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					PHY 201                                           Practice test 2                            Summe r 05

1. Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one
ball’s initial speed was 2.00 m s , and the other’s was 3.00 m s in the opposite direction,
what will be their speeds after the collision?

Let A represent the ball moving at 2.00 m/s, and call that direction the positive direction. Let B
    represent the ball moving at 3.00 m/s in the opposite direction. So vA  2.00 m s and
     vB  3.00 m s . Use Eq. 7-7 to obtain a relationship between the velocities.
                 vA  vB    vA  vB   vB  5.00 m s  vA
                                                         
     Substitute this relationship into the momentum conservation equation for the collision,
     noting that mA  mB .
                                                                
                   mA vA  mB vB  mA vA  mB vB  vA  vB  vA  vB 
                   1.00 m s  vA   vA  5.00 m s   2vA  6.00 m s  vA  3.00 m s
                                                                        

                                   
                   vB  5.00 m s  vA  2.00 m s
     The two balls have exchanged velocities. This will always be true for 1-D elastic collisions
     of objects of equal mass.


2. A 95-kg fullback is running at 4.0 m s to the east and is stopped in 0.75 s by a head-on
tackle by a tackler running due west. Calculate (a) the original momentum of the
fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler,
and (d) the average force exerted on the tackler.

Call east the positive direction.
        (a)         poriginal  mvoriginal   95 kg  4.0 m s   3.8 102 kg m s
                     fullback              fullback

        (b)        The impulse on the fullback is the change in the fullback’s momentum.

                                                        
         pfullback  m  vfinal            vfinal         95 kg  0  4.0 m s   3.8 10 kg m s
                                                                                                2

                               fullback        fullback 

        (c)        The impulse on the tackler is the opposite of the impulse on the fullback, so
 3.8  10 kg m s
         2


         (d)    The average force on the tackler is the impulse on the tackler divided by the time
of interaction.
                                             p           3.8  10 2 kg m s
                                   F                                         5.1 102 N
                                             t                0.75 s




                                                                                                          1
3. A wheel of diameter of 68.0 cm slows down uniformly from 8.40 m/s to rest over a
distance of 115 m. (a) What is the total number of revolutions the wheel rotates in
coming to rest? (b) What is the angular acceleration? (c) How long does it take for the
wheel to come to the stop?

See Example 8-7 p. 202-203



4. A hollow cylinder (hoop) is rolling on a horizontal surface at speed v  3.3 m s when it
reaches a 15º incline. (a) How far up the incline will it go? (b) How long will it be on the
incline before it arrives back at the bottom?

(a) Assuming that there are no dissipative forces doing work,
        conservation of energy may be used to find the final
        height h of the hoop. Take the bottom of the incline to be
        the zero level of gravitational potential energy. We
        assume that the hoop is rolling without sliding, so that                                h
           v R . Relate the conditions at the bottom of the                      
        incline to the conditions at the top by conservation of
        energy. The hoop has both translational and rotational
         kinetic energy at the bottom, and the rotational inertia of the hoop is given by I  mR 2 .
                                                                             v2
              Ebottom  Etop  1 mv 2  1 I  2  mgh  1 mv 2  1 mR 2 2  mgh 
                                  2        2                  2        2
                                                                             R
                              3.3 m s 
                                           2
                    v2
               h                              1.111 m
                    g        9.8 m s 2
                                                            h          1.111 m
        The distance along the plane is given by d                              4.293 m  4.3 m
                                                           sin        sin15o

(b) The time can be found from the constant accelerated linear motion. Use the relationship
                                          2x     2  4.293 m 
              x  1  v  vo  t  t                          2.602 s .
                                         v  vo   0  3.3 m s
                     2


         This is the time to go up the plane. The time to come back down the plane is the same,
          and so the total time is 5.2 s .




                                                                                                     2
5. A 140-kg horizontal beam is supported at each end. A 320-kg piano rests a quarter of
   the way from one end. What is the vertical force on each of the supports?


     Let m be the mass of the beam, and M be the mass of the                                 L
                                                                                 L/4
     piano. Calculate torques about the left end of the beam, with
     counterclockwise torques positive. The conditions of
     equilibrium for the beam are used to find the forces that the
     support exerts on the beam.
                                                                                                       FR
           FR L  mg  12 L  Mg  14 L   0                           FL         Mg   mg

                                                                       
          FR   1 m  1 M  g   1 140 kg   1  320 kg  9.80 m s 2  1.47  103 N
                 2     4           2             4


          F   y
                    FL  FR  mg  Mg  0

                                                       
          FL   m  M  g  FR   460 kg  9.80 m s 2  1.47  103 N  3.04  103 N
        The forces on the supports are equal in magnitude and opposite in direction to the above
two results.
                   FR  1.5  103 N down             FL  3.0  103 N down



6. (a) An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water.
What is the specific gravity of the object?

7.84 N – 6.86 N = water g V ------> 0.98 N = (1x103 kg/m3 ) (9.80 m/s2 ) V

V = 1x10-4 m3 Mass of the object M = 7.84 N/(9.8 m/s2 ) = 0.8 Kg

Density of object is object = M/V = 0.8 Kg/(1x10-4 m3 ) = 8x103 kg/m3 .

Specific gravity = object /water = 8


   (b) How much pressure does it take for a pump to supply a drinking fountain with
300 kPa, if the fountain is 30.0 m above the pump?
            1                    1 2
        P  v12  gy1  P2  v2  gy2
         1
            2                    2
With v1 = v2 , y1 = 0 , y2 = 30.0 m , P2 = 300 kPa ,  = 1x103 kg/m3

You calculate P1 = 594 kPa




                                                                                                   3
7. A mass sitting on a horizontal, frictionless surface is attached to one end of a spring;
the other end is fixed to a wall. 3.0 J of work is required to compress the spring by 0.12
m. If the mass is released from rest with the spring compressed, the mass experiences a
maximum acceleration of 15 m s 2 . Find the value of (a) the spring stiffness constant and
(b) the mass.

(a)       The work done to compress a spring is stored as potential energy.
                                              2W      2  3.0 J 
                          W  1 kx 2  k  2                       416.7 N m  4.2 102 N m
                                                     0.12 m 
                                2                                2
                                               x
      (b) The distance that the spring was compressed becomes the amplitude of its motion. The
                                                       k
           maximum acceleration is given by amax  A . Solve this for the mass.
                                                       m
                 k                 k      4.167 10 N m 
                                                     2

          amax  A  m                A                      0.12 m   3.333 kg  3.3 kg
                 m                amax        15 m s 2       




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