This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 120  Homework 4 1.RQ.65  It is reasonable to use the approximate (classical) formula for momentum in each of these situations except for (d). 1.RQ.67 (a)  =  f   i = m vx , 0, 0  m vx , 0, 0 = m 2vx , 0, 0 p p p
(b)     = mvx  mvx = 0 pf pi m = 0.4 kg ,  = 38, 0, 27 m/s v  = m = (0.4 kg) 38, 0, 27 m/s = 15.2, 0, 10.8 kg m/s p v   = p (15.2)2 + (0)2 + (10.8)2 kg m/s = 18.6 kg m/s 1.HW.95  1.HW.99    = 0.9999c, m = 1.67 1027 kg v =
1 1 = 1 = 70.7 1(0.9999)2  2 v
c   = m   = (70.7) (1.67 1027 kg) (0.9999) (3 108 m/s) = 3.54 1017 kg m/s p v 1.HW.104 (a)  CB =  C   B = 2.55, 0.97, 0 kg m/s  3.03, 2.83, 0 kg m/s = p p p 0.48, 1.86, 0 kg m/s  DC =  D   C = 2.24, 0.57, 0 kg m/s  2.55, 0.97, 0 kg m/s = p p p 0.31, 1.54, 0 kg m/s
1  ED =  E   D = 1.97, 1.93, 0 kg m/s  2.24, 0.57, 0 kg m/s = p p p 0.27, 1.36, 0 kg m/s  F E =  F   E = 1.68, 3.04, 0 kg m/s  1.97, 1.93, 0 kg m/s = p p p 0.29, 1.11, 0 kg m/s
(b) (c) The magnitude of the change in momentum is greatest between points B and C. 2 ...
View
Full Document
 Winter '08
 Labrake
 Mass, Momentum, Work, kg, m/s

Click to edit the document details