# hw10 - karna (pk4534) – HW 10 – coker – (58245) 1...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: karna (pk4534) – HW 10 – coker – (58245) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 5 kg steel ball strikes a wall with a speed of 10 . 5 m / s at an angle of 50 . 2 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 . 5 m / s 5 kg 1 . 5 m / s 5 kg 50 . 2 ◦ 50 . 2 ◦ If the ball is in contact with the wall for . 184 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 365 . 28 N. Explanation: Let : M = 5 kg , v = 10 . 5 m / s , and θ = 50 . 2 ◦ . 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 (5 kg) (10 . 5 m / s) cos 50 . 2 ◦ . 184 s bardbl vector F bardbl = 365 . 28 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 002 10.0 points A 0.41 kg soccer ball approaches a player horizontally with a velocity of 17 m / s to the north. The player strikes the ball and causes it to move in the opposite direction with a velocity of 24 m / s. What impulse was delivered to the ball by the player? Correct answer: − 16 . 81 kg · m / s. Explanation: Let north be positive: Let : m = 0 . 41 kg , v i = 17 m / s , and v f = − 24 m / s . Δ vectorp = mvectorv f − mvectorv i − (0 . 41 kg)(17 m / s) = − 16 . 81 kg · m / s , which is directed to the south. 003 (part 1 of 2) 10.0 points A railroad car of mass 12800 kg moving at 3 . 96 m / s collides and couples with two cou- pled railroad cars, each of the same mass as the single car and moving in the same direc- tion at 2 . 23 m / s. What is the speed of the three coupled cars after the collision? Correct answer: 2 . 80667 m / s. Explanation: Given : m = 12800 kg , v 1 = 3 . 96 m / s , and v 2 = 2 . 23 m / s . By conservation of momentum, mv 1 + (2 m ) v 2 = (3 m ) v f karna (pk4534) – HW 10 – coker – (58245) 2 v f = mv 1 + 2 mv 2 3 m = (12800 kg) (3 . 96 m / s) 3 (12800 kg) + 2 (12800 kg) (2 . 23 m / s) 3 (12800 kg) = 2 . 80667 m / s . 004 (part 2 of 2) 10.0 points How much kinetic energy is lost in the colli- sion? Correct answer: 12769 . 7 J. Explanation: The change in kinetic energy is Δ KE = KE f − KE i = 1 2 (3 m ) v 2 f − bracketleftbigg 1 2 mv 2 1 + 1 2 (2 m ) v 2 2 bracketrightbigg = 3 2 (12800 kg) (2 . 80667 m / s) 2 − 1 2 (12800 kg) (3 . 96 m / s) 2 − (12800 kg) (2 . 23 m / s) 2 = − 12769 . 7 J , which represents a loss of 2 . 80667 m / s . 005 (part 1 of 2) 10.0 points Given two masses, m 1 = 18 kg and m 2 = 12 kg, m 1 is moving with a velocity v 1 collid- ing with m 2 , which is suspended by a string of length 7 m....
View Full Document

## This note was uploaded on 10/29/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 13

hw10 - karna (pk4534) – HW 10 – coker – (58245) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online