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hw8 solutions - homework 08 VENNES ROSS Due 1:00 am...

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homework 08 – VENNES, ROSS – Due: Oct 26 2007, 1:00 am 1 Question 1, chap 11, sect 4. part 1 of 1 10 points An open cart on a level surface rolls without frictional loss through a downpour of rain. The rain falls vertically downward as shown below. As the cart rolls, an appreciable amount of rain water accumulates in the cart. v rain rain water cart The speed of the cart will 1. remain the same because the raindrops are falling perpendicular to the direction of cart’s motion. 2. decrease because of conservation of mo- mentum. correct 3. increase because of conservation of me- chanic energy. 4. decrease because of conservation of me- chanic energy. 5. increase because of conservation of mo- mentum. Explanation: This is an inelastic collision in the direction along which the cart is rolling. Only mo- mentum vectorp along that direction is conserved. Because the raindrops fall vertically, they do not carry momentum horizontally. Assume Δ m of rain water accumulates on the cart: p i = p f m v = ( m + Δ m ) v . Therefore v = m m + Δ m v v < v . The speed of the cart will decrease because of conservation of momentum. Question 2, chap 11, sect 1. part 1 of 2 10 points A 59 kg pole vaulter falls from rest from a height of 5.5 m onto a foam rubber pad. The pole vaulter comes to rest 0.37 s after landing on the pad. a) Calculate the athlete’s velocity just be- fore reaching the pad. Correct answer: 10 . 388 m / s (tolerance ± 1 %). Explanation: Let : m = 59 kg , Δ y = 5 . 5 m , t = 0 . 37 s , and a = 9 . 81 m / s 2 . Since v i = 0 m/s, v 2 f = 2 a Δ y v f = ± radicalbig 2 a Δ y = ± radicalBig 2 ( 9 . 81 m / s 2 ) ( 5 . 5 m) = 10 . 388 m / s , which is 10 . 388 m / s directed downward. Question 3, chap 11, sect 1. part 2 of 2 10 points b) Calculate the constant force exerted on the pole vaulter due to the collision. Correct answer: 1656 . 46 N (tolerance ± 1 %). Explanation: Since v f = 0 m/s, vector F Δ t = m Δ vectorv = mvectorv f mvectorv i = mvectorv i F = m v i Δ t = (59 kg) ( 10 . 388 m / s) 0 . 37 s = 1656 . 46 N
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homework 08 – VENNES, ROSS – Due: Oct 26 2007, 1:00 am 2 directed upward. Question 4, chap 11, sect 1. part 1 of 1 10 points A 4 kg steel ball strikes a wall with a speed of 10 . 7 m / s at an angle of 32 . 9 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 10 . 7 m / s 4 kg 10 . 7 m / s 4 kg 32 . 9 32 . 9 If the ball is in contact with the wall for 0 . 185 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 388 . 494 N (tolerance ± 1 %). Explanation: Let : M = 4 kg , v = 10 . 7 m / s , and θ = 32 . 9 . The y component of the momentum is un- changed. The x component of the momentum is changed by Δ P x = 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t = 2 M v cos θ Δ t = 2 (4 kg) (10 . 7 m / s) cos 32 . 9 0 . 185 s bardbl vector F bardbl = 388 . 494 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign.
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