PHY101 Final Term Solved Papers

Your Name:

PHYSICS 101 FINAL EXAM

January 20, 2005 3 hours

Please Circle your section 9 am Nappi 10 am Bergli 11 am Hasan 12:30 pm McBride 12:30 pm Ziegler Problem Score /40 /16 /16 /14 /19 /16 /20 /10 Total /151

Instructions: When you are told to begin, check that this examination booklet contains all the numbered pages from 2 through 20. The exam contains 8 problems. Read each problem carefully. You must show your work. The grade you get depends on your solution even when you write down the correct answer. BOX your final

Answer: Do not panic or be discouraged if you cannot do every problem; there are both easy and hard parts in this exam. If a part of a problem depends on a previous answer you have not obtained, assume it and proceed. Keep moving and finish as much as you can!

Possibly useful constants and equations are on the last page, which you may want to tear offand keep handy.

Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the Honor Code during this examination.

Signature Problem 1

Miscellaneous Questions

1. (6 points) A 3.0 kg mass attached to a spring oscillates on a frictionless table with an amplitude A = 8.0·10−2 m. Its maximum acceleration is
2. 5 m/s2. What is the total energy of the system?

For a spring F = −kx = ma k = mamax A = 130 kg/s2

Therefore Epot = 1 2 kA2 = 0.42 J

Q2. (4 points) What is the speed of a satellite of mass m = 1.0 · 104 kg on a circular orbit around the moon at a distance of 4Rmoon from the moon’s surface?

The radius is actually r = 5Rmoon as the satellite is 4Rmoon above the moon surface.

FG = G · m · M r2 = m · v2 r = Fc

from there we get

v = r G M r = r G M 5Rmoon = 762 m s

3. A man of mass M = 75 kg lowers himself down from the top of a building by using a rope wound on a drum (a hollow cylinder of radius r = 0.50 m and mass 2M = 150 kg), as shown in the picture. The man and the drum start at rest.

a) (6 points) Find the angular acceleration of the drum.

The force for the man is F = Mg − T = Ma T = M (g − a) For the torque we can write τ = Tr = Iα with Ihollow cylinder = mr2 = (2M )r2 and a = αr therefore Iα = M (g − αr)r and thus

α = Mgr Mr2 + I = Mgr 3Mr2 = g 3r = 6.5 s−2

b) (4 points) What is the velocity of the man when he has dropped 20 m?

With v = at and s = 1 2 at2 we get

v = 2sa = 2sαr = 11.4 m s