1) A car traveling at 20 m/s rounds a curve so that its centripetal 
  acceleration is 5 m/s^2. What is the radius of the curve?
         a = v^2/r  ===> r=v^2/a = 80 m

2) A 1500 kg car travels at a constant speed of 22 m/s around a circular 
   track which has a radius of 85 m. What is the magnitude of acceleration 
   of the car?
        a = v^2/r = 5.7 m

3) A concrete block is pulled 7 m across a frictionless surface by means 
   of a rope. The tension in the rope is 40N and the net work done on 
   the block is 247 J. What angle does the rope make with the horizontal?
     W = F*cos(theta)*s ===> cos(theta)=W/(F*s) = cos(28 degrees)

4) Two balls of equal size are dropped from the same hight from the roof 
   of the building. One ball has twice the mass of the other. When the
   balls reach the ground, how do the kinetic energies of the two balls 
   compare?
    Energy is conserved, Potential energy is translated into kinetic energy.
    Potential energy E= m*g*h is twice as big for the heavier as for 
    the lighter ball. 
 ===>   The lighter one has one half as much kinetic energy as the other.

5)  A roller coaster car is moving at 20 m/s along a straight horizontal           track. What will its speed after climbing the 15 m hill?
    Initial energy (kinetic only) = mv^2/2
    Final energy (potential+kinetic) = mgh+mv_f^2/2
    ====> v_f = sqrt(v^2 - 2gh) = 10 m/s

6) A 100 kg cannon at rest contains a 10 kg cannon ball. When fired, 
   the cannon ball leaves the cannon with a speed of 90 m/s. What is 
   the recoil speed of the cannon?
      Momentum is conserved: Initial momentum = 0; 
      Final momentum = momentum of the ball+momentum of the cannon
      ===> v(cannon)= -m(ball)*v(ball)/m(cannon)= - 9 m/s

7) A stone of mass 2 kg falls 100 m near the surface of the earth.
   It strikes the ground without any rebound. Approximately how much 
   kinetic energy is transferred to earth in this process?
   For completely inelastic collision total energy is lost. The answer is zero.

8) A 3 kg ball and 1 kg ball are placed on opposite ends of massless beam.
   The system is in equilibirum. What is the ratio of the total lenght 
   of the beam to the length of the beam segment from the heavier ball 
   to the pivot?
     Equilibirium ==> torques are equal
     3 kg * L1 = 1 kg * L2   ===> (L2+L1)/L1= L2/L1 + 1 = 3+1 =4 

9) What is the angular spped of the second hand of watch?
    Angular speed = circumference/time 
     if you assume watch hand indicating seconds ==> 2*pi/60 = 0.1 rad/s
     if you assume a second hand i.e. the one indicating minutes
                                                ===> 2*pi/3600 rad/s

10) A 3 kg ball is placed a distance a away from a pivot of a massless
   beam. 1 kg ball is placed at the other end of the beam. The total length 
   the beam is b. The system is in equilibrium. Waht is the ratio of b/a ?
   Equilibrium==> torques are equal
   3 kg*g*a = 1 Kg*g*(b-a)   ===> 4 kg*a = 1Kg*b  ===> b/a = 4