Solutions to real  Test 2

 Student Perfomance


1. (25 points)

A 2.10 x 103 kg box starts at the bottom of a long driveway. The driveway makes an angle of 20-degrees with the horizontal. At the bottom of the incline, the speed of the box is 3.8 m/s. The coefficient of kinetic friction between the block and the surface is µ.
The block rises a vertical distance h = 0.400 m and a corresponding distance L along the incline before coming to rest.

Please use the conservation of energy methods of Ch. 5 in the presence of friction.

What is the coefficient of kinetic friction µ between the block and the incline?


Solution Outline: v1= 3.8 m/s  and  h = 0.400

K1 + U1 = K2 + U2 + Heat
0 + ½ mv12  = 0 +  mgh +µmgcos20*L
0 +  ½ mv12  =    mgh + µmgcos20*h/sin20

Note that L =h/sin20 since h = Lsin20

Solve for µ = 0.306


2. (30 points)

A ball (of mass m1 = 1. 00 kg) rolling across a table at velocity +1.20 m/s makes an elastic head on collision with an different second ball (of mass m2 = 1. 40 kg) that moves left with velocity - 1.00 m/s. Assume motion to the right means the velocity is positive! Thus, motion to the left means velocity is negative !

 

(a) (5 points) What is the direction of motion of the first ball of mass m1 after the collision, to the right or left? Show all work!!

(b) (5 points) What is the direction of motion of the second ball of mass m2 after the collision, to the right or left? Show all work!!

 

(c) (10 points) What is the numerical value of the velocity of the first ball that you calculated in part (a) ?

(d) (10 points) What is the numerical value of the velocity of the second ball that you calculated in part (b) ?


Solution outline:

Note that the masses do not cancel, so
(1)(1.2)   + (1.4) (-1.0) =  (1) v1f  +  (1.4)v2f  or
1.2 - 1.4 =  (1)v1f  + (1.4)v2f 
Thus,
-0.2 = (1)v1f + (1.4) v2f    
Also,  
v1i  -  v2i = - (v1f  - v2f ) or
1.2 - (-1) =  - (v1f  - v2f )  
Thus,
2.2 = - (v1f  - v2f ) 

Note that v1i = 1.2 m/s and v2i = -1.0 m/s
Solve the two equations in bold above  for the final velocity of each block.  Note if the velocity is negative, then the object moves left.   
Answers: v1f = -1.37 m/s and v2f =+0.83 m/s


3. (25 points) Three masses are situated on the rectangle as shown below. The dimensions are given. m1 = 3 kg and m2 = 4 kg.

(a) (13 points) Find the magnitude of the net force on m3 = 1 kg.

(b) (12 points) Find the direction of the net force on m3. Please draw the net force vector on the axis shown. Draw the vector in the correct quadrant! (Hint: The vector does NOT point in the first quadrant.) Please try to indicate the direction by showing the reference angle that the vector makes with the x-axis. Thank you.

(a) Click here to see problem 33 of Quiz 6 for the methods used to solve this test problem.

FNET = 1.6x10-9 N after plugging in the numbers.

(b)
 
 Tan (angle) = (F31/F32)

50o  =  tan-1(F31/F32)
Vector points in the third quadrant ! 

 4. ( 11 points) Extra Credit


A spherical ball with I =

starts from rest at point 1 at the top of the incline. The vertical distance of the ball is h = 1.0 m above the ground. The ball mass m = 1.0 kg . What is the linear speed v of the ball when it reaches the bottom at point 2 ?

Solution Outline:
Use conservation of energy,

K1 + U1 = K2 + U2
0 + mgh = (1/2)mv2 + KROT + 0
KROT  = ½ I (v2/r2) =  ½(2/5*mr2)(v2/r2)
 mgh = (1/2)mv2 + ½(2/5*mr2)(v2/r2)

Note that  the radius r  and the mass m cancel out. (What ?!) We have use the fact that angular speed = v/r  . Solve for the speed v.