmer's velocity relative to the shore is .
v x = v r − u s sin(θ)
ANSWER: d2 v r −u s sin(θ) d2 v r +u s sin(θ) d2 v r −u s
d2 v r +u s
Correct Note that this is the same time that you found when you considered only the motion in the y direction, even though the expressions are different. If you set both expressions equal to each other, you can solve for u s .
Hint 3. Solve for us Find a symbolic expression for u s . If you used the previous hints, solve for u s by eliminating t from the two equations you derived. Express your answer in terms of v r , d 1 , d 2 , and θ . ANSWER:
us
=
d 1 vr d 1 sin(θ)+d 2 cos(θ)
Answer Requested
ANSWER: us
= 4.04 km/hour
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Correct Another way to do this problem is to add the swimmer's and the river's velocities vectorially, and set the angle that the resultant vector makes with AC or the river bank equal to that which AB makes with the same.
Problem 4.27 The radius of the earth's very nearly circular orbit around the sun is 1.5 × 10 11 m .
Part A Find the magnitude of the earth's velocity. Assume a year of 365 days. Express your answer to two significant figures and include the appropriate units. ANSWER: v
= 3.0×104
m s
All attempts used; correct answer displayed
Part B Find the magnitude of the earth's angular velocity. Express your answer to two significant figures and include the appropriate units. ANSWER: ω
= 2.0×10−7
rad s
Answer Requested
Part C Find the magnitude of the earth's centripetal acceleration as it travels around the sun. Express your answer to two significant figures and include the appropriate units. ANSWER: ar
m = 6.0×10−3 s2
Answer Requested https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Problem 4.9 A rocketpowered hockey puck moves on a horizontal frictionless table. The figure below shows the graphs of v x and v y , the x and ycomponents of the puck’s velocity. The puck starts at the origin.
Part A In which direction is the puck moving at t = 3 s ? Give your answer as an angle from the xaxis. Express your answer using two significant figures. ANSWER: θ
= 51 ∘ above the xaxis
Answer Requested
Part B How far from the origin is the puck at 5 s ? Express your answer to two significant figures and include the appropriate units. ANSWER: s
= 180 cm
Answer Requested
Problem 4.11 A physics student on Planet Exidor throws a ball, and it follows the parabolic trajectory shown in the figure . The ball's position is shown at 1 s intervals until t =
3 s
. At t =
1 s
, the ball's velocity is v ⃗ = ( 2.0^i
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^ + 2.0 j ) m/s
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Part A Determine the ball's velocity at t =
.
0 s
ANSWER: ^ 2.0 i m/s ^ ^ (2.0 i + 4.0 j ) m/s ^ ^ (2.0 i + 2.0 j ) m/s ^ ^ (2.0 i − 2.0 j ) m/s
.
Correct
Part B Determine the ball's velocity at t =
2 s
.
ANSWER: ^ 2.0 i m/s ^ ^ (2.0 i + 4.0 j ) m/s ^ ^ (2.0 i + 2.0 j ) m/s ^ ^ (2.0 i − 2.0 j ) m/s
.
Correct
Part C t = 3 s https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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Determine the ball's velocity at t =
3 s
.
ANSWER: ^ 2.00 i m/s ^ ^ (2.00 i + 4.00 j ) m/s ^ ^ (2.00 i + 2.00 j ) m/s ^ ^ (2.00 i − 2.00 j ) m/s
.
Correct
Part D What is the value of g on Planet Exidor? Express your answer with the appropriate units. ANSWER: 2.00
m s2
Correct
Part E What was the ball's launch angle? Express your answer with the appropriate units. ANSWER: 63.4 ∘
Correct
Problem 4.13 A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point.
Part A What was the bullet's flight time? Express your answer with the appropriate units. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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ANSWER: 7.69×10−2 s
Correct
Part B What was the bullet's speed as it left the barrel? Express your answer with the appropriate units. ANSWER: 650
m s
Correct
Problem 4.15 A boat takes 2.90 hours to travel 30.0 km down a river, then 6.30 hours to return.
Part A How fast is the river flowing? Express your answer with the appropriate units. ANSWER: 2.79
km h
Correct
Problem 4.20 The figure shows the angularvelocityversustime graph for a particle moving in a circle.
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Part A How many revolutions does the object make during the first 3.4 s ? Express your answer using two significant figures. ANSWER: n
= 8.6
Correct
Problem 4.22 An oldfashioned singleplay vinyl record rotates on a turntable at 45.0 rpm. What are (a) the angular velocity in rad/s and (b) the period of the motion?
Part A Express your answer with the appropriate units. ANSWER: 4.71
rad s
Correct
Part B Express your answer with the appropriate units. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=4057470
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ANSWER: 1.33 s
Correct
Problem 4.32 A 4.8mdiameter merrygoround is initially turning with a 3.8 s period. It slows down and stops in 23 s .
Part A Before slowing, what is the speed of a child on the rim? Express your answer to two significant figures and include the appropriate units. ANSWER: v
m = 4.0 s
Correct
Part B How many revolutions does the merrygoround make as it stops? Express your answer using two significant figures. ANSWER: Δθ
= 3.0 rev
Answer Requested
Problem 4.48 You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.
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Part A What minimum speed will allow the ball to clear the roof? ANSWER: 13 m/s
Answer Requested
Part B At what angle should you toss the ball? ANSWER: 48 ∘ to the horizontal.
All attempts used; correct answer displayed
Problem 4.59 Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest to its top speed of 1 rotation every 1.50 s . The astronaut is strapped into a seat 4.20 m from the axis.
Part A What is the astronaut's tangential acceleration during the first 40.0 s ? Express your answer with the appropriate units. ANSWER: 0.440
m s2
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Correct
Part B How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 2 m/s of acceleration is 1 g. Express your answer with the appropriate units. ANSWER: 7.52 g
Correct Score Summary: Your score on this assignment is 69.1%. You received 67.46 out of a possible total of 108 points, plus 7.17 points of extra credit.
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