P11    

         mgh = (1/2)mV**2  so that V = (2gh)**(1/2) = 4.43 m/s

         mV = 50 kg x 4.43 m/s = 221 kg m/s

         Fav = Dp/Dt = 221 kgm/s /10 ms 22100 N

P27   

         Pi = Pf     0 = 50 kg x 10 m/s + 150 kg x Vb    

         Vb = -500 / 150 = 3.3 m/s (negative sign indicate the boat moves in opposite                                              direction of the person walking northward.)

P39   

         Conservation of momentum:

         10,000 kg x 20 = (90 x 10 + 10,000) V so that V = 18.35 m/s

P59   

         Let m = bullet mass, M = block mass, Vb = muzzle speed, V = block speed

         Conservation of momentum:    mVb = (m+M)V    ----------------- (1)

         Conservation of energy: (m+M)gh = (1/2) (m+M)V**2 ----------- (2)

         (1) leads to V = [m/(m+M)] Vb   

         (2) leads to V = (2gh)**(1/2) so that Vb = [(m+M)/m] (2gh)**(1/2)