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					Extra exercises to lectures 3
           and 4
Exercise 7.20

A small particle of mass m is pulled to the top of a frictionless half-cylinder (of
radius R ) by a cord that passes over the top of the cylinder, as indicated in the
figure.
a- if the particle moves at constant speed, show that F = mgcos
b- by directly integrating W   F.dr      find W when the particle is moving from
the bottom to the top. Can you explain your results?



                                               F
  Solution 7.20                                                      x F
                                                                                n
a)- The radius to the object makes angle  with
the horizontal, so its weight makes the same                                
                                                                     R
angle with the negative side of the x-axis, when
we take the x-axis in the direction of motion                        
tangent to the cylinder.                                                   mg


   F   x    max  0  F  mg cos  0  F  mg cos

b)- We use radian measure to express the next bit of displacement as Rd
                f          f      f          f

            W   F .d   Fd   FRd  R  Fd
                i          i      i          i
                     /2
               R     mg cos d  mgR[sin( / 2)  sin(0)]  mgR
                     0
 This is the same if we calculate the difference in potential energy between the
 bottom and the top, and the result is independent of the pass, thus along the
 quarter of the circle or along the blue pass we get the same results. The weight
 is a conservative force.
Exercise 7.32

A 2.00kg block is attached to a spring of force constant 500N/m, along the x-
   direction. The block is pulled xm= 5cm to the right of equilibrium and released
   from rest. Find the speed of the block as it passes through equilibrium if:
a- the horizontal surface is frictionless,
b- the coefficient of friction between the block and surface is 0.350
                                                                    xm




                                                                0

                                                           Equilibrium
Solution 7.32
a)- The conservation of total mechanical energy in absence of friction gives


   K  U  0 
                              1 2 1 2 1 2 1 2
   Ki  U i  K f  U f        mvi  kxi  mv f  kx f
                              2       2     2      2
                                          1      1 2          k
   xi  xm ; x f  xequil    0; vi  0;  mv 2  kxm  v f 
                                              f                 xm =0.791
                                          2      2            m    m/s
 b)- The presence of friction destroy the conservation of energy

    K  U   f .x ;         f  k N  k mg 
                                   1 2 1 2 1 2 1 2
    K i  U i  K f  U f  f .x  mvi  kxi  mv f  kx f  f .x
                                    2        2        2      2
                                           1 2 1 2
    xi  xm ; x f  xequil  0; vi  0;  mv f  kxm  k mgxm
                                           2         2
          k 2                        k 2
    v 2  xm  2 k gxm  v f 
      f                                 xm  2 k gxm  0.531m / s
          m                          m
Exercise 8.23
A force acting on a particle moving in the xy plane is given by

             F  (2 yx  x 2 y ) N ;
                     ˆ       ˆ                    x, y in meters
The particle moves from the origin to a final
position having coordinates x = 5.00m and                     y
y = 5.00m. Calculate the work done by F
along the three paths indicated. Explain
your results.                                             B           C


Solution 8.23

a)- along OAC                                                             x

         5                        5
                                                          O       A
WOA   dxx.(2 yx  x 2 y )   2 y y 0 dx  0
          ˆ     ˆ       ˆ
         0                        0
         5                        5
WAC   dyy.(2 yx  x y )   x 2
          ˆ     ˆ     ˆ   2
                                              dy 125 J
         0                        0    x 5

WOAC  WOA  WAC  0  125 J  125 J
b)- along OBC
                       5                        5
               WOB   dyy.(2 yx  x y )   x 2
                         ˆ     ˆ     ˆ 2
                                                           dy 0
                       0                        0   x 0
                       5                        5
               WBC   dxx.(2 yx  x y )   2 y y 5 dx 50 J
                         ˆ     ˆ     ˆ 2
                       0                        0

               WOBC  WOB  WBC  0  50 J  50 J

c)- along OC
                   5                                5
         WOC   (dxx  dyy ).(2 yx  x y )   (2 ydx  x 2 dy )
                    ˆ     ˆ       ˆ     ˆ   2
                   0                                0
                                                           5
         along OC, x  y  dy  dx  WOC   (2 xdx  x 2 dx)
                                                           0

          WOC  66.7 J


d)- the force F is a non-conservative force since the work is path dependent.
Exercise 8.33
   A 5.00kg block is set into motion up inclined plane with an initial speed of 8.00
   m/s. The block comes to rest after traveling 3.00m along the plane, which is
   inclined at an angle of 30o to the horizontal. For this motion determine:
1. The change in the block’s kinetic energy,
2. The change in the potential energy of the block-Earth system,
3. The friction force exerted on the block ( assumed to be constant)
4. What is the coefficient of kinetic friction?



                                           8m/s
                                                        3m


                                                  30o
Solution 8.33
                                                   1                  1
a) The change in kinetic energy is        K        m(v 2  vi2 )   mvi2  160 J
                                                         f
                                                   2                  2

b) The change in potential energy is          U  mg (3m) sin 30o  73.5 J


c) The mechanical energy converted due to friction is K+U= - f x= 86.5 J
Hence f =86.5/3 = 28.8N
d) f= k n =k mgcos30o = 28.8 J
Hence, the coefficient of friction is k =0.679N
Exercise 8.48

A block slides down a curved frictionless track and then up an inclined plane as
in the figure. The coefficient of kinetic friction between the block and incline is
k = 0.325. Use energy methods to show that the maximum height reached by
the block is
                                       h
                         ymax                  5.1m
                                  1  k cot 




                                                                                h = 8m

              ymax
                                        30o
Solution 8.48

The potential energy of the block-Earth system is mgh


 The amount of energy         k mgd cos         is converted into internal energy
 due to friction on the incline. Therefore the final height ymax     is found from
          K  U   fd  0  U f  U i   fd
                                                        ymax
           mgymax  mgh   k mgd cos           ; d
                                                       sin 

Solving                                 h
                          ymax   
                                   1  k cot 

Using the data we get the answer given
Exercise 8.70

A ball is tied to one end of a string. The other end of the string is held fixed. The
ball is set moving around a vertical circle without friction, and with speed v  Rg
at the top of the circle, as it is shown in the figure. At what angle  should the string
be cut so that the ball will then travel through the center of the circle?

Solution 8.70                                                    vo            v  Rg
 At the top position, if the tension is zero                    
 then
                   v2                                           R     
             mg  m  v  Rg                                            C
                   R                                                  
 At any angle , the conservation of energy gives
 us the speed vo . Take the origin of potential
 energy at the bottom position , ( =- )
       Ki  U i  K f  U f
 1 2               1 2
   mvi  mg  2 R  vo  mg ( R  R cos  )  vo  (3  2cos  ) gR
 2                 2
What we need is now to solve the projectile problem with initial velocity
components vxi  vo cos  ; v yi  vo sin 
The requirement is that the trajectory ( parabola ) have to pass through the center
C. the origin is taken the position of the point when we cut the string
                          1     g
                      y              x 2  tan  x  yo
                          2 vo cos 2 
                             2                              0



The center has coordinates in the x,y axis with origin at the position of the mass
with speed vo :
                                                               y
           xC  R sin  ;            yC   R cos                                  x
                                                                 O
Substituting and using     sin 2   cos 2   1                         R   
                                                                                 C
Simplifying, to get
                                           6  36  12
        3cos   6 cos   1  0  cos  
              2

                                               6
The only physical solution is with the negative sign . Thus

 cos   0.1835    79.43o    100.6o
Exercise 8.72

A pendulum, comprising a string of length L and a small sphere, swings in the
vertical plane. The string hits a peg located a distance d below the point of
suspension.
a- show that if the sphere is released from a height below that of the peg, it will
return to this height after striking the peg.
b- show that if the pendulum is released from the horizontal position (  = 90o )
and is to swing in a complete circle centered on the peg, then the minimum
value of d must be 3L/5.




                                                                d
                                               L

                                                       peg
Solution 8.72

a) Energy is conserved in the swing of the pendulum, and the stationary peg
   does no work. So the ball’s speed doesn’t change when the string hits the
   peg, and the ball swings equally high on both sides.
b) Relative to the point of suspension,

                       U i  0, U f   mg[d  ( L  d )]

   From this we find                       1 2
                             mg (2d  L)  mvi  0
                                           2
   Also for centripetal motion

                vi2
         mg  m          where R  L  d  vi2  Rg  ( L  d ) g
                R
   Solve these equations to get
                          L d         3
                 2d  L    0  d  L
                          2 2         5
Exercise 9.4

Two blocks of masses M and 3M are placed on a horizontal, frictionless surface.
A light spring is attached to one of them , and the blocks are pushed together
with the spring between them. A cord initially holding the blocks together is
burned; after this, the blocks of masse 3M moves to the right with a speed
2.00m/s.
a- What is the speed of the block of mass M?
b- Find the original elastic potential energy in the spring if M = 0.35 kg
Solution 9.4

For the system of two blocks


                 p  0
                 Mvm  3m  2  0
                  vm  6.00     Motion to the left



b)

               1 2 1       1
                 kx  MvM  3Mv3M  8.40 J
                        2      2

               2     2     2
Exercise 9.12
A professional diver performs a dive from a platform 10m above the water
surface. Estimate the order of magnitude of the average impact force she
experiences in her collision with the water. The impact time take it 1.0s and the
mass of the diver is 55 kg.
Solution 9.12

If the diver starts from rest and drops vertically into the water, the velocity just
before the impact is found from
                             1 2
       Ki  U i  K f  U f  vimpact  0  0  mgh  vimpact  2 gh
                             2
 With the diver at rest after an impact time t , the average force during impact
 is given by the second Newton’s law

                               m(0  vimpact )        m 2 gh
                       Fav                                  
                                     t                t
Using the given numbers:


                           Fav  770N
Exercise 9.24
As shown in the figure, a bullet of mass m and speed v passes completely
through a pendulum bob of mass M. The bullet emerges with a speed of v/2.
The pendulum bob is suspended by a stiff rod of length l and negligible mass.
What is the minimum value of v such that the pendulum bob will barely swing
through a complete vertical circle?




                                                           l
                                          m
                                                      M

                                              v                       v/2
Solution 9.24


Energy is conserved for the bob-Earth system between bottom and top of swing.
At the top the stiff rod is in compression and the bob nearly at rest
                                  1
            Ki  U i  K f  U f  Mvb  0  0  Mg 2 
                                     2

                                  2
            vb  g 4  vb  2 gf
             2



Momentum of the bob-bullet system is conserved in the collision
                   v                 4M
             mv  m  M (2 g )  v                g
                   2                  m
Exercise 9.27

a) Three carts of masses 4 kg, 10 kg and 3 kg move on a frictionless horizontal
track with speeds of +5m/s, +3m/s, and - 4m/s respectively. The carts stick
together after collision. Find the final velocity of the train of three carts.
b) Does your answer require that all the carts collide and stick together at the
same time? What if they collide in a different order?


                                            4m/s        3m/s
                                                                          -4m/s
Solution 9.27

a)- Using conservation of momentum for a perfect inelastic collision:

   p     before
                     p after  m1v1  m2v2  m3v3  (m1  m2  m3 )vG

                     v  2.24m / s          To the right

b)- No.
For example, if m2 and m3 were to stick first, they would move with a speed

                       m2 v2  m3v3  (m2  m3 )vG 23

Then, when this 13 kg =(m2 + m3) collides with the 4 kg , we have


  m1v1  (m2  m3 )vG 23  m1v1  m2v2  m3v3  (m1  m2  m3 )vG
Thus coupling order makes no difference
Exercise 9.58

A bullet of masse m is fired into a block        vi       M
                                             m
of mass M initially at rest at the edge of
a frictionless table of height h.
The bullet remains in the block, and
                                                      h
after impact the blocks lands a distance
d from the bottom of the table.
Determine the initial speed of the bullet.
                                                              d
Solution 9.58

                               M m
     mvi  ( M  m)v f  v f       vi
                                m 
                    1 2        2h           d    g              gd 2
     d  v f t ; h  gt  t  
                               g  and v f   d    
                    2                       t    2h             2h

                         M  m  gd
                                     2
This gives         vi         
                         m  2h

				
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