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03 homework VENNES, ROSS Due: Sep 21 2007, 1:00 am Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 7.6 m/s. Also, it has an acceleration in the direction parallel to the walls of 2.6 m/s2 . It hits the opposite wall at the same height. a 1 7.6 m/s 6.29474 m/s 7.61994 m 6 9. 8 83 2m /s 2.6 m/s2 d The horizontal motion will carry the particle to the opposite wall, so d = vx tf and 7.6 m/s 18.4 m tf = d vx (18.4 m) = (7.6 m/s) = 2.42105 s . a) What will be its speed when it hits the opposing wall? Correct answer: 9.86832 m/s (tolerance 1 %). Explanation: is the time for the particle to reach the opposite wall. Horizontally, the particle reaches the maximum parallel distance when it hits the oppod , so the final site wall at the time of t = vx parallel velocity vy is vy = a t ad = vx (2.6 m/s2 ) (18.4 m) = (7.6 m/s) = 6.29474 m/s . The velocities act at right angles to each other, so the resultant velocity is vf = = 2 2 vx + vy Let : d = 18.4 m , vx = 7.6 m/s , a = 2.6 m/s2 , Basic Concepts Kinematics equations v = vo + g t (7.6 m/s)2 + (6.29474 m/s)2 = 9.86832 m/s . 1 2 gt 2 Question 2, chap 4, sect 3. part 2 of 2 10 points s = so + vo t + 50. 366 6 homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am b) At what angle with the wall will the particle strike? Correct answer: 50.3666 (tolerance 1 %). Given: y1 = -15.0 yards x = 13.0 yards y2 = 40.0 yards Solution: ytot = -15 yards + 40 yards = 25 yards vx vy 7.6 m/s 6.29474 m/s Thus d = (13 yards)2 + (25 yards)2 = 28.178 yards xtot = 13 yards 2 Explanation: When the particle strikes the wall, the vertical component is the side adjacent and the horizontal component is the side opposite the angle, so tan = vx , vy so = arctan = arctan = 50.3666 . Note: The distance traveled parallel to the walls is y= 1 2 at 2 1 = (2.6 m/s2 ) (2.42105 s)2 2 = 7.61994 m . Question 4, chap 4, sect 4. part 1 of 1 10 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q P R Question 3, chap 4, sect 3. part 1 of 1 10 points A quarterback takes the ball from the line of scrimmage, runs backward for 15.0 yards, then runs sideways parallel to the line of scrimmage for 13.0 yards. At this point, he throws a 40.0 yard forward pass straight down the field. What is the magnitude of the football's resultant displacement? Correct answer: 28.178 yards (tolerance 1 %). Explanation: Basic Concepts: xtot = x ytot = y1 + y2 The displacements are perpendicular, so dtot = (xtot )2 + (ytot )2 How do the speeds of the ball at the three points compare? 1. vP = vR = vQ 2. vP < vQ < vR 3. vQ < vR < vP 4. vP = vR < vQ homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am 5 2 5. v = m/s. correct 5. vR < vQ < vP 2 5 6. vQ < vP = vR correct 6. v = m/s. 5 7. v = 5 2 m/s. Explanation: The speed of the ball in the x-direction is 3 8. v = m/s. constant. Because of gravitational accelera5 tion, the speed in the y-direction is zero at 2 point Q. Since points P and R are located at m/s. 9. v = 5 the same point above ground, by symmetry we see that they have the same vertical speed 10. v = 5 3 m/s. component (though they do not have the same Explanation: velocity). The answer is then "vQ < vP = vR ". Question 5, chap 4, sect 4. part 1 of 1 10 points A target lies flat on the ground 5 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m/s2 . Air resistance is negligible. A student rolls a 5 kg ball off the horizontal roof of the building in the direction of the target. v m = 5 kg , not required h = 10 m , x = 5 m , and g = 10 m/s2 . 3 Observe the motion in the vertical direction only and it is a purely 1-dimension movement with a constant acceleration. So the time need for the ball to hit the ground is t= 2h g 10 m and the horizontal speed should be x v= t for the ball to hit the target. Therefore v=x g 2h 10 m/s2 2 (10 m) 5m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1. v = 5 5 m/s. 5 3 m/s. 2. v = 3 3. v = 3 5 m/s. 4. v = 2 5 m/s. 10 m = (5 m) 5 = m/s 2 5 2 m/s . = 2 Question 6, chap 4, sect 4. part 1 of 1 10 points For the ballistic missile aimed to achieve the maximum range of 17500 km, what is the maximum altitude reached in the trajectory? homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am Correct answer: 4375 km (tolerance 1 %). 4 Explanation: For the maximum range R, the launch angle is 45 and the range R is 2 v0 sin(2 ) g 2 sin[2 (45 )] v = 0 g 2 v = 0. g The height H is given by How fast was it rolling on the table before it fell off? Correct answer: 1.9611 m/s (tolerance 1 %). Explanation: Basic Concept: Horizontally, x = vi,x t + since ax = 0. Vertically, y = vi,y t + since vi,y = 0. Given: 1 1 g (t)2 = g (t)2 , 2 2 1 ax (t)2 = vx t , 2 R= v 2 sin2 () H= 0 2g 2 v = 0 . 4g So dividing H by R gives 1 H= R 4 1 = (17500 km) 4 = 4375 km . Question 7, chap 4, sect 4. part 1 of 2 10 points A 81 g autographed baseball slides off of a 1.2 m high table and strikes the floor a horizontal distance of 0.97 m away from the table. The acceleration of gravity is 9.81 m/s2 . See the figure below. x = 0.97 m , y = -1.2 m , and g = -9.81 m/s2 . vx y 1.2 m x vy 0.97 m Note: Drawn to scale Solution: From the horizontal motion, we have t = x . vi (1) homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am From the vertical motion, we have t = 2 y g x = vx = 0.494619 s . 5 A baseball is thrown at an angle of 26 relative to the ground at a speed of 29.7 m/s. The ball is caught 70.9 m from the thrower. The acceleration of gravity is 9.81 m/s2 . a) How long is it in the air? Correct answer: 2.65436 s (tolerance 1 %). Explanation: Basic Concept: Horizontally, x = vx t = vi (cos )t Given: Using Eqs. 1 and 2 and solving for vx , we have vx = = g x 2y -9.81 m/s2 (0.97 m) 2(-1.2 m) = 1.9611 m/s . Question 8, chap 4, sect 4. part 2 of 2 10 points What was the direction of the ball's velocity just before it hit the floor? That is, at what angle (in the range -90 to +90 relative to the horizontal directed away from the table) did the ball hit the floor? Correct answer: -67.9931 (tolerance 1 %). Explanation: The vertical velocity component with which the mug hits the floor is vy = vy0 + ay t = 0-gt = -(9.81 m/s2 ) (0.494619 s) = -4.85222 m/s . Hence, the angle at which the mug strikes the floor is given by = tan-1 vy vx -4.85222 m/s = tan-1 1.9611 m/s = -67.9931 . R= Solution: = 26 vi = 29.7 m/s x = 70.9 m g = 9.81 m/s2 t = x vi (cos ) 70.8559 m = (29.7 m/s) (cos 26 ) = 2.65436 s 2 vi sin[2 ] g (29.7 m/s)2 sin[2 (26 )] = 9.81 m/s2 = 70.8559 m . Question 10, chap 4, sect 4. part 2 of 2 10 points b) How high is the tallest spot in the ball's path? Correct answer: 8.63968 m (tolerance 1 %). Explanation: Basic Concept: Question 9, chap 4, sect 4. part 1 of 2 10 points homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am Vertically, 2 2 2 vy,f = vy,i - 2gy = vi (sin )2 - 2gy 6 x = vx0 t y = vy0 t - 1 2 gt . 2 Solution: At maximum height, vy,f = 0 m/s, so 2 0 = vi (sin )2 - 2gymax v 2 (sin )2 ymax = i 2g (29.7 m/s)2 (sin 26 )2 = 2(9.81 m/s2 ) = 8.63968 m Let : = 47 , v0 = 10.4 m/s , x = 17.5 m , and m = 0.28 kg . Solution: The flying time can be determined by: x = v0x t , or x t= v0x x = . v0 cos From the point where the rock was projected (set to be the origin O), the y-coordinate of the point where the rock struck the ground is 1 2 gt 2 1 x x - g = v0 sin v0 cos 2 v0 cos 2 gx , so = x tan - 2 (v0 cos )2 h = |y| g x2 = - x tan 2 (v0 cos )2 (9.8 m/s2 ) (17.5 m)2 = 2 [(10.4 m/s) cos 47 ]2 - (17.5 m) tan 47 y = v0y t - = 11.0626 m . Question 11, chap 4, sect 4. part 1 of 1 10 points A 0.28 kg rock is projected from the edge of the top of a building with an initial velocity of 10.4 m/s at an angle 47 above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 17.5 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s2 . .4 m /s 2 10 47 Building h 17.5 m How tall, h, is the building? Correct answer: 11.0626 m (tolerance 1 %). Explanation: Basic Concepts: vx = vx0 = constant vy = vy0 - g t Question 12, chap 4, sect 4. part 1 of 3 10 points A projectile of mass 0.316 kg is shot from a cannon. The end of the cannon's barrel is at height 7 m, as shown in the figure. The initial velocity of the projectile is 12 m/s . The projectile rises to a maximum height of y above the end of the cannon's barrel and homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am strikes the ground a horizontal distance x past the end of the cannon's barrel. The acceleration of gravity is 9.8 m/s2 . m/ s 7 y = 12 y 59 x Determine the time it takes for the projectile to reach its maximum height. Correct answer: 1.04959 s (tolerance 1 %). Explanation: Basic Concepts: 1 2 gt 2 1 1 = (9.8 m/s2 ) (1.04959 s)2 2 = 5.39806 m , or 2 vy y = i 2g 2 vy sin2 = 2g (12 m/s)2 sin2 (59 ) = 2 (9.8 m/s2 ) = 5.39806 m . 7m Question 13, chap 4, sect 4. part 2 of 3 10 points (1) (2) (3) How long does it take the projectile to hit the ground? Correct answer: 2.64026 s (tolerance 1 %). Explanation: From Eq. 3, the time t2 for the projectile to decend from the top to where the projectile hits the ground is 2 [y + yi ] g v = vi sin - g t , 2 vy y = i , and 2g 1 y = vyi - g t . 2 Let : vi xi yi = 12 m/s , = 0 m, = 7 m , and = 59 . t2 = = Solution: For the horizontal and vertical components of vector of vi , we have vxi = vi cos = 6.18046 m/s , vyi = vi sin = 10.286 m/s . and 2 [(5.39806 m) + (7 m)] (9.8 m/s2 ) = 1.59067 s . Or symbolically, we have 1/2 2 vyi 2 yi + g2 g (10.286 m/s)2 = (9.8 m/s2 )2 Using Eq. 1 (|vyi | = g t), the time t1 to reach the maximum height, (i.e., vtop = 0 , the velocity the at top) is vy t1 = i g (10.286 m/s) = (9.8 m/s2 ) = 1.04959 s . t2 = 2 (7 m) + (9.8 m/s2 ) = 1.59067 s . 1/2 homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am The total time of the projectile's trajectory is t = t1 + t2 = (1.04959 s) + (1.59067 s) = 2.64026 s . Question 14, chap 4, sect 4. part 3 of 3 10 points What is the horizontal range x of the shot? Correct answer: 16.318 m (tolerance 1 %). 8 Calculate the horizontal velocity of the projectile. Correct answer: 1.16279 m/s (tolerance 3 %). Explanation: 100 90 Vertical Distance (m) 80 70 60 50 40 30 20 10 0 0 ymax xmax x 1 Explanation: x = vxi t = vi cos t = (12 m/s) cos(59 ) (2.64026 s) = 16.318 m . Question 15, chap 4, sect 4. part 1 of 2 10 points Before coming to Earth, your teacher was on planet Krypton where there was almost no atmosphere. Your teacher observed and plotted the position of a projectile at constant time intervals of 0.86 s , as shown in the figure. Projectile Position on Planet Krypton 100 90 Vertical Distance (m) 80 70 60 50 40 30 20 10 0 0 2 3 4 5 6 7 8 9 10 Horizontal Distance (m) Note: From the figure above, we observe the following measurements. The position of the maximum height (xmax , ymax) = (5 m, 75 m) and the minimum height (xlef t , ymin) = (0.5 m, 5 m) . The right-hand end point occurs at xright = 9.5 m . There are (9.5 m - 0.5 m) 1 interval 1m =9 time intervals between the two end points. Basic Concept: x = x0 + vx t . (1) Solution: Using Eq. 1, we have vx = x t xright - xlef t = tright - tlef t (9.5 m) - (0.5 m) = (9) (0.86 s) = 1.16279 m/s . Question 16, chap 4, sect 4. part 2 of 2 10 points 2 3 4 5 6 7 8 9 10 Horizontal Distance (m) Figure: Drawn to scale. 1 homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am Calculate the vertical acceleration due to the gravity of planet Krypton on the projectile. Correct answer: 9.34773 m/s2 (tolerance 3 %). Explanation: Basic Concept: y = y0 + v0y t + 1 2 gt . 2 (2) 9 rest down the incline with a constant acceleration of 6.06 m/s2 and travels 25.5 m to the edge of the cliff. The cliff is 23.7 m above the ocean. The acceleration of gravity is 9.8 m/s2 . Find the car's position relative to the base of the cliff when the car lands in the ocean. Correct answer: 33.0067 m (tolerance 1 %). Explanation: Given : a = 6.06 m/s2 , l = 25.5 m , h = 23.7 m , and = 10.1 . The speed of the car when it reaches the edge of the cliff is given by 2 v 2 = vi + 2 a l = 2 a l Solution: The vertical height covered and time lapse from the first data point to the maximum vertical height is ymax = ymax - ymin = (75 m) - (5 m) = 70 m , and xmax - xlef t tmax = vx (5 m) - (0.5 m) = (1.16279 m/s) = 3.87 s . Using Eq. 2 and solving for g, we have 2 ymax g= (t)2 2 (70 m) = (3.87 s)2 = 9.34773 m/s2 . Note: The equation for the trajectory of the projectile, as plotted in the figure, is y = ymax - ymax x - xmax (xmax - xlef t )2 2 2 since vi = 0 m/s, so v= = 2a 2 (6.06 m/s2 ) (25.5 m) (3) = 17.5801 m/s . Now, consider the projectile phase of the car's motion. The initial vertical velocity is viy = v sin and the vertical velocity with which the car strikes the water is 2 2 vy = viy + 2 g h = (7.5) - (0.345679) x - (5 m) where g is defined in Eq. 3. Question 17, chap 4, sect 4. part 1 of 2 10 points , = v 2 sin2 + 2 g h = (17.5801 m/s)2 sin2 10.1 + 2 9.8 m/s2 (23.7 m) = 474.025 m2 /s2 , so that vy = 474.025 m2 /s2 = 21.7721 m/s. A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 10.1 below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am To find the time of flight, vy = viy + g t vy - viy g The initial horizontal velocity is vix = v cos and the horizontal motion of the car during this time is t = x = vix t vy - viy g = (17.5801 m/s) cos 10.1 21.7721 m/s - 3.08297 m/s 9.8 m/s2 = (v cos ) = 33.0067 m . Question 18, chap 4, sect 4. part 2 of 2 10 points Find the length of time the car is in the air. Correct answer: 1.90706 s (tolerance 1 %). Explanation: Using the vertical motion, 10 Consider the following statements: Ball A has greater launch speed than ball B. Ball B has greater launch speed than ball A. Ball A and ball B have the same launch speed. Ball A spends more time in the air that ball B. Ball B spends more time in the air that ball A. Ball A and ball B spend the same time in the air. Which of the statements are correct? 1. and 2. and 3. and 4. and 5. and 6. and correct 7. and 8. and 9. and Explanation: Basic Concept: The time of flight in all cases depends only on the vertical component of the initial velocity. The launch speed depends on both the vertical component and horizontal component of 2 2 the initial velocity: v = vx + vy . Solution: In this case ball A has a greater maximum height than ball B so its vertical component of the initial velocity is greater. For a fixed launch speed, a trajectory has its maximum range when projected with a launch angle of 45 , so ball B has the least possible initial velocity for the given range R. Therefore, ball A has greater launch speed than ball B and ball A spends more time in vy = viy + g t vy - viy t= g 21.7721 m/s - 3.08297 m/s = 9.8 m/s2 = 1.90706 s . Question 19, chap 4, sect 4. part 1 of 2 10 points Two identical balls (A and B) are tossed through the air as shown in the figure below. A y B 61.8 45 x R homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am the air that ball B. Question 20, chap 4, sect 4. part 2 of 2 10 points Two identical projectiles (A and B) are tossed through the air as shown in the figure below. A B y 9. and 11 61.8 75 x R/2 R Explanation: In this case the balls have the same maximum heights, so the time of flight is the same for both. Since ball A has a shorter range, the horizontal component of its initial velocity must be less than that of ball B. Therefore, ball B has greater launch speed than ball A and ball A and ball B spend the same time in the air. See Paul Hewitt's books for more examples of "Figuring Physics". Question 21, chap 4, sect 4. part 1 of 1 10 points A daredevil decides to jump a canyon of width 15.7 m. To do so, he drives a motorcycle up an incline sloped at an angle of 17.4 . The acceleration of gravity is 9.8 m/s2 . What minimum speed must he have in order to clear the canyon? Correct answer: 16.4193 m/s (tolerance 1 %). Explanation: The time to reach the opposite bank is found from x = v0x t . as t= x v0x (15.7 m) . = v0 cos 17.4 Consider the following statements: Ball A has greater launch speed than ball B. Ball B has greater launch speed than ball A. Ball A and ball B have the same launch speed. Ball A spends more time in the air that ball B. Ball B spends more time in the air that ball A. Ball A and ball B spend the same time in the air. Which statements are correct? 1. and 2. and correct (1) Now use 3. and 4. and 5. and 6. and 7. and 8. and 1 2 at , 2 with y=0 at the opposite bank. Thus, this equation reduces to y = v0y t + 0 = v0y t - 1 2 gt , 2 1 gt 2 or 0 = t v0y - v0y = 1 1 x gt = g 2 2 v0x homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am v0 sin = gx 2 v0 cos gx 2 v0 = 2 sin cos gx 2 sin cos gx sin(2 ) or the sine function sin = vwind v 12 v0 = = = v= vwind 1.5 m/s = = 5.89406 m/s sin sin 14.7436 Question 24, chap 4, sect 6. part 1 of 1 10 points (9.8 m/s2 ) (15.7 m) sin 34.8 = 16.4193 m/s . Question 22, chap 4, sect 6. part 1 of 2 10 points An American bald eagle heads south at 5.7 m/s, but encounters a wind blowing from the east at 1.5 m/s. By how many degrees will she be blown off course? Correct answer: 14.7436 (tolerance 1 %). Explanation: The eagle's velocity is directed due south, with the wind directed toward the west. The two velocities form a right triangle with vwind the side opposite the angle , and veagle the side adjacent, so tan = = arctan vwind 1.5 m/s = veagle 5.7 m/s 1.5 m/s 5.7 m/s = 14.7436 A river flows with a uniform velocity v. A person in a motorboat travels 1.57 km upstream, at which time a log is seen floating by. The person continues to travel upstream for 69.7 min at the same speed and then returns downstream to the starting point, where the same log is seen again. Find the flow velocity of the river. Assume the speed of the boat with respect to the water is constant throughout the entire trip. (Hint: The time of travel of the boat after it meets the log equals the time of travel of the log.) Correct answer: 0.187709 m/s (tolerance 1 %). Explanation: Let u be the boat's speed in still water and let be the distance traveled upstream in t = 69.7 min and T be the time of return. Then, for the log, d = v (t + T ) , and for the boat, = (u - v) t + d = (u + v) T . Combining the above gives +d d = + . v u-v u+v Substituting = (u - v) t from the equation above yields the equation for v (u - v) t + d d =t+ . v u+v Question 23, chap 4, sect 6. part 2 of 2 10 points How fast will she be moving relative to the ground? Correct answer: 5.89406 m/s (tolerance 1 %). Explanation: The resultant velocity is the hypotenuse of the right triangle, so you can use the Pythagorean theorem v= 2 2 vwind + veagle homework 03 VENNES, ROSS Due: Sep 21 2007, 1:00 am Simplifying this equation, we obtain v= d 2t 1570 m = (2) (4182 s ) = 0.187709 m/s . 13 Digression: The answer can be readily understood in the following way. Imagine you are at rest on the log observing the motion of the boat. You need to convince yourself that, since both you and the boat are affected in the same way by the water flow, the motion of the boat relative to you is independent of the water flow. In other words, if the boat is moving away from you with a speed say u, for a time t, it will be moving back toward you with the same speed as you see it. So the time interval for the entire trip, i.e. moving away from you and coming back to you is 2t. During this time, the distance which the log has d drifted is d. So the drift-speed of the log is 2t . ... View Full Document