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HW8 - homework 08 GAUTHIER JOSEPH Due 1:00 am Question 1...

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homework 08 – GAUTHIER, JOSEPH – 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. increase because of conservation of mo- mentum. 2. decrease because of conservation of me- chanic energy. 3. increase because of conservation of me- chanic energy. 4. remain the same because the raindrops are falling perpendicular to the direction of cart’s motion. 5. decrease because of conservation of mo- mentum. correct 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 52 kg pole vaulter falls from rest from a height of 4.9 m onto a foam rubber pad. The pole vaulter comes to rest 0.27 s after landing on the pad. a) Calculate the athlete’s velocity just be- fore reaching the pad. Correct answer: 9 . 805 m / s (tolerance ± 1 %). Explanation: Let : m = 52 kg , Δ y = 4 . 9 m , t = 0 . 27 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 ) ( 4 . 9 m) = 9 . 805 m / s , which is 9 . 805 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: 1888 . 37 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 = (52 kg) ( 9 . 805 m / s) 0 . 27 s = 1888 . 37 N

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homework 08 – GAUTHIER, JOSEPH – Due: Oct 26 2007, 1:00 am 2 directed upward. Question 4, chap 11, sect 1. part 1 of 1 10 points A 1 kg steel ball strikes a wall with a speed of 4 . 05 m / s at an angle of 51 . 1 with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 4 . 05 m / s 1 kg 4 . 05 m / s 1 kg 51 . 1 51 . 1 If the ball is in contact with the wall for 0 . 156 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 32 . 6058 N (tolerance ± 1 %). Explanation: Let : M = 1 kg , v = 4 . 05 m / s , and θ = 51 . 1 . 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 (1 kg) (4 . 05 m / s) cos 51 . 1 0 . 156 s bardbl vector F bardbl = 32 . 6058 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign.
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HW8 - homework 08 GAUTHIER JOSEPH Due 1:00 am Question 1...

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