Answers to homework problems:

Chapter 6 part 1.
 8) momentum p = m * v = m * d/t = 65.5 kg * 200 m / 19,8 s = 661.62 kg*m/s
10) momentum = mass * velocity
    momentum of the truck: p = 2 tons * 30 km/h =
                               2000 kg * 30000/3600 (m/s) = 16666.67 kg*m/s
    velocity of a car = momentum/mass = 16666.67 kg*m/s /1000kg =
                      = 16.67 m/s = 60 km/h
27) momentum = massv * velocity and is conserved for a system
    initially the pair has a total momentum = 0 therefore after the
    separation the "father's momentum" = "daughter's momentum"
                    70 kg * 0.5 m/s = 40 kg * v
              ----> v = 0.88 m/s westward
35) The distance from the gun 
    d = distance to maximum of trajectory + distance from maximum of trajectory
      = d1 + d2
    The time of going up to a maximum trajectory = time of descent (does not
    depend on the mass)
    We can get the time from the equation for the vertical component of
    velocity since at the top of the trajectory this component is zero.
    v_y = 0 = v * sin(theta) - g * t ---> t =v * sin(theta)/g
    ----> d1 = v * cos(theta) * t = v^2 * sin(theta) * cos(theta)/g

    The explosion at the top of trajectory separates the projectile into two
    equal masses. The momentum must be conserved. Before the explosion each
    half of the projectile had horizontal momentum:  p = m/2 * v_x. 
    After the explosion, the horizontal momentum of one of the pieces is zero. 
    Therefore, in order to conserve momentum, the other piece has to have its
    momentum twice as large and for the same mass it means that it has to have
    twice as large horizontal velocity.
    ----> d2 = 2 * (v * cos(theta)) * t = 2 * v^2 * sin(theta) * cos(theta)/g

    Total distance d = 3 * v^2 * sin(theta) * cos(theta)/g =
                     = 3 * (90 km/h)^2 * sin(60) * cos(60)/ 9.8m/s^2 =
                     = 82.84 m (convert km to m and h to s)