Goldstein_31_32_7

# Goldstein_31_32_7 - Homework 5 3.31 3.32 3.7a Michael Good...

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Homework 5: # 3.31, 3.32, 3.7a Michael Good Sept 27, 2004 3.7a Show that the angle of recoil of the target particle relative to the incident direction of the scattered particle is simply Φ = 1 2 ( π - Θ). Answer: It helps to draw a ﬁgure for this problem. I don’t yet know how to do this in L A T E X, but I do know that in the center of mass frame both the particles momentum are equal. m 1 v 0 1 = m 2 v 0 2 Where the prime indicates the CM frame. If you take equation (3.2) Gold- stein, then its easy to understand the equation after (3.110) for the relationship of the relative speed v after the collision to the speed in the CM system. v 0 1 = μ m 1 v = m 2 m 1 + m 2 v Here, v is the relative speed after the collision, but as Goldstein mentions because elastic collisions conserve kinetic energy, (I’m assuming this collision is elastic even though it wasn’t explicitly stated), we have v = v 0 , that is the relative speed after collision is equal to the initial velocity of the ﬁrst particle in the laboratory frame ( the target particle being stationary). v 0 1 = m 2 m 1 + m 2 v 0 This equation works the same way for v 0 2 v 0 2 = m 1 m 1 + m 2 v 0 From conservation of momentum, we know that the total momentum in the CM frame is equal to the incident(and thus total) momentum in the laboratory frame. ( m 1 + m 2 ) v cm = m 1 v 0 We see 1

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v cm = m 1 m 1 + m 2 v 0 This is the same as v 0 2 v 0 2 = v cm If we draw both frames in the same diagram, we can see an isosceles triangle where the two equal sides are v 0 2 and v cm . Φ + Φ + Θ = π Φ = 1 2 ( π - Θ) 3.31 Examine the scattering produced by a repulsive central force f + kr - 3 . Show that the diﬀerential cross section is given by σ (Θ) d Θ = k 2 E (1 - x ) dx x 2 (2 -- x ) 2 sin πx where x is the ratio of Θ and E is the energy. Answer: The diﬀerential cross section is given by Goldstein (3.93): σ (Θ) = s sin Θ ± ± ± ± ds d Θ ± ± ± ± We must solve for s , and ds/d Θ. Lets solve for Θ( s ) ﬁrst, take its derivative with respect to s , and invert it to ﬁnd ds/d Θ. We can solve for Θ( s ) by using Goldstein (3.96): Θ( s ) = π - 2 Z r m sdr r q r 2 (1 - V ( r ) E ) - s 2 What is V ( r ) for our central force of f = k/r 3 ? It’s found from - dV/dr = f .
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## This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.

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Goldstein_31_32_7 - Homework 5 3.31 3.32 3.7a Michael Good...

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