Complete F09 Midterm 2 answers for archive PDF

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Physics 150/140 Second Mid-Term Exam November 10, 2009

Check your section: MWF 10 MWF 11 MWF 12 TR 12

Name:

Prof. Lipeles Prof. Hollebeek Prof. Fortune Prof. Thomson

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Signature: _____________________________________ Penn ID: _____________________________________

Instructions: 1.This is a closed book exam. 2. A formula sheet is attached. You may tear it off and keep it. 3. Each problem is worth 15 points. 4. For full credit, show all your work. 5. Numeric results should be written with three significant digits. 6. Draw a box around your final answer. 7. The exam will end at 5:50 pm.

1 2 3 4 Total

1. Two masses slide on a frictionless surface. Initially, m1=2.00 kg is moving to the right with speed 10.0 m/s and m2=5.00 kg is also moving to the right but with speed 3.00 m/s. a) Find the velocity of the center of mass.

If you differentiate the formula for the position of the center of mass, you will get the formula for the velocity of the center of mass.

vcm 

m1 v1  m2 v2  5.00m / s m1  m2

b) Find the initial kinetic energy.

K

1 1 m1v12  m 2 v 22  122.5 J 2 2

c) A massless spring (k=1120 N/m) is attached to m2 on the left as shown in the figure. Find the total kinetic energy when the spring is maximally compressed.

The maximum compression occurs when both objects are moving at the same velocity. Otherwise they are either getting closer of farther apart and thus the compression is changing. For the two objects to be moving at the same velocity, momentum conservation would tell you that that velocity must be the velocity of the center of mass.

v1  v 2  vcm 1 2 K  (m 1  m 2 )v cm  87.5J 2

d) What is the maximum compression of the spring? The energy of the spring plus the kinetic energy from part c must equal the total energy (part b)

Ei  E f 1 1 (m  m2 ) 2  kx 2 2 1 2 1 122.5  87.5  kx2 2 E  K U 

x  35.0 * 2  0.250meters 1120

2. An object of mass m=3.00 kg starts from rest at the top of a frictionless curved surface of height h1 =6.00m. It slides down the curved surface, then slides up a rough slope to a height h2 =4.00 m. At the top of the rough slope it encounters another frictionless surface and a spring with k=400. N/m. The rough slope is inclined at 30.0 degrees above the horizontal and has the coefficients of friction shown in the figure.

a) What is the kinetic energy of the object when it is at the bottom of the curved surface for the first time? Equate the energy at the top to the energy at the bottom

Etop  Ebottom 1 mgh1  mv2  K bottom 2 K  mgh1  (3.00)(9.81)(6.00)  176.58  177 J Remember to use the more accurate number (176.58) in subsequent calculations. b) What is the maximum compression of the spring? First find out how much energy the object has left after travelling up to the spring by calculating the energy loss on the slope. The free body diagram on the slope is shown to the right from which we have

N  mg cos( ) f k   k N   k mg cos( ) The energy lost to friction is fk D where the distance travelled up the slope is h2 /sin(θ)

E 

k mg cos( )h2 (0.250)(3.00)(9.81)(.866)(4.00)   50.97  51.0 J sin( ) .500

Remember to use the more accurate number (50.97) in subsequent calculations. Now equate the energy remaining at the top to the object’s energy

Ei  Elost  Etop  mgh2 

1 2 kx 2

176.58  50.97  (3.00)(9.81)(4.00) 

1 ( 400)x2 2

x  0.1986  0.199meters c) Where does the object eventually come to rest forever? If it ever stops on the slope, then it will stay there since fs max =µ smgcos(θ) is greater than the net force down the slope mgsin(θ) . [ 20.4 > 14.7] When the object reaches the bottom for the second time, the remaining energy is 176.58 – 2(ΔElost )=74.64J You might think that it could make it up the slope again since ΔE is only 50.97J but this does not account for the gain in potential energy required to get to the top which is almost 120J! So imagine that it stops at a height h. The energy lost to friction would be (h/4)*50.97 and the potential energy would be mgh so

E  E  mgh h 74.64  (50.97)  (3)(9.81) h 4.00 74.64  1.7699  1.77meters h 3(9.81)  12.74...