MID TERM II

Part I. Fill the following blanks (2 points for each blank):

1. A 200-g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75 rpm. The tension in the string is 6.17 N.
2. In the figure as shown, the blocks are moving with constant speed, the coefficient of kinetic friction between 2-kg block and the table is 0.5. Assume that the pulley is light and frictionless.
   
3. A 5000-kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s speed is 20 m/s. The mass of water that has been collected in the car is 500 kg.
4. A 200-g puck slides on a 3.0-m-long frictionless 20? ramp. The minimum initial speed of the puck is 4.5 m/s that the puck can reach the top of the ramp.
5. You must stretch a spring with spring constant k = 1000 N/m for 63 cm to store 200-J energy.

Part II. Problems (5 points for each problem)
You must give the solution processes.

1. As shown in the figure, two blocks are released from rest. (a) What is the acceleration of the blocks? (b) How long does it take for the 100-kg block to reach the floor? Assume that the pulley is light and frictionless.
   
   

   Solution. Draw free-body diagrams for the two blocks as shown. (a) Use Newton's second for 90-kg block:      ΣFy  = T - m1g = m1a1;      Use Newton's second for 90-kg block      ΣFy  = T - m2g = m2a2;      Acceleration constraint is a2 = -a1.     Combining the three equations we have      a1 = (m2 - m1)g/(m1 + m2) = (100 kg - 90 kg)(9.8 m/s2)/(100 kg + 90 kg) = 0.52 m/s2.       a2 = -a1 = -0.52 m/s2.  (b) We find the time:      y = ? a2t2;      -1.0 m = ? (-0.52 m/s2)t2,     Solve t = 1.96 s.

2. A 25-g bullet traveling horizontally at 1200 m/s goes through a 350-kg stationary target and emerges with a speed of 900m/s. The target is free to slide on a smooth horizontal surface. (a) What is the impulse exerted on the target during the collision? (b) What is the target’s speed just after the bullet emerges?
   Solution.
   (a) Find the impulse:
        J_x = Δp_x = m₁(v_f1 - v_i1) = (0.025 kg)(9900 m/s - 1200 m/s) = -7.5 kgm/s
        The impulse on the target is 7.5 kgm/s.
   (b) Use conservation of momentum:
         m₁v_f1 +  m₂v_f2 =  m₁v_i1 +  m₂v_i2;
        (0.025 kg)(900 m/s) + (350 kg)v_f2 = (0.025 kg)(1200 m/s) + (350 kg)(0),
        Solve, v_f2 = 0.0214 m/s.
3. A 50-g ice cube slides down a 30? slope without friction. The ice cube is pressed against a spring of spring constant 25 N/m at the bottom of the slope, compressing the spring 10 cm. (a) What is the energy the ice cube transfers to the spring? (b) What is the distance that the ice cube will travel up the slope before reversing direction?
   Solution.
   (a) The energy transferred to the spring is
           U_s = ? kx² = ?(25 N/m)(0.10 m)² = 0.125 J.
   (b) Before  reversing direction, the energy of the spring converts to the gravitational potential energy. Use conservation of energy without external forces:
           E_f = E_i;
           mgh = ? kx²;
          (0.050 kg)(9.8 m/s²)h = 0.125 J, solve h = 0.255 m.
        Find the distance from
          d = h/sin θ = (0.255 m)/3in30? = 0.51 m = 51 cm.