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oldhw 19 - alexander(jra2623 oldhomework 19 Turner(92510...

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alexander (jra2623) – oldhomework 19 – Turner – (92510) 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 Two particles of masses m and 3 m are moving toward each other along the x -axis with the same speed v . They undergo a head- on elastic collision and rebound along the x - axis. m v 3 m v Determine the final speed of the heavier object. 1. v 3 m = 4 v 2. v 3 m = 3 2 v 3. v 3 m = 4. v 3 m = 2 3 v 5. v 3 m = 1 2 v 6. v 3 m = 0 correct 7. v 3 m = v 8. v 3 m = 2 v 9. v 3 m = 1 3 v 10. v 3 m = 3 v Explanation: Let : v 1 = v , v 2 = - v , m 1 = m and m 2 = 3 m . Basic Concepts: Conservation of momentum Solution: Denote the smaller mass initially moving to the right as m 1 and the larger mass initially moving to the left as m 2 . Since no external forces act on the two masses, even during the collision, the total momentum is conserved. Also, since the masses collide elastically, en- ergy is conserved. Since this is a head on, elastic collision we can use v 2 = 2 v cm - v 2 , where v cm = m 1 v 1 + m 2 v 2 m 1 + m 2 , so v 2 = 2 m 1 v 1 + 2 m 2 v 2 m 1 + m 2 - v 2 ( m 1 + m 2 ) m 1 + m 2 = 2 m v - 6 m v + m v + 3 m v m + 3 m = 0 . Alternative Solution: From the relative velocities formula, v 1 - v 2 = v 2 - v 1 . (1) From conservation of momentum, m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 . (2) Combining Eqs. 1 and 2 gives m 1 v 1 + m 2 v 2 = m 1 ( v 2 + v 2 - v 1 ) + m 2 v 2 , v 2 = m 1 v 1 + m 2 v 2 - m 1 ( v 2 - v 1 ) m 1 + m 2 = m v + 3 m ( - v ) - m ( - v - v ) m + 3 m = m ( v - 3 v + v + v ) 4 m = 0 . 002 10.0 points Two ice skaters approach each other at right angles. Skater A has a mass of 76 . 8 kg and travels in the + x direction at 2 . 03 m / s. Skater B has a mass of 61 . 1 kg and is moving in the + y direction at 2 . 38 m / s. They collide and cling together. Find the final speed of the couple.
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alexander (jra2623) – oldhomework 19 – Turner – (92510) 2 Correct answer: 1 . 54602 m / s. Explanation: From conservation of momentum Δ p = 0 m A v A ˆ ı + m B v B ˆ = ( m A + m B ) v f Therefore v f = radicalbig ( m A v A ) 2 + ( m B v B ) 2 m A + m B = radicalbig (155 . 904 kg m / s) 2 + (145 . 418 kg m / s) 2 76 . 8 kg + 61 . 1 kg = 1 . 54602 m / s 003 10.0 points A(n) 79 g particle moving with an initial speed of 49 m / s in the positive x direction strikes
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