HW3__Chap3

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 1/20 HW3: Chap3 Due: 11:59pm on Friday, October 6, 2023 You will receive no credit for items you complete after the assignment is due. Grading Policy 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 x-y plane), the definition of acceleration is → a ( t ) = → v ( t + Δ t ) − → v ( t ) Δ t in the limit Δ t → 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 Δ t ) approximate the acceleration as → a ( t ) ≈ → v ( t + Δ t ) − → v ( t ) Δ t . Obviously the difference between → v ( t + Δ t ) and → v ( t ) is another vector that can lie in any direction. If it is longer but in the same direction, → a ( t ) will be parallel to → v ( t ) . On the other hand, if → v ( t + Δ t ) has the same magnitude as → v ( t ) but is in a slightly different direction, then → a ( t ) will be perpendicular to → v . In general, → v ( t + Δ t ) can differ from → v ( t ) in both magnitude and direction, hence → a ( 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, → a a_vec . 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 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 → a ( t ) ≈ ( → v ( t + Δ t ) − → v ( t ))/Δ t given above.) Part A Which direction best approximates the direction of → a a_vec 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:

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 2/20 Correct Part B Which direction best approximates the direction of → a a_vec 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 v_vec is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of → v v_vec is changing there is an acceleration. ANSWER: 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 → a a_vec 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 v_vec is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of → v v_vec is changing there is an acceleration. ANSWER: straight up downward to the left downward to the right straight down straight up upward to the right straight down downward to the left upward to the right to the right straight down downward to the right

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 4/20 Part A Express your answer using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answer using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Quantitative Prelecture Video: Projectile Motion Click Play to watch the video below. Answer the ungraded questions in the video and the graded follow-up questions at right. Part A During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. What was the ball’s speed 2.2 s after she kicked it? ANSWER: v x , v y = 0,7.2 m/s r x , r y = 4.3,5.7 m 26 m/s 21 m/s 13 m/s 20 m/s

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 5/20 Correct The speed is the magnitude of velocity, which is determined using the kinematic equations v x = v 0 cos θ 0 and v y = v 0 sin θ 0 − gt . Part B During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. At t = 2.2 s after she kicked the ball, was it moving upward in its trajectory, downward in its trajectory, or was it at the apex of its trajectory? ANSWER: Correct The sign of the y -component of the velocity indicates how the ball is moving on its trajectory. When the velocity’s y -component is positive, the ball is moving upward and when the y -component is negative the ball is moving downward. When the ball is at the apex of its trajectory, the y -component of the velocity is zero. Part C During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. Max, the goalie, caught the ball 60 m away from Maya and 1.0 m off the ground. Max caught the ball ______ s after Maya kicked it. ANSWER: Correct To find the time between Maya kicking the ball and Max catching the ball, you can use the kinematic equations x = ( v 0 cos θ 0 ) t and y = ( v 0 sin θ 0 ) t − 1 2 gt 2 . Either set x equal to 60 m and solve for t or set y equal to 1 m and solve for t . Part D During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. Max, the goalie, caught the ball 60 m away from Maya and 1.0 m off the ground. What was the velocity of the ball when Max caught it? ANSWER: Downward At the apex Upward 3.0 v x = –20 m/s, v y = 0 m/s v x = 0 m/s, v y = –14 m/s v x = 20 m/s, v y = –14 m/s v x = 20 m/s, v y = 14 m/s v x = 20 m/s, v y = 0 m/s v x = 0 m/s, v y = 14 m/s

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 7/20 ANSWER: Hint 2. Specify the shape of the x ( t )x(t) graph Does the x position change at a constant rate or a changing rate? You can determine this by looking at the change in x coordinate from one dot to the next. ANSWER: ANSWER: positive negative zero constant changing No elements selected Select the elements from the list and add them to the canvas setting the appropriate attributes. Press TAB to get to the main menu. 0 1 2 3 4 5

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 8/20 Correct Part B Construct a possible graph for the y position versus time, y ( t )y(t) . Hint 1. Determine the initial value of y ( t )y(t) Is the initial value of the y position positive, negative, or zero? ANSWER: Hint 2. Specify the shape of the y ( t )y(t) graph Does the y position change at a constant rate or a changing rate? ANSWER: ANSWER: positive negative zero constant changing

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 10/20 Does the x velocity remain constant or does it change? You can determine this by comparing the x components of the arrows. ANSWER: ANSWER: Correct Part D Construct a possible graph for the y velocity versus time, v y ( t )v_y(t) . Hint 1. Determine the initial value of v y ( t )v_y(t) Is the initial value of the y velocity positive, negative, or zero? Look at the y component of the first arrow. It remains constant. It changes. No elements selected Select the elements from the list and add them to the canvas setting the appropriate attributes. Press TAB to get to the main menu. 0 1 2 3 4 5

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 11/20 ANSWER: Hint 2. Specify the shape of the v y ( t )v_y(t) graph Does the y velocity remain constant or does it change? You can determine this by comparing the y components of the arrows. ANSWER: Hint 3. Specify the rate of change of v y ( t )v_y(t) Does the y velocity change at a constant rate or a changing rate? It may be helpful to write down the y components of successive arrows. See if the differences between successive arrows' y components stay the same. ANSWER: ANSWER: positive negative zero It remains constant. It changes. constant changing

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 13/20 independently, they are related by the fact that they occur for exactly the same amount of time, namely the time t t the projectile is in the air. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t 0 = 0 s corresponds to the moment just after the ball is launched from position x 0 = 0 m and y 0 = 0 m . Its launch velocity, also called the initial velocity, is → v 0 v_vec_0 . Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time t 1 t_1 with velocity → v 1 v_1_vec . Its position at this moment is denoted by ( x 1 , y 1 ) or ( x 1 , y max ) since it is at its maximum height. The other point, at time t 2 t_2 with velocity → v 2 v_2_vec , corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is ( x 2 , y 2 ) , also known as ( x max , y 2 ) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y 2 = y 0 = 0 m . Part A How do the speeds v 0 v_0 , v 1 v_1 , and v 2 v_2 (at times t 0 t_0 , t 1 t_1 , and t 2 t_2 ) compare? ANSWER: Correct Here v 0 v_0 equals v 2 v_2 by symmetry and both exceed v 1 v_1 . This is because v 0 v_0 and v 2 v_2 include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time t 0 t_0 . Recall that t 0 t_0 refers to the instant just after the ball has been launched, so it is still at ground level ( x 0 = y 0 = 0 m ). However, it is already moving with initial velocity → v 0 v_0_vec , whose magnitude is v 0 = 30.0 m/s and direction is θ = 60.0 degrees counterclockwise from the positive x direction. v 0 v_0 = v 1 v_1 = v 2 v_2 > 0 v 0 v_0 = v 2 v_2 > v 1 v_1 = 0 v 0 v_0 = v 2 v_2 > v 1 v_1 > 0 v 0 v_0 > v 1 v_1 > v 2 v_2 > 0 v 0 v_0 > v 2 v_2 > v 1 v_1 = 0

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 14/20 Part B What are the values of the intial velocity vector components v 0 , x and v 0 , y (both in m/s ) as well as the acceleration vector components a 0 , x and a 0 , y (both in m/s 2 )? Here the subscript 0 means "at time t 0 t_0 ." Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector → v 0 v_0_vec points up and to the right. Since "up" is the positive y axis direction and "to the right" is the positive x axis direction, v 0 , x and v 0 , y will both be positive. As shown in the figure, v 0 , x , v 0 , y , and v 0 v_0 are three sides of a right triangle, one angle of which is θ theta . Thus v 0 , x and v 0 , y can be found using the definition of the sine and cosine functions given below. Recall that v 0 = 30.0 m/s and θ = 60.0 degrees and note that sin( θ ) = length of opposite side length of hypotenuse = v 0 , y v 0 , cos( θ ) = length of adjacent side length of hypotenuse = v 0 , x v 0 . What are the values of v 0 , x and v 0 , y ? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: Correct Also notice that at time t 2 t_2 , just before the ball lands, its velocity components are v 2 , x = 15 m/s (the same as always) and v 2 , y = − 26.0 m/s (the same size but opposite sign from v 0 , y by symmetry). The acceleration at time t 2 t_2 will have components (0, -9.80 m/s 2 ), exactly the same as at t 0 t_0 , as required by Rule 2. The peak of the trajectory occurs at time t 1 t_1 . This is the point where the ball reaches its maximum height y max y_max . At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. 15.0,26.0 m/s 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 16/20 ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range R R of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x 2 − x 0 , or just x 2 x_2 in this particular situation since x 0 = 0 . Range can be calculated as the product of the flight time t 2 t_2 and the x component of the velocity v x v_x (which is the same at all times, so v x = v 0 , x ). The value of v x v_x can be found from the launch speed v 0 v_0 and the launch angle θ theta using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from v 0 v_0 and θ theta using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position ( x 0 = y 0 = 0 ) at time t 0 = 0 with initial speed v 0 v_0 and launch angle θ theta measured from the horizontal. As was the case above, t 2 t_2 refers to the flight time and R R refers to the range of the projectile. flight time: t 2 = 2 v 0 , y g = 2 v 0 sin ( θ ) g range: R = v x t 2 = v 2 0 sin ( 2 θ ) g In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. t 0 t_0 t 1 − t 0 t 2 t_2 t 2 − t 1 t 2 − t 0 2

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 17/20 Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration g g is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. Problem 3.32 A tiger leaps horizontally from a 7.0 m -high rock with a speed of 3.4 m/s . Part A How far from the base of the rock will she land? Express your answer using two significant figures. ANSWER: Correct Prelecture Concept Question 3.21 Part A A swimmer is determined to cross a river that flows due south with a strong current. Initially, the swimmer is on the west bank desiring to reach a camp directly across the river on the opposite bank. In which direction should the swimmer head? ANSWER: Increase v 0 v_0 above 30 m/s . Reduce v 0 v_0 below 30 m/s . Reduce θ theta from 60 degrees to 45 degrees . Reduce θ theta from 60 degrees to less than 30 degrees . Increase θ theta from 60 degrees up toward 90 degrees . x = 4.1 m

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 19/20 Part A What minimum speed is required for the ball to clear the 0.90- m -high net about 15.0 m from the server if the ball is "launched" from a height of 2.30 m ? Express your answer with the appropriate units. ANSWER: Correct Part B Where (relative to server) will the ball land if it just clears the net? Express your answer with the appropriate units. ANSWER: Correct Part C Will it be "good" in the sense that it lands within 7.0 m of the net? Express your answer with the appropriate units. ANSWER: Correct v min = 28.1 m s Δ x = 19.2 m Yes, it will. No, it will not.

12/6/23, 3:48 PM HW3: Chap3 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=10825379 20/20 Part D How long will it be in the air? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 99.0%. You received 10.89 out of a possible total of 11 points. t = 0.685 s