(1) p  =  22×10-3×8.1 m/s  =  0.18 kg m/s
(3) F  =  4×104 m/s×1300 kg/s  =  5.2×107 N
(5) p  =  constant ® m1 v1  =  (m1 + m2) v2 or v2  =  m1 v1/(m1 + m2)  =  18 m/s×[12500/(12500 + 5750)]  =  12.3 m/s
(7) Conservation of momentum gives m v  =  (m + M)V, and conservation of energy (after the bullet embeds in the block) gives h  =  V2/(2 g). Thus h  =  m2 v2/[2 g (m + M)2]  =  0.0212×2102/[2×9.8×(0.021 + 1.4)2] m  =  0.49 m
(15) The change in momentum is Dp  =  pf - pi  =  0.145 kg×(52 - (-39)) m/s  =  13.2 kg m/s. The average force between the ball and bat is given by F  =  Dp/Dt  =  13.2/10-3 N  =  1.3×104 N
(17) The impulse is the change in momentum of the ball in the direction perpendicular to the wall. The perpendicular component of the velocity is given by v  =  25 m/s×cos(45°)  =  17.7 m/s. The perpendicular change in momentum is equal to twice this value times the mass, so the impulse given the wall is 2×17.7×0.060 kg m/s  =  2.1 N·s
(21) Conservation of momentum and kinetic energy yield text eq. 7-7: v1 - v2  =  v1¢ - v2¢ which will combine with the conservation of momentum equation m1 v1  =  m1 v1¢ + m2 v2¢. Since v2  =  0, and v1  =  3.7 m/s, we get from the first equation v1¢ - v2¢  =  -3.7 m/s. The momentum conservation equation can be rewritten as 3.7 m/s  =  v1¢ + m2 v2¢/m1. Combining the last two expressions yields 3.7 m/s  =  -3.7 m/s + v2¢(1 + m2/m1), so that v2¢  =  2×3.7 m/s/(1 + .22/.44)  =  4.93 m/s. Since v1¢  =  v2¢ - 3.7 m/s, we get finally v1¢  =  1.23 m/s.
(31) (a) Conservation of momentum gives m v  =  (m + M) V, so that the velocity of the block plus embedded bullet is V  =  m v/(m + M). The kinetic energy before impact is m v2/2, and after collision (m + M)V2/2. By combining the expressions above, the final kinetic energy can be expressed in terms of v as 0.5 v2/(m + M). Thus ratio of KEf to KEi is given by m/(m + M) after a little algebra. The quantity sought is the negative of DKE/KE  =  (KEf - KEi)/KEi  =  m/(m + M) - 1  =  - M/(m + M).
(b) For m  =  14 g and M  =  380 g, the last equation of (a) yields |DKE|/KE  =  380/(14 + 380)  =  0.964.
(46) The center of mass obeys the relation m1 r1  =  m2 r2 and additionally r1 + r2  =  r, where r is the separation distance between the atoms, and r1 is the distance of the c.m. from Carbon and r2 the distance of c.m. from Oxygen. Combining the two equations gives r1  =  r/(1 + m1/m2)  =  1.13×10-10m/(1 + 12/16)  =  6.46×10-11 m.
(49) The mass of the raft (at the center) and the two masses at NE and SW are along a line (diagonal) of the raft. They are equivalent to a single M1  =  2 m + M  =  (2×1200 + 6200)kg  =  8600 kg.
The c.m. will lie between the center and the SE corner, with the distance from the center being r1  =  (1.414×9)/(1+8600/1200)  =  1.56 m.

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On 13 Oct 1999, 12:45.