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1.6. MULTIPLE-CHOICE ANSWERS                                              17

                  answer this, you will need to invoke the fact that Q is inversely proportional to L. (This
                  is believable; doubling the length of the pipe will double the effect of friction between the
                  fluid and the walls, and thereby halve the flow rate.)

              1.7. 1-D collision
                  If a mass M moving with velocity V collides head-on elastically with a mass m that is
                  initially at rest, it can be shown (see Problem 6.3) that the final velocities are given by

                                        (M − m)V                  2MV
                                  V M =              and    v m =      .             (1.8)
                                         M + m                   M + m
                  Check the M = m, M ≪ m, and M ≫ m limits of these expressions.
              1.8. Atwood’s machine
                  Consider the Atwood’s machine in Fig. 1.6, consisting of three masses and two frictionless
                  pulleys. It can be shown that the acceleration of m 2 , with upward taken to be positive, is  m 1
                  given by (just accept this)

                                              4m 2 m 3 + m 1 (m 2 − 3m 3 )
                                       a 2 = −g                   .                  (1.9)            m 2  m 3
                                               4m 2 m 3 + m 1 (m 2 + m 3 )
                                                                                                 Figure 1.6
                  Find a 2 for the following special cases:

                   (a) m 1 = 2m 2 = 2m 3
                   (b) m 2 much larger than both m 1 and m 3
                   (c) m 2 much smaller than both m 1 and m 3
                   (d) m 1 ≫ m 2 = m 3
                   (e) m 1 = m 2 = m 3
              1.9. Dropped ball
                  In Section 1.1.4, we looked at limiting cases of the velocity, given in Eq. (1.3), of a beach
                  ball dropped from rest. Let’s now look at the height of the ball. If the ball is dropped from
                  rest at height h, and if the drag force from the air takes the form F d = −bv, then it can be
                  shown that the ball’s height as a function of time equals
                                               mg  (  m  (   −bt/m  )  )
                                      y(t) = h −   t −   1 − e      .               (1.10)
                                                b      b
                  Find an approximate expression for y(t) in the limit where t is very small (or more pre-
                  cisely, in the limit where bt/m ≪ 1).


             1.6 Multiple-choice answers

              1.1. e This is the only choice with dimensions of length.

              1.2. d This is the only choice with dimensions of time.

                                                       2
              1.3. c This is the only choice with units of kg m/s .
              1.4. b There are 60 · 60 = 3600 seconds in an hour, so one mile per hour equals

                                        mile  1609 meters
                                      1     =             = 0.447 m/s.              (1.11)
                                        hour  3600 seconds
                  A common automobile speed of, say, 60 mph is therefore about 27 m/s. The inverse rela-
                  tion, going from m/s to mph, is 1 m/s = 2.24 mph.
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