Physics 103 Hour Exam #3 Solution

Point values are given for each problem. Within a problem, the points are not necessarily evenly divided between the parts. The total is 50 points.

1. [15 points]

(a) What does the metric preﬁx Giga mean?

109

(b) What is 100 µs in seconds?

(100 µs) 1 s = 10−4 s 106 µs

(c) The Tevatron is a particle accelerator at Fermilab, near Chicago. It accelerates protons up to energies of 1 TeV. What is this in nanoJoules?

(1 TeV) 1012 eV 1.6 × 10−19 J 109 nJ = 160 nJ TeV eV J

1 2. [10 points] Mars has two moons, Phobos and Deimos, both of which have nearly circular orbits. Phobos has mass 1.07 × 1016 kg; it orbits Mars once every 27,550 seconds following an orbit with radius 9.377 × 106 m. Deimos has mass 2.244 × 1015 kg; it orbits Mars once every 109,100 seconds following an orbit with radius 2.346 × 107 m. Use this information to calculate the mass of Mars. (Hint: there is much more information here than you need.) The solution here is written using the orbital radius and period of Phobos. You could do the same calculation using the orbital radius and period of Deimos, and you would get the same answer. The speed of Phobos in its orbit is 2ˇr v = T In order to keep Phobos in a circular orbit, the gravitational force must v2 equal m : p r m m v2 G p m = m r2 p r m m (2ˇr)2 G p m = m r2 p T 2r 4ˇ2r3 m = m GT 2 2ˇ(9.37 × 106 m)3 = 6.67×10−11 N m2 (27 550 s)2 kg m kg2 N s2 = 6.42 × 1023 kg

2 3. [10 points] Positrons are particles which are identical to electrons (for example, they have have the same mass as electrons), except that positrons are positively charged. Positrons were first discovered in a 1933 study of cosmic rays by Carl Anderson. He detected the positrons by using a cloud chamber in the presence of a magnetic field. Charged particles leave visible tracks as they travel through a cloud chamber. In the presence of a magnetic field, charged particles follow curved paths, so their tracks are curved. One of the first images ever taken of a positron is shown below. The positron track is the curved line in the left-center part of the figure. It is partly above and partly below the horizontal band across the middle of the figure.

This problem is continued on the next page....

Figure from Anderson, C. D., Physical Review, vol. 43, p. 491 (1933).

3 (a) How could Anderson tell that this particle was a positron rather than an electron? Give a brief answer. (Hint: another way to ask this question is: how would the track be di erent if it were due to an electron? Make a reasonable assumption about how the apparatus was set up.) If the particle were an electron, the force on it would have had the opposite sign (because of the q in F~ = q~v × B~ ), and hence the track would have curved in the opposite direction. In order for the positively charged positron to curve in the direction shown in the picture, the magnetic ﬁeld must have been pointed into the page.

Carl Anderson was awarded the Nobel Prize in 1936 for this discovery.

(b) The dark horizontal band across the center of the figure is a 6 mm thick lead plate. As the positron passed through the lead plate, it lost energy. Thus, the positron had more energy on one side of the plate (on whichever side it entered the chamber) and less energy on the other side. By considering the di erences between the curved tracks above and below the lead plate, determine whether the positron had more energy in the upper half of the chamber or the lower half of the chamber. In other words, did it start near the top and go down to the bottom, or vice versa? (Assume that the magnetic field is uniform throughout the apparatus.) A charged particle traveling perpendicular to a uniform magnetic field experiences circular motion. We can relate the speed of the particle to the radius of its circular path:

F = ma v 2 qvB = m r qBr = v m

Hence its kinetic energy is

1 1 qBr 2 q2B2r2 KE = mv 2 = m = 2 2 m 2m

When the particle has higher energy, it has a larger radius of curvature, r; that is, its path is “less curved.” In the ﬁgure on the previous page, the path in the lower half of the chamber is less curved, hence the positron had more energy in the lower half.

4 4. [15 points]

d=0.01m V=5000V

Consider a Millikan-type oil drop experiment. The drop has density 900 kg/m3 , radius 1.5×10−6 m, and mass 1.272×10−14 kg. The air in the oil-drop experiment has temperature 273 K, pressure 1.013 × 105 Pa, and viscosity = 1.83 × 10−5 Pa s.

(a) Initially the voltage is turned o , so there is no electric ﬁeld. What is the terminal velocity of the falling oil drop?

6ˇav = mg mg v = 6ˇa η 2 2 (1.272 × 10−14 kg) (9.8 m Pa m N s = s2 6ˇ(1.5 × 10−6 m)(1.83 × 10−5 Pa s) N kg m m = 2.41 × 10−4 s

(b) Now the voltage is turned on, with a voltage of 5000 V applied across a distance of d=0.01 m. The lower plate is at a higher voltage than the upper plate, so the electric ﬁeld points in an upward direction. The oil drop is seen to move upward. What is the smallest possible upward velocity that the oil drop could have? (Hint: what are the possible values of charge that the oil drop could have? Further hint: what is the fundamental thing that was learned from the Millikan oil drop experiment?)

The exam originally had an error in the figure above, giving the voltage as 500 V rather than 5000 V as specified in the problem. The figure has been corrected, and the solution written assuming 5000 V.

The solution is given on the next page

Use the next page if you need extra work space....

5 This blank page is extra work space for problem 4.

mg 6πavη

V 5 V J N m 5 N The electric ﬁeld is E = d = 5 × 10 m( V C ( J = 5 × 10 C . The particle is moving upward, so the Stokes force must be acting downward (since it acts in the direction which slows the motion). Gravity also acts downward. The electric force must be acting upward, and must equal the sum of the other forces.

6ˇav + mg = qE 6ˇav = qE − mg qE − mg v = 6ˇa

The smallest possible charge that any particle could have is q = e = 1.602 × 10−19 . Plug in this value along with the known values of E, m, g, a, and :

qE − mg v = 6ˇa

2 ( −19 � 5 N � −14 � m N s 2 1.602 × 10 C 5 × 10 C − 1.272 × 10 kg 9.8 s2 kg m Pa m = 6ˇ(1.5 × 10−6 m)(1.83 × 10−5 Pa s) N m = −8.62 × 10−5 s

That answer is no good, because we set the problem up in such a way that we’re supposed to get a positive velocity. The negative velocity here means that the particle is still moving downward. The electric force is not strong enough to overcome gravity. More charge is needed, so try the next-smallest possibility, q = 2e = 3.204 × 10−19: qE − mg v = 6ˇa

2 � −19 � 5 N � −14 � m N s 2 3.204 × 10 C 5 × 10 C − 1.272 × 10 kg 9.8 s2 kg m Pa m = 6ˇ(1.5 × 10−6 m)(1.83 × 10−5 Pa s) N m = 6.86 × 10−5 s

That answer is positive, so it is acceptable. Any larger charge (3e, 4e, etc.) would give a higher speed.