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Unformatted text preview: homework 27 – YOO, HEE – 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 = 4 . 1 kg and m 2 = 20 . 1 kg are moving toward each other along the x axis with equal speeds 5 . 59 m / s. Specifically, v 1 x = +5 . 59 m / s (particle 1 moves to the right) and v 2 x = 5 . 59 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: 8 . 51868 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 , v 2 = v , eq. (1) gives v ′ 2 cos θ 2 = parenleftbigg 1 m 1 m 2 parenrightbigg v (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 + 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 = m 1 2 v ′ 2 1 + m 2 2 bracketleftBigg parenleftbigg 1 m 1 m 2 parenrightbigg 2 v 2 + 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 (7) and hence (after a bit of algebra) v ′ 2 1 v 2 = 3 m 2 m 1 m 2 + m 1 . (8) Consequently, v ′ 1 = v × radicalbigg 3 m 2 m 1 m 2 + m 1 = 8 . 51868 m / s . (9) Question 2, chap 10, sect 99....
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass, Work

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