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Assignment 17: Chap 6: CQ 6.4 and Prob 6.30, 40, 48 Conceptual Question CQ 6.4 If two objects collide and one is initially at rest, is it possible for both objects to be at rest after the collision? Explain. answer: No. Only in a precise head-on collision with equal and opposite momentum can both balls wind up at rest. Yes. In the second case, assuming equal masses for each ball, if Ball 2, originally at rest, is struck squarely by Ball 1, then Ball 2 takes off with the velocity of Ball 1. Then Ball 1 is at rest. Problems: Problem 6.30 An 8 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of height 1.0 m The bullet remains lodged in the block after impact and the block lands 2.0 m from the bottom of the table. Determine the initial speed of the bullet. Solution: First, we will find the horizontal speed, v0x , of the block and embedded bullet just after impact. After this instant, the block-bullet combination is a projectile, and we find the 1 time to reach the floor by use of y v0yt ay t , which becomes 2 2 1. m 0 00 1 2 9. m s2 t , giving 80 2 t = 0.452 s x 2. m 00 Thus, v0x 4. m s 43 452 t 0. s Now use conservation of momentum for the collision, with vb = speed of incoming bullet: 8. 10 00 3 kg vb 0 258 10-3 kg 4. m s , so 43 vb 143 m s (about 320 mph) Problem 6.40 A billiard ball rolling across the table at 1.5 m/s makes a head on collision with an identical ball. Find the speed of each ball after the collision a) when the second ball is initially at rest, b) when the second ball is moving toward the first ball at a speed of 1.0 m/s and c) when the second ball is moving away from the first ball at a speed of 1.0 m/s. Solution: First, consider conservation of momentum and write m 1v1i m 2v2i m 1v1 f m 2v2 f Since m 1 m 2 , this becomes v1i v2i v1 f v2 f . (1) For an elastic head-on collision, we also have v1i v2i v1 f v2 f , which may be written as v1i v2i v1 f v2 f (2) Subtracting Equation (2) from (1) gives v2 f v1i (3) Adding Equations (1) and (2) yields v1 f v2i (4) Equations (3) and (4) show us that, under the conditions of equal mass objects striking one another in a head-on elastic collision, the two objects simply exchange velocities. Thus, we may write the results of the various collisions as (a) v1f 0 , v2 f 1. m s 50 (b) v1 f 1. m s , v2 f 1. m s 00 50 (c) v1 f 1. m s , v2 f 1. m s 00 50 Problem 6.48 Consider the frictionless track shown in figure p6.48. A block of mass m1 = 5 kg is released from A. It makes a head on collision at B with a block of mass m2 = 10 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision. Solution: First, we use conservation of mechanical energy to find the speed of m 1 at B just before 1 collision. This gives m 1 v1 0 0 m 1ghi, 2 2 or v1 2 2 ghi 2 9. m s2 5. m 9. m s 80 00 90 Next, we apply conservation of momentum and knowledge of elastic collisions to find the velocity of m 1 at B just after collision. From conservation of momentum, with the second object initially at rest, we have m 1v1 f m 2v2 f m 1v1i 0 , or v2 f m1 m2 v1i v1 f (1) For head-on elastic collisions, v1 f v1i v2 f v2i . Since v2i 0 in this case, this becomes v2 f v1 f v1i and combining this with (1) above we obtain m m2 5. 10. 00 0 v1 f 1 v1i 5. 10. 9. m s 3. m s m1 m2 00 0 90 30 Finally, use conservation of mechanical energy for m 1 after the collision to find 1 the maximum rebound height. This gives 0 m 1ghm ax m 1v1 f 0 2 2 or hm ax 2 v1 f 3. m s 30 2 0. m 556 2g 2 9. m s2 80 Assignment 18: Chap 7: Prob 2, 4, 6, 10 Problems: Problem 7.2: A wheel has a radius of 4.1 m. How fart does a point on the circumference travel if the wheel is rotated through angles of 30 degree, 30 rad, and 30 rev respectively? Solution: The distance traveled is s r , where is in radians. r ad For 30°, s r 4. m 30 1 2. m 1 180 For30 radians, s r 4. m 30 r 1. 102 m 1 ad 2 2 r ad For30 revolutions, s r 4. m 30 r 1 ev 7. 102 m 7 1r ev Problem 7.4: A potter's wheel moves from rest to an angular speed of 0.2 rev/s in 30 s. Find its angular acceleration in radians per second per section. Solution: f i We use and find t 0. r s 0 2 r 20 ev ad 4. 10-2 r s2 2 ad 30 s 1r ev Problem 7.6 A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates through 50 revs before coming to rest. Find the constant angular acceleration of the centrifuge. Solution: r 2 r 1 m i ev ad n i 3600 377 r s ad m i 1 r 60. s n ev 0 2 r ad 50. r 0 ev 314 rad 1r ev w i 0 377 r s 2 2 ad Thus, 226 r s2 ad 2 2 314 r ad Problem 7.10 The tub of a washer goes into it spin dry cycle starting from rest and reaching an angular speed of 5 rev/s in 8 s. At this point, a person doing the laundry opens the lid, and the washer turns off. The tub slows to rest in 12 s. Through how many revolutions does the tub turn during the entire 20 s interval? Assume constant acceleration. Solution: We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period. The angular displacement during the acceleration period is f i 5. r s 2 r 1 r 0 0 ev ad ev 1 avt t 8. s 126 r 0 ad 2 2 and while decelerating, f i 0 5. r s 2 r 1 r 0 ev ad ev 2 t 12 s 188 rad 2 2 1r ev The total displacement is =1 2 126 188 r ad 50 rev 2 r ad Assignment 19: Chap 7: Conceptual Question 13 Problems 19, 23, 43 CQ 7.13 A pail of water can be whirled in a vertical circular path such that no water is spilled. Why does the water remain in the pail even when the pail is upside down above your head? Answer: The tendency of water is to move in a straight line path tangent to the circular path followed by the container. As a result, at the top of the circular path, the water presses against the bottom of the pail, and the normal force exerted by the pail on the water provide the radial force required to keep the water moving in its circular path. Problem 7:19 A 55 kg skater is moving at 4 m/s when she grads the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.8 m around the pole. a) determine the force exerted by the horizontal rope on her arms. b) Compare this force with her weight. Solution: (a) From Fr m ac , we have v2 55. kg 4. m s 2 0 00 T m t 1. 103 N 1. kN 10 10 r 800 0. m (b) The tension is l ger than her weight by a factor of ar T 1. 103 N 10 2. tm es 04 i 0 m g 55. kg 9. m s2 80 Problem 7:23 A 50 kg child stands at the rim of a merry go round of radius 2 m rotating with an angular speed of 3 rad/s a) What is the child's centripetal acceleration? B) What is the minimum force between her feet and the floor of the carousel that is required to keep her in the circular path? c) What minimum coefficient of static friction is required? Is the answer you found reasonable? In other words, is she likely to stay on the merry go round? Solution: ac r 2 2. m 3. r s 18. m s2 2 (a) 00 00 ad 0 c (b) F m ac 50. kg 18. m s2 900 N 0 0 (c) We know the centripetal acceleration is produced by the force of friction. Therefore, the needed static friction force is f 900 N . Also, the normal s force is n m g 490 N . Thus, the minimum coefficient of friction required f s m ax 900 N is s 84 = 1. n 490 N So large a coefficient of friction is unreasonable, and she will not be able to stay on the merry-go-round. Problem 7:43 An athlete swings a 5 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.8 m at an angular speed of 0.5 rev/s. What are the a) tangential speed of the ball and b) its centripetal acceleration? c) If the maximum tension the rope can withstand before breaking is 100 N, what is the maximum tangential speed the ball can have? Solution: 500 r s r s The angular velocity of the ball is 0. ev ad (a) vt r 0. m r s 2. m s 800 ad 51 2 vt r 2 0. m r s 7. m s2 2 (b) ac 800 ad 90 r (c) During parts a) and b) the ball hangs down at some angle as it swings around. T FBD AD L θ m 0.8 m ac W From summing the forces horizontally and vertically F mac F m(0) T cos mac T sin W 0 T cos mac T sin mg T cos 5kg *7.9m / s 2 39.5 N T sin mg 5kg *9.81m / s 2 49 N Solve for the angle: T sin 49 N tan 1.24 arc tan(1.24) 51.1o T cos 39.5 N From this the length of the rope can be found. 0.8m 0.8m cos L 1.27m L cos 51.1o When swinging the rope faster, this angle will change and the radius at which the ball swings will be farther out…Assume the length of the rope remains the same. Then, 100N FBD aD 1.27m θ=? m r ac W From summing the forces vertically and horizontally F m(0) F mac T sin W 0 T cos mac mg 49 v2 sin 0.49 T cos m T 100 r T cos arcsin(0.49) 29.3o v2 r m so r cos 29.3o 1.27 (100 N ) cos(29.3o ) r 1.11m v2 (1.11m) 12m 2 / s 2 5kg v 12m2 / s2 3.47m / s Assignment 20: Chap 7 Conceptual Questions 4 and 6 Problems 29, 36 CQ 7:4 Explain why the CQ 7:6 Problem 7:29 Solution: Problem 7:36 Solution:

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posted: | 4/30/2011 |

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