Chapter 9:  Do listed problems--the first two problems deal with impulse: 7, 8

Also: 17, 18, 21, 22, 25 (classic !), 28 (impulse) , 31, 37 ( 2-D--see example 9.9), 39 ( 2-D see example 9.10), 43, 47, 50

Chapter 10: 25, 26, 27, 28, 42, 51*, 56*, 57*

Some problems deal with earlier chapters, i.e. 1-D motion, force, projectile motion, etc, which is why I assigned them i.e. to give you an integrated view. You should find many of them fun if you get past the time it takes to solve them. However, this quiz should not be to time consuming.
Simulations
Quiz  9
Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column.
error  log #1: See the corrected version of #27, Chapter 10.
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7 and 8 were done in discussion; were you there? If not ask a student who was !
17. Momentum in the x-direction is conserved.
Initial momentum = Final momentum
MBVBi + MIVIi = MBVBf + MIVIf   
i = initial and f = final.
B = bird and I = insect.
 VBi   and VIi are given. Assume bird's initial rightward direction is  positive. Thus, the initial insect velocity  is negative.  Note that VBf  = VIf   = Vf .   Find VIf   .
18.  Momentum in the x-direction is conserved.
Initial momentum = Final momentum
MVVVi + MCVCi = MVVVf + MCVCf   
i = initial and f = final.
V = volkswagon and C =cadillac.
 VVi   and VCi are given. Assume Volkswagon's initial rightward direction is  positive. Thus, the initial cadillac velocity  is negative.  Note that VVf  = VCf   = Vf   = 0.   Find VVi .
21. Momentum in the x-direction is conserved.
Initial momentum = Final momentum
MBVBi + MRVRi = MBVBf + MRVRf   
i = initial and f = final.
B = Bob and R =rock.
 VBi   and VRi are given. They are both zero. Assume Bob's final leftward direction is  negative. Thus, the final rock velocity  is positive.     Find VBf .
 22. Momentum in the x-direction is conserved.
Initial momentum = Final momentum
MDVDi + MSVSi = MDVDf + MSVSf   
i = initial and f = final.
D = Dan and S = skateboard.
 VDi   and VSi are given. They are the same. Assume Dan's and skateboard's  initial rightward direction is  positive.  Note that VSf   is given.    Find VDf   .
25. This classic problems reveals momentum's properties and how the external force acts to change it. The momentum's properties mimic the velocity's  during projectile motion. Review Chapter 4, and realize the 2 only differ by the mass factor. 

Just after launch, Px = M*Vx = M*V*cos30    ad Px = M*Vy = M*V*sin30   . Since the velocity x-component is constant, so is the momentum x-component, since it only differs by the mass factor.   The reason is this:
Since the gravitational force is vertically down, the x-component of force Fx = 0 = dPx/dt means the x-component is constant. 

At the top, the motion is horizontal, so the y-component vanishes (is zero) .  Remember the x-component is constant throughout the motion. 

Just before  impact, the x-component has the same value it had at  launch. The y-component = M*Vfy. Prove that Vfy =
 - V*sin 30. Consult chapter 4.  

Finally, only the y-component changes. Thus, the total change in momentum is M*Vfy   -     M*V*sin30 = 
 -     M*V*sin 30 -     M*V*sin 30 = -2* M*V*sin 30. Show that this expression equals Mg*(total time of flight). You can look up the formula for the total time of flight.  
28.
 Momentum in the x-direction is conserved.
Initial momentum = Final momentum

Use the same methods as the previous problems. With letters, label your masses iand your velocities to indicate the object  and    initial or final. Let R = racket and B = ball. i = initial and f = final.

Let the initial racket direction of motion be positive. Thus the initial ball velocity is negative; the given final ball velocity is positive. Find the final racket velocity. 
31. This problem is just like #7 and #8. This hint will help you if you missed out on the in-class-hint given  in discussion. 

First find the area under the curve of the shown triangle. This area equals the cart's change in momentum in the x-direction.

However, there's a trick here: The triangle  shown represents the force on the ramp ! The force on the cart has the same magnitude but opposite direction; thus you must take the negative of the area under the curve !

Once that is done, write the following equation:

Pfx - Pix = mVfx  -   mVix  =  negative of area under the curve. Now, Vix  = velocity just before the collision with the ramp and   Vfx = velocity just after the collision with the ramp.

To find  Vix , you must use the methods of Chapter 6, i.e.  problem 46 (b) , but without friction.  After substituting Vix  into the previous equation relating the momentum  change to  area,  find Vfx  .  Than use the method of Chapter 6, i.e.  problem 46 (a) without friction to find the distance the cart rebounds up the ramp.

37. 2-D--see example 9.9. Once you get the final velocity, both magnitude and direction, of the coupled players  just after the collision, use simple kinematics to relate the distance to the  time. The collision occurs at the center of the 50.0 m diameter rink.  Give the angle ( see example 9.9) for the traveled  distance.  
39. 2-D see example 9.10, almost a carbon copy. 
43. This problem is very close to 50. See hints below. The collision replicates example 9.9. Solve for the initial velocity of the rocket before the collision.   After the collision, this is a very simple projectile motion problem. Here are the key hints. Find  the time it takes the stuck rocket and package  to hit the ground from H = (1/2)*g*t2.  Using projectile motion principles,  find  the horizontal  distance traveled by the stuck rocket and package from the  product of the time and magnitude of the  horizontally directed speed  Vf  just after the collision.   
47. 
(a) Use chapter 2 kinematics to find the time from the two velocities and displacement traveled. You should also find the acceleration of the bullet inside the block since  average force on the bullet is its mass times its acceleration. 
(b) Use conservation of momentum. m*V1i  + M*V2i =  m*V1f    + M*V2f      .   
m*V1i  =   m*V1f    + M*V2f    .  Given one of the final velocities, find the other.   
50.  The collision replicates example 9.9. After the collision, this is a very simple projectile motion problem. Here are the key hints. Find  the time it takes the tangled vehicles to hit the ground from H = (1/2)*g*t2.  Using projectile motion principles,  find  the horizontal  distance traveled by the free falling tangled vehicles  from the  product of the time and the  horizontally directed speed  Vf   just after the collision.   
More hints posted later; stay tuned !
Chapter 10 
25. Use equations (10.43)
26. Use equations (10.43)
27. 
(a) Do NOT use equations (10.43). This collision is perfectly inelastic ! 
That means  m1v1i + m2v2i = (m1 + m2)Vf , where v2i   = 0 and v1i   =  vo .   Final the final common velocity Vf . 
(b) Compute ½mvo2  -  ½(m1 + m2)Vf2 , where Vf  is the final common velocity of both masses. Divide the above difference by ½mvo2  .  By the way, I basically did this problem in class. 
28. 
(a) Use equations (10.43)
(b)  See problem 27 (a) .  
42. In both  (a) and (b) you must use energy conservation  to find the speed  of the smaller mass  just before it collides with the block on level ground.  

(a) Find the  common  velocity of the stuck blocks just after the collision . See my class notes  or the collision of figure 9.13. 
(b) Using equations (10.43), find the velocity of the smaller mass just after the collision.  It will move leftward after  colliding with the larger one. Then use conservation of energy to compute how high it rises before coming momentarily to rest.
51. m*vB = (m + M)*Vf . 
(1/2)*(m + M)*Vf2 = (1/2)*k*d2. 
56. Using equations (10.43), find an expression for the  velocity V2f of the heavier  ball just after the collision in terms of vo and  other symbols.    The heavier ball   will swing upward in a circular arc after the   collision. Use conservation of energy to relate  the velocity V2f  of the heavier   ball just after the collision and the height h it rises before momentarily coming to rest. From class notes ,  h = L*( 1 - cos 50).  From these two expressions, solve for vo in terms of the other symbols. 
57. 

(a)  You will have to use the other set of equations besides equation 10.43. That is because both balls are moving before the collision . 
Use M1V1i + M2V2i = M1V1f + M2V2f   
and V1i  -  V2i   = V 2f  -  VIf   .

(b) See example 9.5
error  log #1: See the corrected version of #27, Chapter 10.
Check back for more hints .