Homework 11 Solutions

Homework 11 Solutions - lim (kl9356) Homework 11 Weathers...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 = 32 . 9 . The y component of the momentum is un- changed. The x component of the momentum is changed by P x = 2 M v cos . Therefore, using impulse formula, F = P t = 2 M v cos 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 , = 27 . 1 , and t = 3 . 8 ms . Note that the range of the golf ball is given by L = v 2 sin2 g , so the initial velocity of the ball is v = radicalBigg L g sin(2 ) = 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....
View Full Document

Page1 / 4

Homework 11 Solutions - lim (kl9356) Homework 11 Weathers...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online