Ch2-HW4-solutions

# Ch2-HW4-solutions - schopper(as55864 – Ch2-HW4 – haley...

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Unformatted text preview: schopper (as55864) – Ch2-HW4 – haley – (56465) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A tennis ball has a mass of 0 . 058 kg. A profes- sional tennis player hits the ball hard enough to give it a speed of 52 m / s. The ball hits a wall and bounces back with almost the same speed. As indicated in the following figure, high-speed photography shows that the ball is crushed 2 cm at the instant when its speed iss momentarily zero, before rebounding. 2 cm vectorp i Making the very rough approximation that the large force that the wall exerts on the ball is approximately constant during contact, determine the approximate magnitude of this force. Hint: Think about the approximate amount of time it takes for the ball to come momentarily to rest. (For comparison note that the gravitational force on the ball is quite small, only about (0 . 058 kg)(9 . 8 N / kg) ≈ . 6 N . A force of 5 N is approximately the same as a force of one pound.) Correct answer: 3920 . 8 N. Explanation: First we want to find the time interval as the ball is slowing down during the collision. vectorv avg = vectorv i + vectorv f 2 v avg ,x = v i,x + v f,x 2 = 52 m / s + 0 2 = 26 m / s ⇒ v avg ,x = Δ x Δ t ⇒ Δ t = Δ x v avg ,x = . 02 m 26 m / s = 0 . 000769231 s . Now apply the momentum principle. De- fine the system to be the tennis ball. The only significant force on the ball during the colli- sion is due to the wall. Thus F net ,x = F wall ,x . | F net ,x | = vextendsingle vextendsingle vextendsingle vextendsingle Δ p x Δ t vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle F wall ,x vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle p f,x − p i,x Δ t vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle − (0 . 058 kg)(52 m / s) . 000769231 s vextendsingle vextendsingle vextendsingle vextendsingle = 3920 . 8 N . 002 (part 1 of 9) 2.0 points A 0 . 5 kg block of ice is sliding by you on a very slippery floor at 2 . 6 m / s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for . 003 s. The block eventually slides at an angle of 22 ◦ from its original direction. The overhead view shows in the following diagram is approximately to scale. The red arrow represents the average force your toe applies briefly to the block of ice. II I III x z Which of the possible paths shown in the diagram corresponds to the correct overhead view of the block’s path? schopper (as55864) – Ch2-HW4 – haley – (56465) 2 1. III 2. I 3. II correct Explanation: Image II is correct because the kick is brief, and after the kick, vectorv is constant so the block will move with a constant velocity....
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## This note was uploaded on 09/12/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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Ch2-HW4-solutions - schopper(as55864 – Ch2-HW4 – haley...

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