hw20 - Overton, Mays Homework 20 Due: Mar 22 2005, 4:00 am...

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Unformatted text preview: Overton, Mays Homework 20 Due: Mar 22 2005, 4:00 am Inst: Turner 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points Consider the general case of collisions of two masses m 1 = m with m 2 = 2 m along a fric- tionless horizontal surface. Denote the initial and the final center of mass momenta to be p i cm = p 1 + p 2 , and p f cm = p 1 + p 2 . And the initial and final kinetic energies to be K i = K 1 + K 2 , and K f = K 1 + K 2 . m 1 m 2 v 1 v 2 For an elastic collision which pair of state- ments is correct? 1. p i cm = p f cm , K i < K f 2. p i cm > p f cm , K i = K f 3. p i cm < p f cm , K i = K f 4. p i cm < p f cm , K i > K f 5. p i cm > p f cm , K i > K f 6. p i cm = p f cm , K i = K f correct 7. p i cm > p f cm , K i < K f 8. p i cm = p f cm , K i > K f 9. p i cm < p f cm , K i < K f Explanation: Momentum and energy are always con- served in elastic collision. 002 (part 2 of 3) 10 points Consider a perfectly inelastic head-on colli- sion, with initial velocity of m 1 to be v 1 and that of m 2 to be v 2 = 0. Note: Perfectly inelastic means that the two particles stick together after the collision. Find the final speed v cm of the ( m 1 + m 2 ) system. 1. v cm = 3 5 v 1 2. v cm = 1 4 v 1 3. v cm = 1 2 v 1 4. v cm = 1 3 v 1 correct 5. v cm = 3 4 v 1 6. v cm = 2 3 v 1 7. v cm = 2 v 1 8. v cm = 1 5 v 1 9. v cm = v 1 10. v cm = 2 7 v 1 Explanation: In a collision the linear momentum is always conserved. From conservation of linear momentum we get m 1 v 1 + m 2 0 = ( m 1 + m 2 ) v f v cm = v f = m 1 ( m 1 + m 2 ) v 1 = m 3 m v 1 = v 1 3 003 (part 3 of 3) 10 points Now consider an elastic head-on collision again with initial velocity of m 1 to be v 1 and of m 2 to be v 2 = 0. Find the final speed v f 2 of m 2 . 1. v f 2 = 2 7 v 1 Overton, Mays Homework 20 Due: Mar 22 2005, 4:00 am Inst: Turner 2 2. v f 2 = 1 4 v 1 3. v f 2 = 1 2 v 1 4. v f 2 = 2 3 v 1 correct 5. v f 2 = 1 5 v 1 6. v f 2 = 3 4 v 1 7. v f 2 = 2 v 1 8. v f 2 = 3 5 v 1 9. v f 2 = 1 3 v 1 10. v f 2 = v 1 Explanation: The CM velocity is given by V CM = i m i v i i m i = m 1 v 1 m 1 + m 2 as v 2 = 0. Thus, in the CM frame, m 2 has a velocity v ( CM ) 2 = v 2- V CM =- V CM =- m 1 v 1 m 1 + m 2 . After a one dimensional elastic collision, each body changes the direction of its momentum with the magnitude of the momentum con- served. Note: In the CM frame the total momen- tum is still zero if each velocity changes sign. Hence, the final velocity of m 2 in the CM frame is v ( CM ) 2 ,f = V CM = m 1 v 1 m 1 + m 2 The velocity v lab 2 ,f ( v f 2 ) in the lab frame is v lab 2 ,f = v ( CM ) 2 ,f + V CM = 2 V CM = 2 m 1 v 1 m 1 + m 2 = 2 3 v 1 ....
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hw20 - Overton, Mays Homework 20 Due: Mar 22 2005, 4:00 am...

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