Kaplunovsky - practice 04

Kaplunovsky - practice 04 - practice 04 Due: Feb 12 2007,...

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practice 04 – – Due: Feb 12 2007, noon 1 Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18 . 4 m. It moves toward the opposing wall at a constant rate of 8 . 6 m / s. Also, it has an acceleration in the direction parallel to the walls of 2 . 2 m / s 2 . It hits the opposite wall at the same height. 18 . 4 m 2 . 2 m / s 8 . 6 m / s a) What will be its speed when it hits the opposing wall? Correct answer: 9 . 80386 m / s (tolerance ± 1 %). Explanation: Let : d = 18 . 4 m , v x = 8 . 6 m / s , a = 2 . 2 m / s 2 , Basic Concepts Kinematics equations v = v o + g t s = s o + v o t + 1 2 g t 2 d a 4 70698 m 8 . 6 m / s 9 80386 m 61 3071 5 . 03537 m The horizontal motion will carry the parti- cle to the opposite wall, so d = v x t f and t f = d v x = (18 . 4 m) (8 . 6 m / s) = 2 . 13953 s . is the time for the particle to reach the oppo- site wall. Horizontally, the particle reaches the maxi- mum parallel distance when it hits the oppo- site wall at the time of t = d v x , so the ±nal parallel velocity v y is v y = a t = a d v x = (2 . 2 m / s 2 ) (18 . 4 m) (8 . 6 m / s) = 4 . 70698 m / s . The velocities act at right angles to each other, so the resultant velocity is v f = r v 2 x + v 2 y = r (8 . 6 m / s) 2 + (4 . 70698 m / s) 2 = 9 . 80386 m / s . Question 2, chap 4, sect 3. part 2 of 2 10 points
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practice 04 – – Due: Feb 12 2007, noon 2 b) At what angle with the wall will the particle strike? Correct answer: 61 . 3071 (tolerance ± 1 %). Explanation: When the particle strikes the wall, the ver- tical component is the side adjacent and the horizontal component is the side opposite the angle, so tan θ = v x v y , so θ = arctan p v x v y P = arctan p 8 . 6 m / s 4 . 70698 m / s P = 61 . 3071 . Note: The distance traveled parallel to the walls is y = 1 2 a t 2 = 1 2 (2 . 2 m / s 2 ) (2 . 13953 s) 2 = 5 . 03537 m . Question 3, chap 4, sect 4. part 1 of 4 5 points A ball of mass 0 . 4 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below. The velocity of the ball as it leaves the kicker’s foot is 18 m / s at angle of 60 above the horizontal. The top of the fence is 9 m high. The kicker’s foot is in contact with the ball for 0 . 05 s. The ball hits nothing while in ±ight and air resistance is negligible. The acceleration due to gravity is 9 . 8 m / s 2 . b 20 m 9 m 18 m / s 60 b b b b b b b b b b b b b b b b b b b b Determine the magnitude of the average net force exerted on the ball during the kick. Correct answer: 144 N (tolerance ± 1 %). Explanation:
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Kaplunovsky - practice 04 - practice 04 Due: Feb 12 2007,...

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