TEST 3 FALL '11 SOLUTIONS BELOW; SCROLL DOWN WITHIN EACH PROBLEM TO SEE THEM.  
1.  (80 POINTS)  A 85.0-kg speed skier has finished a long down hill race and reaches a final  slope (fig. 1 below) designed to slow  her down. At the bottom of this slope her speed is 29.0 m/s.  She slides up the inclined plane of snow on  her skis and at a certain vertical height h has speed 1.95 m/s. The force of friction between her skis and the snow does  work of magnitude 3995.0 J .  (Ignore air friction.) 
(a) (44  point)  What is the vertical height h? For full credit you must use the work – energy theorem or conservation of energy. 

(b) (8 points)    What is the  work W
g done by gravity during her trip up the inclined plane? Is this work positive or negative ? Give a brief explanation of the correct sign you choose (positive or negative.)  

(c)  (4  points)  What is the work done by the normal force ? Explain your answer. 

(d)  (8 points)  Use basic trigonometry to find the distance L moved  along the incline.  See fig. 2 for reference.
(e)  (16 points) Assume the force of friction between the skis and snow is constant with  magnitude f. What is  f? 

SOLUTIONS: (a) (1/2)*mvi2 + mgyi = (1/2)*m*vf2 + mgh + 3995.0 J, where the initial y value is zero, the initial speed is 29.0 m/s and the final  speed 1.95 m/s. This gives h = 37.9 m.
(b) Wg = -mgh = = -31600 J. This work is negative since the the gravitational force acts downward and the vertical component of  displacement points up in the opposite direction.  
(c) ZERO since the normal force and motion are perpendicular
(d)  L  = h/sin 30 = 75.8 (m) 
(d) -3995.0 J = -f*L, where L = 78.5 (m); thus f = 52.7(N). 

2. (80 POINTS)  On a frictionless  air track, a 0.5000 kg-glider  moving to the right at 2.950 m/s collides with a 0.4000 kg glider moving to the left at 1.950 m/s.  The collision is elastic. Assume the right  direction is the positive direction of motion .

(a) (30 points) What is the velocity  of the  0.5000 kg-glider after the collision?

(b) (30  Points) What is the velocity  of the  0.4000 kg-glider after the collision?

 

(c) (8 points) Using the velocities of parts (a) and (b), compute the total kinetic energy of the system after the collision.  Do not round off during intermediate computations.


(d) (8 points) Using the initial velocities before the collision,  compute the total kinetic energy of the system before the collision.  Do not round off during intermediate computations.

 


(e) (4 points) Are the answer to parts (c)  and (d) equal? Explain.  
 

SOLUTIONS:
(a) (0.500 kg)*(2.950 m/s) + (0.400 kg)*(-1.950 m/s) = (0.500 kg)*V + (0.400 kg)*V ' and 4.9 m/s = V' - V. 
V = -1.406 m/s and 
(b) V' = + 3.499 m/s. 
(c) (1/2)*(0.500 kg)*(-1.406 m/s)2 + (1/2)*(0.400 kg)*(-1.950 m/s)2 = 2.9360 J 
(d) (1/2)*(0.500 kg)*(2.950 m/s)2 + (1/2)*(0.400 kg)*(-1.950 m/s)2 = 2.9360 J 
(e) Answers are the same because collision is elastic . 

3. EXTRA CREDIT.  (20 POINTS)   

An airplane propeller is rotating at a steady 1900 RPM (revolutions per minute.)  

(a) (4  points) Compute the angular velocity ω in rad/s. (Hint:  How many radians are there in one revolution? )

(b) (6 points)  How many seconds does it take the propeller to turn through 35 degrees? Remember to convert to radians .


(c) (6 points)  What is the period of this propeller?  Note: The period is the time for one revolution.

(d) (4 points)  If the propeller were  turning at 18 rad/s, at how many
RPM would it be turning?

SOLUTIONS: 
(a) (1900)*(6.28)*(1/60) RAD/S = 199 RAD/S

(b) THETA = 35*3.14/180 RAD = 0.610 RAD = 199*t; thus t = 0.003 seconds.   
(c) period  = 1/f = 6.28/w = 6.28/198.97 seconds = 0.0316 seconds.  
(d)  18*(1/6.28)*( 60)  RPM = 172 RPM. 


 

 

 

 

 

 

 

 

 

 

 

 

 




 

 



 

 

 

 

 

 

 

 

 

 




 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 (  points)

(a)  (  points)   
(b)  (  points)  

(c)  (  points)  

(d)  (  points) 

(e)  (  points)

(f)  (  points)