Homework15 PHY 303K

Homework15 PHY 303K - homework 15 – FIERRO JEFFREY –...

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Unformatted text preview: homework 15 – FIERRO, JEFFREY – Due: Mar 22 2008, 11:00 pm 1 Question 1, chap 11, sect 4. part 1 of 1 10 points A 0 . 246 kg puck, initially at rest on a hor- izontal, frictionless surface, is struck by a . 169 kg puck moving initially along the x axis with a speed of 2 . 2 m / s. After the colli- sion, the 0 . 169 kg puck has a speed of 1 . 2 m / s at an angle of 38 ◦ to the positive x axis. Determine the velocity of the 0 . 246 kg puck after the collision. Correct answer: 1 . 00011 m / s (tolerance ± 1 %). Explanation: b x m 2 vectorv 2 φ m 1 vectorv 1 θ y m 1 vectorv Let the puck initially at rest be m 2 . Using momentum conservation, we have x : m 1 v = m 1 v 1 cos θ + m 2 v 2 cos φ y : 0 = m 1 v 1 sin θ − m 2 v 2 sin φ From the above two relations, we can get, tan φ = m 2 v 2 cos φ m 2 v 2 sin φ = m 1 v 1 sin θ m 1 v − m 1 v 1 cos θ φ = arctan parenleftbigg v 1 sin θ v − v 1 cos θ parenrightbigg . From momentum conservation vectorp in the y- direction (ˆ ), we may solve for v 2 , v 2 = m 1 v 1 sin θ m 2 sin φ = 1 . 00011 m / s . Question 2, chap 11, sect 4. part 1 of 1 10 points A(n) 2 . 33 kg object moving at a speed of 5 . 58 m / s strikes a(n) 0 . 753 kg object initially at rest. Immediately after the collision, the 2 . 33 kg object has a velocity of 4 . 48 m / s di- rected 17 . 6981 ◦ from its initial line of motion. It is not known whether this collision is per- fectly elastic or not. What is the speed of the 0 . 753 kg object immediately after the collision? Correct answer: 5 . 85162 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 2 . 33 kg , v 1 i = 5 . 58 m / s , v 1 f = 4 . 48 m / s , m 2 = 0 . 753 kg , and θ = 17 . 6981 ◦ . In any collision, the net meonentum is con- served as a vector: vectorp 1 i = vectorp 1 f + vectorp 2 f , or diagrammatically p 1 i p 1 f p 2 f 17 . 6981 ◦ Aplying the cosine theorem to this triangle, we have: p 2 2 f = p 2 1 i + p 2 1 f − 2 p 1 i p 1 f cos θ = (13 . 0014 kg m / s) 2 + (10 . 4384 kg m / s) 2 − 2 (13 . 0014 kg m / s) (10 . 4384 kg m / s) cos17 . 6981 ◦ = 19 . 4152 kg 2 m 2 / s 2 , and p 2 f = radicalBig (19 . 4152 kg 2 m 2 / s 2 ) = 4 . 40627 kg m / s , so p 2 f = m 2 v 2 f . Solving for v 2 f , we have v 2 f = p 2 f m 2 = (4 . 40627 kg m / s) (0 . 753 kg) = 5 . 85162 m / s . homework 15 – FIERRO, JEFFREY – Due: Mar 22 2008, 11:00 pm 2 This collision happens to be perfectly elas- tic. To see this, let’s check the kinetic energies before and after the collision. Before the col- lision, we had K i = m 1 v 2 1 i 2 = 36 . 2739 J ....
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This note was uploaded on 03/25/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework15 PHY 303K - homework 15 – FIERRO JEFFREY –...

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