TIME OF COMPLETION_______________            NAME______SOLUTION_______________________

                                                     

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1111, Exam 2                                                                                                             Section 1

Version 1                                                                                                                    March 29, 2004

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on seven (7) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

                                                                                                                                          

3.         You will have 75 minutes to complete the examination.

 

4.         The total weight of the examination is 100 points.

 

5.         There are six (6) multiple choice and four (4) calculation problems. Work all problems.  Show all work; partial credit will be given for correct work shown.

 

6.         If you have any questions during the examination, see your instructor who will

be located in the classroom.

 

7.         Start:                12:00 p.m.

Stop:                1:15 p.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      15

 

 

 

8

 

                      20

 

 

 

                       9

 

                      15

 

 

 

                      10

 

                      20

 

 

 

                 TOTAL

 

                     100

 

 

 

 

 

           PERCENTAGE

 

 

 

 

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT.

 

1. The units of work are

 

a.       N-m.

 

b.      kg-m2/s2.

(5)

c.       J.

 

d.      All of the above.

 

 

 

 

 

2.  If the angle between the net force and the displacement of an object is greater than 90o,

 

e.       Kinetic energy of the object increases.

 

f.        Kinetic energy of the object decreases.

(5)

g.       Kinetic energy remains the same.

 

h.       The object stops.

 

 

 

 

3. Linear momentum is

a.       Always conserved.

 

b.      A scalar quantity.

 

c.       A vector quantity.

(5)

d.      Unrelated to force.

 

 

 

 

 

4.  Which of the following is NOT conserved in an inelastic collision:

 

a.       Momentum.

(5)

b.      Mass.

 

c.       Kinetic energy.

 

d.      Impulse.

 

 

5.      An object slides down a frictionless incline as shown below.

a.       Total energy of the object at point A is greater than that at point B.

(5)

b.      Total energy of the object at point A is lesser than that at point B.

 

c.       Total energy of the object at point A is equal to that at point B.

 

d.      Impossible to tell without knowing the mass of the object.

 

6.A woman runs up a flight of stairs. She gains 2000 J of gravitational potential energy. If she runs up the same stars with twice the speed, what will be her gain in potential energy?

 

a.       4000 J.

(5)

b.      2000 J.

 

c.       1000 J.

 

d.      500 J.

 

 

7. A 10.0-g bullet moving horizontally at 400 m/s penetrates a 3.00-kg wood block resting on a horizontal surface.

 

a. If the bullet slows down to 300 m/s after emerging from the block, what is the speed of the                block immediately after the bullet emerges?

 

V1i = 400 m/s

V2i = 0 m/s

 

V1f = 300 m/s

 

m1V1i + m2V2i = m1V1f  + m2V2f

 

(m1V1i + m2V2i - m1V1f ) / m2 = V2f

 

V2f = ((0.0100 kg)(400 m/s) + (3.00 kg)(0 m/s) - (0.0100 kg)(300 m/s)) / (3.00 kg) = 0.333 m/s

 

b. Is this an elastic or inelastic collision? EXPLAIN.   

 

KEi  = KE1i + KE2i = ½ m1 V1i2 +  ½ m2 V2i2 = ½ (0.0100 kg)(400 m/s)2 = 800 J

 

KEf = KE1f + KE2f = ½ (0.0100 kg)(300 m/s)2 + ½ (3.00 kg)(0.333 m/s)2 = 450 J

 

DKE = KEf  - KEi  = 450 J – 800 J = -350 J

 

Kinetic energy is lost hence the collision is inelastic.

 

 

 

8. A painter on a scaffold drops a 1.50-kg can of paint from a height of 6.00 m. Neglect air resistance.

 

    1. What is the kinetic energy of the can when the can is at a height of 4.00 m?

 

KEi + PEgi = KEf + PEgf

 

½ mVi2+ m g yi = ½ mVf2+ m g yf

 

Vi = 0 m/s

 

yi = 6.00 m

 

yf = 4.00 m

 

g m (yi - yf) = KEf  

 

KEf = 29.4 J

 

    1. With what speed does it hit the ground?

 

KEi + PEgi = KEf + PEgf

 

½ mVi2+ m g yi = ½ mVf2+ m g yf

 

Vi = 0 m/s

 

yi = 6.00 m

 

yf = 0 m

 

2 g (yi - yf) = Vf2

 

Vf = 10.8 m/s

 

    1. This painter turns out to be physically inclined, so he drops the can again, this time aiming at the spring that is positioned below the scaffold. The can falls onto the spring and compresses it by 0.500 m. What is the spring constant?

.

 KEi + PEgi + PEsi = KEf + PEgf + PEsf

 

½ mVi2+ m g yi + ½ kxi2= ½ mVf2+ m g yf + ½ kxf2

 

Vi = 0 m/s

 

Vf = 0 m/s

 

yi = 6.00 m

 

yf = 0 m (we assume that the spring is compressed to the ground)

 

xi = 0 m

 

xf = 0.500 m (we assume that the spring is compressed to the ground)

 

m g yi = ½ kxf2

 

k = 2 m g yi / xf2

 

k = 706 N/m

 

    1. How much work does the can do while compressing the spring?

 

W =  ½ kxf2

 

W = 88.3 J

 

 

9. The blades of the fan running at low speed turn at 250 rpm. When the fan is switched to high speed, the rotation rate increases to 350 rpm in 5.75 s.

 

  1. What is the magnitude of the angular acceleration of the blades? Assume constant angular acceleration.

 

w0 = (250 rev/min)(2p rad / 1 rev)(1 min / 60s) = 26.2 rad/s

w =  (350 rev/min)(2p rad / 1 rev)(1 min / 60s) = 36.7 rad/s

 

w = w0 + at

a = (w - w0)/ t

a = (36.7 rad/s - 26.2 rad/s)/ (5.75 s)

a = 1.82 rad/s2

 

  1. How many revolutions do the blades go through while the fan is accelerating?

 

q = q0 + w0t + ½ at2

q =  (26.2 rad/s)(5.75 s) + ½ (1.82 rad/s2)(5.75 s)2 = (181 rad )(1 rev / 2p rad ) = 28.8 rev

 

  1. What is the magnitude and the direction of the tangential acceleration of the tip of the blade that is 0.600 m from the center of rotation?

 

at = a R        

 

at = (1.82 rad/s2)( 0.600  m) = 1.09 m/ s2

 

Direction: tangential.        

 

 

10. A 5.00-kg block is held against a spring of spring constant 20.0 N/cm compressing it 3.00 cm. The block is released and the spring extends, pushing the block along a frictionless horizontal surface.

 

 

a. What is the speed of the block when the spring is at its equilibrium position?

 

KEi + PEsi = KEf + PEsf

 

½ mVi2 + ½ kxi2= ½ mVf2 + ½ kxf2

 

Vi = 0 m/s

 

xi = 0.0300 m

 

xf = 0 m

 

k = 2000 N/m

 

kxi2= mVf2

 

kxi2/m = Vf2

 

Vf = 0.600 m/s

 

b.      The block is not attached to the spring, so it continues to move to the right. At some point surface becomes rough. The coefficient of friction between the surface and the block is 0.200. How far will the block slide along the rough surface before coming to rest?

 

 

WNC  = ½ mVf2 - ½ mVi 2

 

WNC = ½ (5.00 kg) (0 m/s)2 – ½ (5.00 kg)(0.600 m/s)2

 

WNC = - 0.900 J

 

WNC  = - fk d = - (mkmg) d

 

d = WNC / - (mkmg)

 

d = 0.0918 m