Phys105A_W09_HW2sols

Phys105A_W09_HW2sols - Physics 105A Homework # 2 Solutions...

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Physics 105A Homework # 2 Solutions Richard Eager Problem (1) Taylor 2.7 (a) Motion of a particle moving in one dimension subject to a force F = F ( v ) that is only a function of the particle’s velocity. Writing Newton’s 2nd Law F ( v )= m dv dt we can separate variables dt = m dv F ( v ) . Integrating both sides yields t = m ± v v 0 dv ± F ( v ± ) For a constant force F ( v )= F 0 the general expressions yields t = m ± v v 0 dv ± F 0 = m F 0 ( v - v 0 ) After solving for v, v = F 0 m t + v 0 or v = at + v 0 where the acceleration a = F 0 m . This is precisely the result for constant acceler- ation. Problem (2) Taylor 2.31 Basketball of mass m = 600 g =0 . 6 kg and diameter D = 24 cm =0 . 24 m is dropped from a height y = 30 m. The linear and quadratic drag are f lin = bv f quad = cv 2 where for a sphere of diameter D, b = β D c = γ D 2 1
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and β =1 . 6 × 10 - 4 N · s/m 2 , γ =0 . 25 N · s 2 /m 4 at STP. From equation (2.7) f quad f lin = cv 2 bv = γ D β v = (1 . 6 × 10 3 s m 2 ) Dv For the basketball diameter
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Phys105A_W09_HW2sols - Physics 105A Homework # 2 Solutions...

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