Homework 11 Solutions

# Homework 11 Solutions - lim(kl9356 – Homework 11 –...

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Unformatted text preview: lim (kl9356) – Homework 11 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 3 kg steel ball strikes a wall with a speed of 13 . 2 m / s at an angle of 32 . 9 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 3 . 2 m / s 3 kg 1 3 . 2 m / s 3 kg 32 . 9 ◦ 32 . 9 ◦ If the ball is in contact with the wall for . 233 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 285 . 399 N. Explanation: Let : M = 3 kg , v = 13 . 2 m / s , and θ = 32 . 9 ◦ . 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 (3 kg) (13 . 2 m / s) cos 32 . 9 ◦ . 233 s bardbl vector F bardbl = 285 . 399 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 002 10.0 points A golf ball ( m = 39 . 1 g) is struck a blow that makes an angle of 27 . 1 ◦ with the horizontal. The drive lands 283 m away on a flat fairway. The acceleration of gravity is 9 . 8 m / s 2 . If the golf club and ball are in contact for 3 . 8 ms, what is the average force of impact? Neglect air resistance. Correct answer: 601 . 689 N. Explanation: Let : g = 9 . 8 m / s 2 , m = 39 . 1 g , θ = 27 . 1 ◦ , and t = 3 . 8 ms . Note that the range of the golf ball is given by L = v 2 sin2 θ g , so the initial velocity of the ball is v = radicalBigg L g sin(2 θ ) = radicalBigg (283 m) (9 . 8 m / s 2 ) sin(2 × 27 . 1 ◦ ) = 58 . 4761 m / s . The average force exerted is the change in its momentum over the time of contact F = mv t = (39 . 1 g) (0 . 001 kg / g) (58 . 4761 m / s) (3 . 8 ms) (0 . 001 s / ms) = 601 . 689 N . 003 10.0 points A 4 . 4 kg particle has a velocity of v x = 4 . 59 m / s and v y = − 3 . 7 m / s....
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## This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

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Homework 11 Solutions - lim(kl9356 – Homework 11 –...

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