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1 Section 53: Collisions Mini Inestigation: Newton s Cradle, page 34 Answers may ary Sample answers: A In Step, releasing one end ball caused the far ball on the other end to swing out at the same speed as the original ball, while the middle balls appeared to remain still Changing the setup did not change the outcome as long as all balls were touching When the middle ball was remoed, momentum was not transferred all the way to the end of the line B Yes, the collisions appear to consere momentum, although the balls slow down after a while C Yes, the collisions appear to consere kinetic energy The end ball moes at the same speed as the beginning ball, so kinetic energy is consered (ignoring external forces) D The deice as a whole does not appear to consere mechanical energy Energy is lost in the sound of the collisions, and friction between the balls and between the string and the supports Tutorial Practice, page 36 Gien: 35 kg; 54 m/s [right] ; m 48 kg; Required: Analysis: The collision is perfectly elastic, which means that kinetic energy is consered Apply conseration of momentum and conseration of kinetic energy to construct and sole a linearquadratic system of two equations in two unknowns First, use conseration of momentum to sole for in terms of, remembering that This is a one-dimensional problem, so omit the ector notation for elocities, recognizing that positie alues indicate motion to the right and negatie alues indicate motion to the left m Then, substitute the resulting equation into the conseration of kinetic energy equation to sole for the final elocity of ball Solution: m (35 kg )(54 m/s) (48 kg ) 35 kg 89 m/s The conseration of kinetic energy equation can be simplified by multiplying both sides of the equation by and noting that Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-

2 m + i m i + m f f + m (35 kg )(54 m/s) ( 35 kg ) (89 m/s 48 ) (35) + (48 kg ) (35)(35)(54 m/s) (89 m/s 48 f ) + (35)(48) f 357 m /s 357 m /s (844 m/s) f f f (844 m/s) f 0 f (3984 f 844 m/s) The factor of means the equation has a solution This solution describes the system before the collision The equation has a second solution describing the system after the collision m/s 844 m/s m/s Statement: The final elocity of ball is 46 m/s [right] Gien: ; ; m; m m Required: ; Analysis: ; + m i + m f Solution: m + m (0 m/s) m + m + m i + m i m + m (0 m/s) m + m f + m ( ) f + ( ) f 0 ( ) Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-

3 0 or The final speed of the first stone cannot be the same as its initial speed, so 0 in the equation for 0 Substitute 0 Statement: The final speed of the first stone is 0 m/s The final speed of the second stone is Tutorial Practice, page 38 Gien: 40 kg; m 0 kg; 60 m/s [forward]; Required: f Analysis: Use f m i + m i Solution: f m i + m i (40 kg )(60 m/s [forward]) + (0 kg )(0 m/s) 40 kg + 0 kg f 40 m/s [forward] Statement: The elocity of the balls is 40 m/s [forward] after the collision (a) Gien: 00 kg; i 600 km/h [E] ; m 300 kg; i 300 km/h [E] Required: f Analysis: Conert the elocities to metres per second and use f m i + m i i 600 km h 000 m km h 3600 s [E] i 67 m/s [E] (one extra digit carried) i 300 km h 000 m km h 3600 s [E] i 833 m/s [E] (one extra digit carried) Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-3

4 Solution: f m i + m i (00 kg )(67 m/s [E]) + (300 kg )(833 m/s [E]) 00 kg kg f 36 m/s [E] (one extra digit carried) Statement: The final elocity of the ehicles is 4 m/s [E] (b) Gien: 00 kg; m 300 kg; f 36 m/s [E] Required: p Analysis: The momentum before the collision is equal to the momentum after the collision Use the answer from (a) to determine the momentum after the collision p ( ) f Solution: p ( ) f (00 kg +300 kg)(36 m/s [E]) p kg "m/s Statement: The momentum before and after the collision is kg m/s (c) Gien: 00 kg; i 600 km/h [E] 67 m/s [E]; m 300 kg; i 300 km/h [E] 833 m/s [E]; f 36 m/s [E] Required: E k Analysis: E k E k f " E k i ; E k m Solution: E k E k f " E k i (m + m ) # " f m + i m & i $ % ' ( (00 kg +300 kg)(36 m/s) " [(00 kg)(67 m/s) + (300 kg)(833 m/s) ] E k "8 )0 4 J Statement: The decrease in kinetic energy is J 3 Gien: 66 kg; y 5 m; m 7 kg; Required: f Analysis: Use conseration of energy to determine the speed of the snowboarder at the bottom of the hill Then, use f m i + m i to calculate the final elocity of both people after the collision gy m i Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-4

5 Solution: gy m i g y f m i + m i (98 m/s )(5 m) m/s (one extra digit carried) (66 kg )( m/s) + (7 kg )(0 m/s) 66 kg + 7 kg f m/s Statement: The final speed of each person after the collision is m/s Section 53 Questions, page 39 Answers may ary Sample answer: (a) Since the boxes stick together after the collision, we know this is an inelastic collision Momentum is consered in an inelastic collision Momentum is always consered if there are no external forces acting on the system (b) Kinetic energy is not consered in an inelastic collision In an inelastic collision, some kinetic energy is absorbed by one or both objects, causing the kinetic energy after the collision to be less than the kinetic energy before the collision Gien: 85 kg; m 80 kg; ; f 30 m/s [forward] Required: Analysis: Use f, rearranged to isolate f ( ) f ( ) f m ( ) f m Solution: ( ) f m (85 kg + 80 kg )(30 m/s) (80 kg )(0 m/s) 85 kg 33 m/s Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-5

6 Statement: The speed of the skateboarder just before he landed on the skateboard was 33 m/s 3 Gien: m T 30 kg; 0 kg; i ; 5 m/s [S] Required: f Analysis: Since the total mass is 30 kg, and the mass of the first cart is 0 kg, the mass of the second cart, m, is 0 kg The carts are at rest before they are released, so the initial momentum of the system is zero The final momentum must equal the initial momentum p m Solution: p f m f + m f 0 (0 kg )(5 m/s [S]) + (0 kg ) f f 50 m/s [S] f 50 m/s [N] Statement: The final elocity of the other cart is 50 m/s [N] 4 (a) Yes Both momentum and kinetic energy are consered in a perfectly elastic collision (b) Gien: 85 kg; i 65 m/s ; m 0 kg; i ; f m/s Required: Analysis: Use the conseration of momentum equation to sole for m i + m i m f + m f Solution: m i + m i m f + m f (85 kg )(65 m/s) + (0 kg )(0 m/s) (85 kg )( m/s) + (0 kg ) 555 m/s 935 m/s +0 f f 54 m/s Statement: The final elocity of the second person is 54 m/s in the direction that the first person was originally traelling 5 (a) It is an inelastic collision because the two skaters stick together after the collision (b) Gien: 95 kg; i 50 m/s; m 30 kg; i Required: f Analysis: Use f Solution: f (95 kg )(50 m/s) + (30 kg )(0 m/s) 95 kg +30 kg f m/s Statement: The final speed of the skaters is m/s [initial direction of the first skater] Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-6

7 6 Gien: m 50 kg; m/s [E]; m/s [W] Required: f Analysis: Use f m i + m i Solution: f m i + m i (50 kg)( m/s [E]) + (50 kg)( m/s [W]) 50 kg +50 kg (50 kg )( m/s [E]) (50 kg )( m/s [E]) 50 kg +50 kg f Statement: The elocity of both cars is 0 m/s after the collision They come to a complete stop 7 Answers may ary Sample answer: No, if a moing object collides with a stationary object in a perfectly elastic collision, is it not possible for both objects to be at rest after the collision Because kinetic energy is consered in a perfectly elastic collision, and there was kinetic energy before the collision, there must also be kinetic energy after the collision Therefore, one of the objects must be moing after the collision 8 (a) Gien: kg; i 90 0 km/h [N]; m 0 3 kg; i 30 0 km/h [N] Required: f Analysis: Conert the elocities to metres per second and then use f m i + m i km i 90 0 h i 5 m/s [N] km i 30 0 h i 833 m/s [N] Solution: f m i + m i 000 m km h 3600 s [N] 000 m km h 3600 s [N] (304 kg )(50 m/s [N]) + (0 3 kg )(833 m/s [N]) 30 4 kg +0 3 kg f 37 m/s [N] Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-7

8 Conert back to kilometres per hour f 37 m s km 000 m 3600 s h f 85 km/h Statement: The elocity of the ehicles after the collision is 85 km/h [N] (b) Gien: kg; i 5 m/s [N]; m 0 3 kg; i 833 m/s [N]; f 37 m/s [N] Required: E ki ; E kf Analysis: E k m Solution: E ki ( ) [(304 kg)(5 m/s) + (0 3 kg)(833 m/s) ] E ki 40 6 J E kf (m + m ) f (3 04 kg kg)(37 m/s) E kf J (one extra digit carried) Statement: The total kinetic energy before the collision was J, and the total kinetic energy after the collision was J (c) Gien: E ki J; E kf J Required: E k Analysis: E k E kf " E ki Solution: E k E kf " E k i (396 #0 6 J) " (4 # 0 6 J) E k "4 # 0 6 J Statement: The decrease in kinetic energy during the collision is J Copyright 0 Nelson Education Ltd Chapter 5: Momentum and Collisions 53-8