Review Energy and Momentum a. a. b. b. c a. b. 4. c a. a. b. 6. b.

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1 Review 6 Energy and Momentum 1. A player pushes a 250 g hockey puck over frictionless ice with a constant force, causing it to accelerate at 24 m/s 2 over a distance of 50 cm. a. Find the work done by the hockey player on the puck. b. What is the change in the kinetic energy of the puck? 2. You exert a horizontal force of 4.6 N on a textbook as you slide it 0.6 m across a library table to a friend. Calculate the work you do on the book. 3. An electric motor lifts an elevator at a constant speed of 15 m/s. The engine must exert a force of 9000 N in order to balance the weight of the elevator and the friction in the elevator cable. What power does the motor produce in kw? 4. Leah is helping to build a water habitat in a neighborhood park. The habitat includes an upper pond connected to a lower pond, 3.2 m below, by a trickling stream with several small cascades. At a home-building store, she finds a 45 W pump that has a maximum circulation rate of 1900 L of water per hour. Can the pump develop enough power to raise the water from the lower to the upper pond? (The mass density of water,, is 1.00 kg/l.) 5. A gardener lifts a 25 kg bag of sand to a height of 1.1 m, carries it across the yard a distance of 15 m and sets it down against the wall. a. How much work does the gardener do when he lifts the bag of sand? b. How much total work is done after the gardener sets down the bag of sand? 6. A 0.15 kg baseball is thrown at a speed of 6.5 m/s. The batter hits the ball and it flies into the outfield at a speed of 19.2 m/s. How much work is done on the baseball? 7. A 6 g block initially at rest is pulled to the right along a frictionless horizontal surface by a constant horizontal force of N for a distance of 3 cm. a. What is the work done by the force? b. What is the change in the kinetic energy of the block? c. What is the speed of the block after the force is removed? 8. Zeke slides down a snow hill on a rubber mat. Zeke s mass is 76 kg and the mass of the mat is 2 kg. Zeke starts from rest at the crest of the hill. You may ignore friction. a. What is the change in the gravitational potential energy of Zeke and the mat when they slide to 1.2 m below the crest? b. What is the change in the kinetic energy of Zeke and the mat when they slide to 1.2 m below the crest? c. How fast are Zeke and the mat moving when they are 1.2 m below the crest? 9. Kuan stands on the edge of a building s roof, 12 m above the ground, and throws a 150 g baseball straight down. The ball hits the ground at a speed of 18 m/s. What was the initial speed of the ball? 10. Meena releases her 10.5 kg toboggan from rest on a hill. The toboggan glides down the frictionless slope of the hill, and at the bottom of the slope it moves along a rough horizontal surface, which exerts a constant frictional force on the toboggan. a. When the toboggan is released from a height of 15 m, it travels 6 m along the horizontal surface before coming to rest. How much work does the frictional force do on the toboggan? b. From what height should the toboggan be released so that it stops after traveling 10 m on the horizontal surface?

2 11. A golfer uses a club to hit a 45 g golf ball resting on a tee so that the golf ball leaves the tee at a horizontal speed of 38 m/s. a. What is the impulse on the ball? b. What is the average force that the club exerts on the golf ball if they are in contact for s? c. What average force does the golf ball exert on the club during this time interval? 12. A kg hollow racquetball with an initial speed of 12 m/s collides with a backboard. It rebounds with a speed of 6.0 m/s. a. Calculate the impulse on the ball. b. If the contact time lasts for 0.04 s, find the average force on the ball. 13. A tennis player receives a shot with the 60 g ball traveling horizontally at 50 m/s, and returns the shot with the ball traveling horizontally. The tennis ball and the tennis racket are in contact for s. The average force exerted on the ball by the tennis racket is 5700 N. Find the speed of the tennis ball after it leaves the racket. 14. A single uranium atom has a mass of kg. It decays into the nucleus of a thorium atom by emitting an alpha particle at a speed of m/s. What is the recoil speed of the thorium nucleus if the mass of an alpha particle is kg? 15. A 10 g bullet is fired into a stationary 5 kg block of wood. The bullet lodges inside the block. The speed of the block plus bullet system immediately after the collision is measured as 0.6 m/s. What was the original speed of the bullet? 16. Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha has a mass of 64 kg, and each boot has a mass of 4.5 kg. Aisha throws both boots forward at the same time, at a velocity of 6 m/s relative to her. What is Aisha s resulting velocity? 17. At 9 s after takeoff, a 250 kg rocket attains a vertical velocity of 120 m/s. a. What is the impulse on the rocket? b. What is the average force on the rocket? 18. In a circus act, a 18 kg dog is trained to jump onto a 3 kg skateboard moving with a velocity of 0.14 m/s. At what velocity does the dog jump onto the skateboard if afterward the velocity of the dog and skateboard is 0.1 m/s? 19. While sitting in a stationary wagon, a girl catches a 2.6 kg medicine ball moving at a speed of 2.7 m/s. If the 11 kg wagon is on frictionless wheels, what is the velocity of the girl, wagon, and ball after she has caught the ball? The girl s mass is 55 kg. Ignore friction between the wheels and the road. 20. A loaded freight car of mass kg, moving at 18 km/h along a straight, level track, collides with a stationary empty freight car of mass kg. At the collision, the two boxcars lock together. a. What is the velocity of the moving pair of boxcars after the collision? b. How much energy is lost during the collision?

3 1a. F = ma F = (0.25 kg)(24 m/s 2 ) F = 6 N W = Fd W = (6 N)(0.5 m) W = 3 J 1b. KE = W KE = 3 J 2. W = Fd W = (4.6 N)(0.6 m) W = 2.8 J 3. P = W t P = Fd t (9000 N)(15m) P = 1 s P = W 4. P = W t P = mgh t P = (1900 kg)(9.8 m/s2 )(3.2 m) 3600 s P = 17 W Since the pump only needs to produce 17 W of power, the 45 W pump will work fine. 5a. W = PE G W = mgh W = (25 kg)(9.8 m/s 2 )(1.1 m) W = 270 J 5b. W = KE + PE G W = 0 6. W = KE W = KE f KE i W = 1 2 mv f mv i 2 W = 1 (0.15 kg)( m/s)2 1 (0.15 kg)(6.5 m/s)2 2 W = 24 J 7a. W = Fd W = (0.012 N)(0.03 m) W = J

4 7b. KE = W KE = J 7c. KE = KE f KE i KE = 1 2 mv f KE m 2( J) kg 0.35 m/s 8a. PE G = PE f PE i PE G = 0 mgh i PE G = (78 kg)(9.8 m/s 2 )(1.2 m) PE G = 920 J 8b. KE + PE G = 0 KE = PE G KE = ( 920 J) KE = 920 J 8c. KE = KE f KE i KE = 1 2 mv f KE m 2(920 J) 78 kg 4.8 m/s 9. KE + PE G = 0 KE f KE i + PE f PE i = mv f mv i mgh i = v i 2 = 1 2 v f 2 gh i v i = v f 2 2gh i v i = (18 m/s) 2 2(9.8 m/s 2 )(12 m) v i = 9.4 m/s

5 10a. The work done by friction must be equal to the kinetic energy of the toboggan at the bottom of the hill and also equal to the gravitational potential energy at the top of the hill. W = PE G W = mgh W = (10.5 kg)(9.8 m/s 2 )(15 m) W = 1500 J 10b. W 1 = PE 1 W 2 PE 2 Fd 1 = mgh 1 Fd 2 mgh 2 d 1 = h 1 d 2 h 2 d 1 15 m = 10 m 6 m d 1 = 25 m 11a. p = m v p = (0.045 kg)(38 m/s) p = 1.7 kg m/s 11b. F = p 1.7 kg m/s F = s F = 860 N 11c. F = 860 N in the opposite direction 12a. p = m v p = (0.042 kg)( 6m/s 12 m/s) p = 0.76 kg m/s 12b. F = p 0.76 kg m/s F = 0.04 s F = 19 N 13. F = p F = p f p i F = mv f mv i F + mv i m ( 5700 N)(0.001 s) + (0.060 kg)(50 m/s) kg 45 m/s

6 14. p Th + p α = 0 p Thf p Thi + p αf p αi = 0 p Thf + p αf = 0 m Thf v Thf + m αf v αf = 0 v Thf = m αf v m αf Thf kg v Thf = kg kg ( m/s) v Thf = m/s 15. p b + p w = 0 p bf p bi + p wf p wi = 0 p bf p bi + p wf = 0 m b v bf m b v bi + m w v wf = 0 m b v f m b v bi + m w 0 m b v bi = m b v f + m w v f m b v bi = (m b + m w )v f v bi = (m b + m w ) m b v f (0.01 kg + 5 kg) v bi = (0.6 m/s) 0.01 kg v bi = 300 m/s 16. p A + p b = 0 p Af p Ai + p bf p bi = 0 p Af + p bf = 0 m A v Af + m b v bf = 0 m A v Af = m b v bf v Af = m b v m bf A 9.0 kg v Af = (6 m/s) 64 kg v Af = 0.84 m/s 17a. p = m v p = (250 kg)(120 m/s) p = kg m/s 17b. F = p kg m/s F = 9 s F = 3300 N

7 18. p d + p s = 0 p df p di + p sf p si = 0 m d v df m d v di + m s v sf m s v si = 0 m d v f m d v di + m s v f m s v si = 0 m d v di = m d v f + m s v f m s v si v di = (m d + m s )v f m s v si m d (18 kg + 3 kg)(0.1 m/s) (3 kg)(0.14 m/s) v di = 18 kg v di = m/s 19. p mb + p gw = 0 p mbf p mbi + p gwf p gwi = 0 p mbf p mbi + p gwf = 0 m mb v f m mb v mbi + m gw 0 (m mb + m gw ) m mb v mbi m mb (m mb + m gw ) v mbi 2.6 kg (2.7 m/s) (2.6 kg + 66 kg) 0.10 m/s 20a. p l + p e = 0 p lf p li + p ef = 0 m l v f m l v li + m e 0 (m l + m e ) m l v li m l v m l + m li e kg (18 km/h) kg kg 14 km/h