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# hw 27 - homework 27 HETTIGER CHRISTOF Due Apr 3 2008 4:00...

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homework 27 – HETTIGER, CHRISTOF – Due: Apr 3 2008, 4:00 am 1 Question 1, chap 10, sect 99. part 1 of 2 10 points Two particles of masses m 1 = 6 . 3 kg and m 2 = 27 . 5 kg are moving toward each other along the x axis with equal speeds 9 . 89 m / s. Specifically, v 1 x = +9 . 89 m / s (particle 1 moves to the right) and v 2 x = - 9 . 89 m / s (particle 2 moves to the left). The particles collide elastically. After the collision, the first particle moves at θ 1 = 90 to its original direction while the second par- ticle is deflected through a smaller angle θ 2 < 90 . x y vectorv 1 vectorv 2 vectorv 1 vectorv 2 θ 1 = 90 θ 2 Find the final speed | v 1 | of the first particle. Correct answer: 14 . 8496 m / s (tolerance ± 1 %). Explanation: In elastic collision, the net momentum vector P and the net kinetic energy K are both con- served. In components, P x = m 1 v 1 - m 2 v 2 = 0 - m 2 v 2 cos θ 2 , (1) P y = 0 = m 1 v 1 - m 2 v 2 sin θ 2 , (2) K = m 1 v 2 1 2 + m 2 v 2 2 2 = m 1 v 2 1 2 + m 2 v 2 2 2 . (3) Given v 1 = v 0 , v 2 = - v 0 , eq. (1) gives v 2 cos θ 2 = parenleftbigg 1 - m 1 m 2 parenrightbigg v 0 (4) while eq. (2) implies v 2 sin θ 2 = m 1 m 2 v 1 ; (5) therefore v 2 2 = ( v 2 cos θ 2 ) 2 + ( v 2 sin θ 2 ) 2 = parenleftbigg 1 - m 1 m 2 parenrightbigg 2 v 2 0 + m 2 1 m 2 2 v 2 1 . (6) Substituting this formula into the energy con- servation eq. (3), we arrive at m 1 + m 2 2 v 2 0 = m 1 2 v 2 1 + m 2 2 bracketleftBigg parenleftbigg 1 - m 1 m 2 parenrightbigg 2 v 2 0 + m 2 1 m 2 2 v 2 1 bracketrightBigg = m 1 ( m 2 + m 1 ) 2 m 2 v 2 1 + ( m 2 - m 1 ) 2 2 m 2 v 2 0 (7) and hence (after a bit of algebra) v 2 1 v 2 0 = 3 m 2 - m 1 m 2 + m 1 . (8) Consequently, v 1 = v 0 × radicalbigg 3 m 2 - m 1 m 2 + m 1 = 14 . 8496 m / s . (9) Question 2, chap 10, sect 99. part 2 of 2 10 points Find the deflection angle θ 2 of the second particle. Correct answer: 24 . 0461 (tolerance ± 1 %).

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hw 27 - homework 27 HETTIGER CHRISTOF Due Apr 3 2008 4:00...

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