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CHAPTER 7 BLM ANSWER KEY BLM 7-1: Section 7.1 Review/ Reinforcement The speed of the cart at point B was 4.6 m/s. W 33 J ( 3.0 N)(4.0 m) W 21 J Answers 1. 2. (a) kinetic W E k (b) work-energy theorem Ek 12 mvB 2 (c) isolated system vB (d) decreases, increases, constant 2 E k m vB 2(21 J) 2.0 kg The boy did 919 J of work on the wagon. vB 4.6 m s The speed of the cart at point C was 0 m/s. W 21 J (3.0 N) 7.0 m W 0 J The question asks only for the work done by the applied force, so the force of friction is not needed for the calculation. In the equation for work used below, F represents the magnitude of the force and d represents the magnitude of the displacement. W Ek v0 4. (a) W F d cos The speed of the skier at the 60 m height (point B) was 34 m/s. ETA ETB W 60.0 N 20.0 m cos 40.0 3. m s W 919.25 J EgA EkA EgB EkB W 919 J mg hA 12 mvA2 mg hB 12 mvB2 vB 2 g hA hB The speed of the cart at point A was 5.7 m/s. m vB 2 9.81 2 120 m 60 m s m vB 34 s W 12 (5 m)(6 N) (3 m)(6 N) W 33 J W Ek Ek 12 mvA 2 2Ek vA m m vA 5.7 s (b) The speed of the skier at the take-off point (point C) was 42 m/s. vB 2 g hA hB m vB 2 9.81 2 120 m 30 m s m vB 42 s CHAPTER 7 BLM ANSWER KEY (c) The skier is moving at half of the take-off speed at a height of approximately 98 m. Half of the takeoff speed is 21 m/s. Since the point is not known, label the position “21.” EgA EkA Eg21 Ek21 2 mg hA 12 mvA2 mg h21 12 mv21 2 g hA 12 v21 h21 g h21 9.81 120 m 21 m s2 1 2 m s 2 9.81 sm2 Ek A Eg B Ek B Eg A Eg B Ek B 1 2 mA v 2 mB g hB 12 mB v 2 mA g hA 0 J 0 J 12 mA 12 mB v 2 mA g hA mB g hB v v 2 mA g hA mB g hB mA mB BLM 7-2: Conservation of Total Energy/Problem Solving 3. (a) Answers 1. (a) Ee 12 kx 2 N Ee 12 70.0 (3.5 m) 2 m 2 ET 4.3 10 J mv 2 F d 2 F d m v (2)(225 N)(1.20 m) 0.30 kg v 42 (b) When the mass is at x = 0, Ee = 0 and Ek 12 mvmax 2 . The maximum velocity is 17 m/s. 1 2 Eg Ek at the arrow's highest point. (b) mg h 12 mv 2 v2 2g 42.426 m 2 s 2 9.81 sm2 h 92 m 2. When the 75.0 kg box hits the floor, both boxes will possess the same speed, assuming that the string does not stretch: 5.4 m/s. 2(428.75 J) 3.0 kg vmax 17 The arrow rises to a height of 92 m. h mvmax 2 428.75 J vmax m s h 2 At x = 3.5 m, all of the energy is elastic potential energy. So, the total energy of the system is 4.3 102 J. Ek W v m (3.0 m) (25.0 kg) 9.81 (3.0 m) s s 2 m s Arrow’s velocity as it leaves the bow: 42 m/s 1 2 m 75.0 kg 25.0 kg v 5.4 h21 98 m 2 (75.0 kg) 9.81 m s The total energy of the system at any point x is given as follows, where A is the amplitude of the system. 1 2 mkA2 12 mv 2 12 kx 2 v k 2 ( A x2 ) m At 0 m displacement v 70.0 mN (3.52 0.02 ) 3.0 kg v 17 m s CHAPTER 7 BLM ANSWER KEY Ee 12 kx 2 At 2.0 m displacement v v 14 m s ET Ek Ee 428.75 J At 3.5 m displacement v v0 4. N m 70.0 (3.52 3.52 ) 3.0 kg m s The ± sign above indicates that the mass could be moving to the right or to the left at this instant. (c) The acceleration when the displacement is 0.0 m is F a m kx a m 70 a 70 N m N m ( 2.0 m) 3.0 kg m a 47 2 s d h vh t vi t cos 5.64 m vi t cos 45 5.64 m (0.7071) t Equation 1 (3.5 m) a 3.0 kg m a 82 2 s N m The ± sign above indicates that the mass could be accelerating to the right or to the left at this instant. The results for part (b) can be used to give Ek 12 mv 2 m Ek 12 (3.0 kg) 13.8744 s Ek 2.9 102 J m 0 1.30 m vi t (sin 45 ) 12 9.81 2 t 2 s 2 Equation 2 Substitute vi from Equation 1 into Equation 2. 5.64 m m t (0.7071) 12 9.81 2 t 2 (0.7071)t s m 2 6.94 m 4.905 2 t s 0 1.30 m t The acceleration when the displacement is 3.5 m is 70 The angle the gun makes with the desktop is 45°. The height of the end of the gun is 1.30 m. When the ball hits the ground, d v 0. d v h vv t 12 g t 2 The acceleration when the displacement is 2.0 m is a If a projectile starts from rest from a point above the ground, then at time t, its horizontal and vertical components of position will be as shown below. The gun’s spring constant is 6.0 N/m. vi (0.0 m) 3.0 kg m a0 2 s (d) N 2 Ee 12 70.0 (2.0 m) m Ee 140 J 70.0 mN (3.52 2.02 ) 3.0 kg 6.94 m 4.905 sm2 t 1.19 s Substitute t = 1.19 s into Equation 1 to find vi, the initial velocity of the ball. 5.64 m (0.7071)1.19 s m vi 6.70 s vi The kinetic energy of the ball as it leaves the gun is equal to the elastic potential energy of the gun’s spring. Solve for the spring constant. CHAPTER 7 BLM ANSWER KEY Ee Ek 1 2 Edme Mariotte kx mv 2 1 2 (1620–1684) 2 mv 2 x2 0.00056 kg(6.70 k (0.065 m) 2 N k 6.0 m k m 2 s ) BLM 7-3: The Law of Conservation of Momentum/Science Inquiry Answers The following is some information on the scientists selected. French physicist Mariotte demonstrated the conclusions of John Wallis on the conservation of momentum by conducting a series of experiments on pendulums allowed to strike each other. The following Internet iste provides more information on Mariotte. www.newadvent.org/cathen/09671a.htm Isaac Newton (1642–1727) The law of conservation of momentum follows directly from Newton’s second and third laws. Newton described the momentum experiments made by Huygens, Mariotte, Wren, and Wallis and his own careful experiments in his book, the Principia. Newton’s experiments consisted of verifying the law of conservation of momentum by carefully correcting for air resistance of colliding pendulums. More information about Newton can be found on the following Internet site. www.arts.richmond.edu/~rubin/pedagogy/131/ 131notes/131notes_130.html Scientist Role John Wallis (1616–1703) John Wallis was one of the founders of the Royal Society. He was also one of the first scientists to suggest the law of the conservation of momentum when, in 1668, the Royal Society accepted the challenge to investigate the behaviour of colliding bodies. Wallis is credited with the discovery of the law of momentum — even though Wren and Huygens sent in similar solutions to the Royal Society — because Wallis considered perfectly inelastic bodies and the other two scientists investigated only elastic bodies. Further information on Wallis can be found on the Internet at www.maths.tcd.ie/pub/HistMath/People/Wallis/ RouseBall/RB_Wallis.html Christopher Wren (1632–1723) Along with John Wallis and Christiaan Huygens, Christopher Wren produced similar results related to the conservation of momentum. More details about Wren’s work can be found on the Internet at www.arct.cam.ac.uk/~Campbell/phd/wren/science.html Christiaan Huygens (1629–1695) Huygens conducted experiments that demonstrated how momentum (mv) is conserved in physical interactions. He also showed that mv 2 (2Ek) was conserved. This was a first step in working out the law of conservation of energy. More information is posted on the Internet at www.clas.ufl.edu/users/rhatch/08sr-hygns-bio.html 2. (a) Speed of the roller coaster at point B: 12.7 m/s Ek Eg Eheat Ek Eg 1 2 mvB 2 mg hB Eheat 12 mvA 2 mg hA 1 2 mvB 2 mg hB Eheat 12 mvA 2 mg hA vB vB 2(mg hB Eheat 12 mvA 2 mg hA ) m 2 (6.1803 10 4 kg m s 2 2 ) (1.24 105 J) (1.1250 10 4 J) (2.4721 105 kg m s 2 2 ) 900.0 kg vB 12.7 m s BLM 7-4: Concept Review/Reinforcement Answers 1. The average force exerted is 2.3 103 N, acting in the opposite direction of your head’s motion. This is a very large force, and illustrates how dangerous it can be to bump your head. F t p p m(vf vi ) m m p (4.5 kg) 0 2.5 s s kg m p 11.25 s p F t 11.25 kgsm F 0.0050 s F 2.3 103 N CHAPTER 7 BLM ANSWER KEY 2. The momentum of the players before the collision equals the momentum of the players after the collision. The defender was moving toward the other player at a velocity of –3.4 m/s. mp vp md vd 0 vd vd mp vp md vd 3.4 3. 85 kg (4.0 m s ) 100 kg m s The initial momentum of the tool and astronaut is equal to the final momentum. The astronaut is moving at 0.093 m/s[forward] toward the spaceship. 0 ma va mt vt mt vt va ma va 0.70 kg (12.0 ms ) 90 kg m va 0.093 [forward] s 4. The law of the conservation of momentum states that the sum of the momenta of two objects before a collision is equal to the sum of their momenta after they collide. As shown below, momentum is conserved in collisions (A) and (B) because the vector sum of the momenta of the balls is equal before and after the collisions. 5. The total momentum before the collision is equal to the total momentum after the collision. Sphere B moves at a velocity of 3.23 m/s[E46.6°S]. Refer to the diagram and calculations below. CHAPTER 7 BLM ANSWER KEY Momentum in the x direction mvA x 0 mvA x mvB x 2. (a) Ek Eg Eheat Ek Eg vB x vA x vA x m m (4.20 cos34.0 ) s s m vB x 2.218 s vB x 5.70 Momentum in the y direction 0 mvA y mvB y 1 2 mvB 2 mg hB Eheat 12 mvA 2 mg hA 1 2 mvB 2 mg hB Eheat 12 mvA 2 mg hA vB vB vB y vA y m s Pythagorean theorem and the tangent function 2 vB vB x 2 vB y 2 (b) mechanical (c) positive, increases (d) conservative (e) third (f) equal (g) impulse 4 3. 5 kg m s 0 12 (900.0 kg)(12.7 ms ) 2 10.0 m F 7.26 103 N The water going over the falls will increase in temperature by approximately 2.3°C. W Eg Qgained W opposite 5 ) (1.24 10 J) (1.1250 10 J) (2.4721 10 F W mg h Answers 2 F d E k f E k i m s BLM 7-5: Chapter 7 Test/Assessment s 2 2 ) The brakes do work on the roller coaster to reduce its kinetic energy to zero after 10.0 m. The frictional braking force is in the opposite direction of the motion and is 7.26 103 N. 2.218 46.6 clockwise from the x- axis kg m m s m s tan 1.0586 4 900.0 kg 2.348 1 1. (a) 2 (6.1803 10 2 W Ek 2 Positive x- and negative y -components place the resultant in the fourth quadrant. tan m (b) vB y 2.348 2 2(mg hB Eheat 12 mvA 2 mg hA ) vB 12.7 vB y 4.20sin 34.0 m m 2 vB 2.218 2.348 s s m vB 3.23 s vB y tan vB x Speed of the roller coaster at point B: 12.7 m/s m Qgained (1.0 kg) 9.81 2 (979.0 m) s Qgained 9603.99 J mcT Qgained T Qgained T 9603.99 J 1.0 kg(4186 kgJ C ) mcw T 2.3 C CHAPTER 7 BLM ANSWER KEY 4. The average force of the racquet on the ball was 1.9 102 N. F t p F F m vf vi t 0.057 kg 10 ms 13 ms 7.00 103 s F 1.9 N 5. Cars are made with bumpers that retract during a collision in order to increase the time of the collision, thus reducing the force of the impact