MIDTERM I

1. A stone is thrown from the top of a building upward at an angle of 30.0° to the horizontal and with an initial speed of 20.0m/s. If the height of the building is 45.0m, (a) how long is the stone "in flight"? (b) What is the speed of the stone just before it strikes the ground?

 Solution. (a) the initial x and y components of the velocity are

                        vx0 = v0cosq0 = (20.0m/s)(cos30.0°) = 17.3m/s

                        vy0 = v0sinq0 = (20.0m/s)(sin30.0°) = 10.0m/s

                     To find t, we can use y = vy0t - 1/2gt2, with y = -45.0m and vy0 = 10.0m/s (we have chosen the top of the building as the origin)

                        -45.0m = (10.0m/s)t - 1/2(9.80m/s2)t2, which give t = 4.22s.

                 (b) The y component of the velocity just before the stone strikes the ground can be obtained using the equation

                          vy = vy0 - gt with t= 4.22s:

                          vy = 10.0m/s - (9.80m/s2)(4.22s) = -31.4m/d

                       Since vx = vx0 = 17.3m/s, the required speed is 

                        v =(vx2 + vy2)1/2 = [(17.3m/s)2 + (-31.4m/s)2]1/2 = 35.9m/s.

  

2. An ice skater moving at 12m/s coasts to a halt in 95m on an ice surface. What is the coefficient of friction between ice and skates?

    Solution. The deceleration of the skater is due to the friction between ice and skaters. We find the deceleration from

                         v2 = v02 + 2a(x - x0)

                         0 = (12m/s)2 + 2a(95m - 0), which gives

                         a = -0.76m/s2.

                  We find friction force from free body diagram

                         x-component:    Ffr = ma = mk FN

                                 y-component:    FN = mg

                   From two equations we get

                         mk = |a|/g = (0.76m/s2)/(9.80m/s2) = 0.077

                       

3. A coin placed 30.0cm from the center of a rotating, horizontal turntable slips when its speed is 50.0 cm/s. (a) What provides the central force when the coin is stationary relative to the turntable? (b) What is the coefficient of static friction between coin and turntable?

 Solution. (a) The static friction between the coin and turntable provides the centripetal force.

               (b) We write SF = ma,   where a is centripetal acceleration, and F = Ffr. We have

                          aR = v2/r = (50.0 x 10-2m/s)2/(30.0 x 10-2m) = 0.83m/s2

                     We find friction force from

                          F = msFN = msmg = maR.    

                          ms = aR/g = (0.83m/s2)/(9.80m/s2) = 0.085

  

4. An astronaut weighs 140N on the Moon's surface. When he is in a circular orbit about the Moon at an altitude h = RM, what gravitational force does the Moon exert on him?

     Solution. The astronaut weighs 140N on the Moon's surface, the distance between the Moon and him is RM. When he is in the circular orbit  the distance between the astronaut and the Moon increases from RM to RM + h = 2RM. We find the gravitational force from

                              F1/F2 = (R2/R1)2

                     (140N)/F2 = (2RM/RM)2,  which gives F = 1/4(140N) = 35N (toward the Moon).