HW7 Solution - choi (hc5878) – hw7 – Demkov – (59910)...

Info iconThis preview shows pages 1–3. 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: choi (hc5878) – hw7 – Demkov – (59910) 1 This print-out should have 48 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A revolutionary war cannon, with a mass of 2170 kg, fires a 23 . 4 kg ball horizontally. The cannonball has a speed of 102 m / s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immedi- ately after it was fired? Correct answer: 1 . 09991 m / s. Explanation: Let : m = 23 . 4 kg , M = 2170 kg , and v = 102 m / s . The cannon’s velocity immediately after it was fired is found by using conservation of momentum along the horizontal direction: M V + mv = 0 ⇒ − V = m M v where M is the mass of the cannon, V is the velocity of the cannon, m is the mass of the cannon ball and v is the velocity of the cannon ball. Thus, the cannon’s speed is | V | = m M | v | = 23 . 4 kg 2170 kg (102 m / s) = 1 . 09991 m / s . 002 (part 2 of 2) 10.0 points The same explosive charge is used, so the total energy of the cannon plus cannonball system remains the same. Disregarding friction, how much faster would the ball travel if the cannon were mounted rigidly and all other parameters re- mained the same? Correct answer: 0 . 548479 m / s. Explanation: By knowing the speeds of the cannon and the cannon ball, we can find out the total kinetic energy available to the system K net = 1 2 mv 2 + 1 2 M V 2 . This is the same amount of energy available as when the cannon is fixed. Let v ′ be the speed of the cannon ball when the cannon is held fixed. Then, 1 2 mv ′ 2 = 1 2 ( mv 2 + M V 2 ) . ⇒ v ′ = radicalbigg v 2 + M m V 2 = v radicalbigg 1 + m M = (102 m / s) radicalBigg 1 + 23 . 4 kg 2170 kg = 102 . 548 m / s . Thus, the velocity difference is v ′ − v = 102 . 548 m / s − 102 m / s = . 548479 m / s . 003 (part 1 of 3) 10.0 points Tony (of mass 70 kg) coasts on his bicycle (of mass 10 kg) at a constant speed of 5 m / s, carrying a 5 kg pack. Tony throws his pack forward, in the direction of his motion, at 2 m / s relative to the speed of bicycle just before the throw. What is the initial momentum of the system (Tony, the bicycle, and the pack)? Correct answer: 425 kg · m / s. Explanation: Let : m T = 70 kg , m B = 10 kg , m P = 5 kg , and v i = 5 m / s . choi (hc5878) – hw7 – Demkov – (59910) 2 Since p = mv . Thus p i = ( m T + m B + m P ) v i = (70 kg + 10 kg + 5 kg)(5 m / s) = 425 kg · m / s . 004 (part 2 of 3) 10.0 points What is the momentum of the system im- mediately after the pack is thrown? Correct answer: 425 kg · m / s. Explanation: Since momentum is conserved, p f = p i = 425 kg · m / s ....
View Full Document

This note was uploaded on 11/01/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.

Page1 / 21

HW7 Solution - choi (hc5878) – hw7 – Demkov – (59910)...

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

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