QUIZ 10 - CH 8 - MOMENTUM AND ENERGY ( (SAMPLE TEST 3, FOR ENERGY)  AND ( SAMPLE TEST 4 SPRING '11, FOR MOMENTUM ) (ANSWERS) 
 8, 13, 22, 24, 28 (TRY 95), 31, 36, 41,  43 (TRY 82 ) , 46, 48, 49,
51, 54, 55 (TRY 51  AND  54  ), 57( TRY 107),  82 (TRY 43), 86.
 DISCUSSIONS POSTED BELOW:
8. (a) IMPULSE = CHANGE IN MOMENTUM.   See Example 8.2 and also note this problem is related to the last lab-- A VERNIER LAB WILL HAVE A MINI LAB ON THIS SUBJECT.  
Note: m*vf - m*vi = change in momentum where  vi and vf represent velocity components along the x - axis.  Let rightward be positive. For example the pitched ball initially  moves to the right, so vi is positive and equals +45.0 m/s.  After impact with the bat, the ball   moves left in the negative direction; thus vf < 0 and equals - 55.0 m/s . 
(b) Clearly the x-component of force  Fx is negative. We have m*vf - m*vi   = Fx *change in time. Note Fx is the average force. Normally,  Impulse =
 m*vf - m*vi   =  area under the force curve,  obtained via integration. But it can be shown that the area  under the curve = average force *(change in time) from the MEAN VALUE THEOREM FOR INTEGRALS (Review Math 1) .   
13.

 (a) IMPULSE = CHANGE IN MOMENTUM. The problem seems to have been  plucked from EXP 20 VERNIER where you had to analyze the shape of the force curve. In the experiment, clearly the shape was not a perfect rectangle like this idealization , but the main concepts apply: The area under the curve represents  the change in momentum resulting from the force. | m*vf - m*vi  |  =  area under the force curve. Note we have given you the magnitude of the change in momentum or impulse since the force curve plots the force magnitude. 

(b) Assume the rightward direction is positive. You are given vi = +5.00 m/s. To get the exact signed component of impulse  you would write:
 m*vf  - m*vi   =  area under the force curve.  

(i)    If the force acts rightward  m*vf  - m*vi   =  area under the force curve = positive area of rectangle, in which case the final velocity component will be greater than the initial velocity component.  
(ii)   If the force acts leftward  m*vf  - m*vi   =  area under the force curve = negative area of rectangle, in which case the final velocity component will be less than the initial velocity component ; indeed vf  could be negative depending on the magnitude of the negative area  but that is up to you to investigate
22.   m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f  .   Let right be positive; thus v1i = + 1.50 m/s and v2i  = -1.10 m/s. Also you are given that v1f = + 0.250 m/s. 
(a) Find  velocity component of the second  (2)  mass. Hint After collision the second mass is moving to the right. 
(b) There are two ways to explore this. One way is to find out if  the collision is elastic or not and  possibly  bypass the kinetic energy computation.  See equation 8.27  and discussions thereof  and  simply find out if the  magnitudes of the relative velocities are equal before and  after the collision. In other words , check if  : v1i  - v2i   = v2f  -  v1f  . 
Note we used this concept in EXP 19 Vernier for the magnetic bumper to magnetic bumper collisions.  If the equality is true, then the collision is elastic and the change  in the combined kinetic energy will be zero. If the equality is false, then you must compute the change in combined kinetic energy in the  following way: 
Change in combined kinetic energy = (1/2)*m1*v1f 2 + (1/2)*m2*v2f 2  - (1/2)* m1*v1i 2 - (1/2)*m2*v2i 2 .
24. (a) This part is essentially a replica of example 8.4. mA*vAi + mB*vBi   =  mA*vAf  + mB*vBf  , where  
 mA*vAi + mB*vBi   = 0, since the closed system starts from rest.  Thus,    mA*vAf  + mB*vBf  = 0  and you can now find the final velocity of A given  B's final  velocity.

(b) Spring potential energy = (1/2)*k*x'2 = combined kinetic energy  after the the spring has expanded = 
 (1/2)*mA*vAf 2 + (1/2)*mB*vBf 2   . From this you can find the spring potential energy. Note if I gave you  the spring's force constant k, you could find the  distance |x'| it was compressed.
28.  This is a subtle variation of #24--- Here the system ( YOU AND THE ROCK) is also at rest initially but the rock has a final velocity that is not collinear with the final velocity of You as you slide in a direction opposite the horizontal component  of rock's final velocity.   
m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f  .  which translates into:
0 = m1*v1f  + m2*v2f  ,  where the written velocities are are x-components. Let 1 = You and 2 = rock,
Clearly v2f = 12*cos 35. Find v1f   .  We are assuming rightward is positive, so you will recoil to the left in the negative direction.

Note we have assumed for this system, rock and You,  the  net external force in the  x direction is zero, resulting in conservation of momentum in the x-direction. In other words, in the x direction, the only relevatn forces acting on  system members You and  rock are action reaction pairs , which sum to zero. 

In the  y direction , momentum not  conserved for the system of You and rock since the rock has an upward y directed velocity of 12*sin 35 and You have NO  y-component of velocity. This will be discussed in greater detail later, but the bottom line is this: The rock transfers its y-component of momentum to the Earth. So momentum is conserved in the y direction if you expand the system  to include the Earth.     In other words,  it can be easily be shown  m2*12*sin 35 = momentum transferred to Earth.  Here is how:

APPENDIX : DERIVATION OF THE PREVIOUS STATEMENT:  "THE EARTH RECEIVES MOMENTUM m2*12*sin 35."

ISOLATE THE ROCK : m2*a = pos - neg = FRH - m2g, where FRH is the magnitude of the  force on the rock from the hand.  We know  m2*a = change in momentum in y-direction/time , where time is the time period your hand exerts a force on the  rock.   But we also know change in momentum in y direction =  m2*12*sin 35. THUS:
m2*a = pos - neg = FRH - m2g or   FRH   = m2g  + m2*a  = m2g  + m2*12*sin 35/time. Note  FRH   is the normal force of the hand ( YOUR HAND) on the rock. 

ISOLATE YOU ( AND YOUR HAND) UNDER THE ROCK : Let's look at the situation before throw  the rock.    Since your  vertical acceleration a is zero we have:
sum of the force in y direction on you = pos - neg = 0 =  N - m1*g  - m2*g since both the rock's weight and your own weight act on you vertically down.  Note N is the magnitude of the normal force on you from the Earth. Thus,  N =
m1g  +  m2*g before you throw the rock.
While you throw the rock, the following forces are acting on you: Upward, the normal force of magnitude N,  and downward, your weight m1g + FHR.  Note:  FHR = FRH = m2*g  + m2*12*sin 35/time
 Since N -   m1*g - FHR  = 0,  N =   m1g +  FHR   = m1*g +  m2*g  + m2*12*sin 35/time. The the normal force has increased by the term  m2*12*sin 35/time. Finally throughout all this discussion, the above normal force of magnitude N is equal and opposite to the force on Earth of magnitude N' due to you in contact with it. Thus N' = N quoted above.

We have the following data to compare with regard to the normal force of magnitude N' on Earth exerted by you:

Before you throw  rock: N' = m1*g  +  m2*g = m1*g  +  FHR  since FHR  = m2*g before throwing.
While you  throw rock:  N' = m1*g  +  FHR  = m1*g +  m2*g  + m2*12*sin 35/time. Now if you consider these forces acting down on the Earth for the time period referred to as "time", the time  the rock takes to gain its momentum
 m2*12*sin 35,  then  we would rewrite the above equation for the total  momentum  transferred to Earth during this interval:
TOTAL MOMENTUM = FORCE *TIME = N'*time =  m1*g*time +  m2*g*time  + m2*12*sin 35. The  last term is the magnitude of  downward momentum transferred to  Earth by  rock  you throw up with vertical momentum
 m2*12*sin 35. QED.

THE proof could repeated  in the case of "tryout" # 87, discussed below. But in future problems a proof will normally not be required.

31.  See example 8.12. In the x direction momentum is conserved:
mA*(40.0 m/s) = mA*VA cos 30 + mB*VBcos 45
 In the y direction momentum is conserved:
0 = mA*VA sin 30 - mB*VBsin 45.

Solve these two equations simultaneously for VA and VB.  Then answer the other questions.
36. CLASS NOTES.
41.  See example 8.9. After they collide, the momentum is (m + M)*Vf* at an angle of 24 degrees with vertical or 66 degrees with the horizontal .  We have (m + M)*Vf cos 66 = m*Vc  and (m + M)*Vf sin 66 = M*VT  where m is the mass of the compact, M is the mass of the Truck and Vf = 16.0 m/s. Find Vc  andVT.  
43. CLASS NOTES.
46.   m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f   and v1i  - v2i =  v2f   -  v1f .   Solve the two equations simultaneously for the two final velocities. Let m1 = 0.150 kg and m2 = 0.300 kg,  where v1i   = +0.80 m/s  and    v2i   = -2.20 m/s. 
48.    m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f   and v1i  - v2i =  v2f   -  v1f .   Solve the two equations simultaneously for the two final velocities. Then answer the other questions. Let m1 = 30.0 g  and m2 = 10.0 g,  where v1i   = +0.200 m/s  and    v2i   = -0.400 m/s. 
49.   m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f   and v1i  - v2i =  v2f   -  v1f .   Solve the two equations simultaneously for the two final velocities.  Let m1 = 1.0 u  and m2 = 2.0 u,  where v1i   = v  and    v2i   = 0.  Your  answers for the 2 velocities will be in terms of v. Then answer the other questions.
51. CLASS NOTES.
54. CLASS NOTES.
55. CLASS NOTES
57.  See example 8.14. Since the closed system starts from rest,  0 =  m1*v1i + m2*v2i   =  m1*v1f  + m2*v2f  .  Thus,      m1*v1f  + m2*v2f    = 0.  What is v1f  if v2f   = -0.70 m/s ?