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F5Bh0C9YYiqO_1210015950_jwk572

F5Bh0C9YYiqO_1210015950_jwk572 - homework 02 – KIM JI –...

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Unformatted text preview: homework 02 – KIM, JI – Due: Sep 12 2007, 4:00 am 1 Question 1, chap 2, sect 6. part 1 of 3 10 points A ball is dropped from rest at point O . After falling for some time, it passes by a window of height 3 . 4 m and it does so during time 0 . 35 s. The acceleration of gravity is 9 . 8 m / s 2 . O A B 3 . 4 m b b b b b b b b b b b x y The ball accelerates all the way down; let v A be its speed as it passes the window’s top A and v B its speed as it passes the window’s bottom B . How much did the ball speed up as it passed the window; i.e. , calculate Δ v down = v B − v A ? Correct answer: 3 . 43 m / s (tolerance ± 1 %). Explanation: Let : h = 3 . 4 m , t AB = 0 . 35 s , and g = 9 . 8 m / s 2 . O A B h t AB b b b b b b b b b b b b t y Assume: Down is positive. The ball falls under a constant acceleration g , so g = Δ v t = v B − v A t and the change of its velocity during time t AB is simply Δ vectorv = vectorv B − vectorv A = g t AB , assuming the downward direction to be positive. Δ v down = (9 . 8 m / s 2 )(0 . 35 s) = 3 . 43 m / s . Question 2, chap 2, sect 6. part 2 of 3 10 points Calculate the speed v A at which the ball passes the window’s top. Correct answer: 7 . 99929 m / s (tolerance ± 1 %). Explanation: Assume: Down is positive. Given: The uniform downward acceleration g , we have y ( t ) = y ( t A ) + vectorv A ( t − t A ) + 1 2 g ( t − t A ) 2 , and hence h = y ( t B ) − y ( t A ) = vectorv A ( t B − t A ) + 1 2 g ( t B − t A ) 2 = vectorv A t AB + 1 2 g t 2 AB . Solving this equation for the velocity vectorv A , we obtain vectorv A = h t AB − g t AB 2 = (3 . 4 m) (0 . 35 s) − (9 . 8 m / s 2 ) (0 . 35 s) 2 = 7 . 99929 m / s . Thus the speed v A = bardbl vectorv A bardbl = 7 . 99929 m / s . Question 3, chap 2, sect 6. part 3 of 3 10 points Now consider a new situation: The ball is thrown upward from the ground with an initial velocity that takes exactly the same time t BA = t AB = 0 . 35 s to pass by the window, with the ball moving up rather than down. Consider the ball’s slowdown during this time: homework 02 – KIM, JI – Due: Sep 12 2007, 4:00 am 2 Let v ′ B be the ball’s speed (do not confuse the speed with the velocity) as it passes the window’s bottom on the way up and let v ′ A be its speed as it passes the window’s top, also in its way up. How does the ball’s slowdown Δ v up = v ′ B − v ′ A compare to its speedup Δ v down on the way down? 1. Δ v up < Δ v down . 2. Δ v up > Δ v down if the mass of the ball is less than 0 . 1 kg and Δ v up < Δ v down if the mass of the ball is greater than 0 . 1 kg 3. Δ v up = Δ v down . correct 4. Δ v up > Δ v down . 5. Δ v up < Δ v down if the mass of the ball is less than 0 . 1 kg and Δ v up > Δ v down if the mass of the ball is greater than 0 . 1 kg Explanation: Δ v up = v ′ A − v ′ B ....
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F5Bh0C9YYiqO_1210015950_jwk572 - homework 02 – KIM JI –...

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