P 37 Lab Notes-Morris

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P 37 Lab Notes-MorrisP 37



     In this experiment we will verify that linear momentum is conserved for both elastic collisions and inelastic collisions.


     One of the most widely used principles in physics is the conservation of linear momentum. The utility of this principle lies in the fact that regardless of the complexity of the actual collision, the total momentum before a collision is equal to the total momentum after a collision:

     mv + mvv + mv(Before Collision) = m(After Collision) 11i22i 11f22f

     Two types of two-body collisions will be investigated using the air track. The first is one in which almost no energy is dissipated; this is called an elastic collision. The second is one in which a fraction of the kinetic energy is dissipated; this is called an inelastic collision.

     Remember that velocities and momenta are vector quantities and consequently, in one dimension, add algebraically. That is to say, momenta in the right-hand direction are positive and momenta in the left-hand direction are negative.


     Air Track & accessories 2 gliders 2 cardboard flags

    Electronic balance 2 photogate timers Masking tape



    1. Level the air track by placing the glider at the middle of the track and adjusting the leveling

    screws until the glider is motionless. Place two independent photogate timers near 70 cm

    and 130 cm.

    2. Place a rubber-band bumper at the front of glider 1 and place a counter mass at the back of

    glider 1. Tape a cardboard flag to the top of each glider, with the longer side horizontal.

    Determine and record the effective length L to the nearest millimeter of each cardboard

    rectangle by measuring the glider’s positions when the LED on top of the photogate turns on,

    then off. Determine the mass of each glider with attachments and record the mass of each.

    3. Set the photogate timers to gate mode and turn on the memory feature. Launch the gliders

    with moderate speed from opposite ends of the air track. The cardboard flags will pass

    through the photogates and a time will be displayed on each photogate. From these times

    the initial velocity of each glider can be calculated. After impact, the gliders will pass back

    through the photogates, adding to the times in memory. Subtract the displayed time from

    the memory time to arrive at the actual time for the final velocities. Be sure to stop the

    - 33 -

    gliders after they have passed through the photogates the second time. Use negative signs

    for the velocities where appropriate.

    4. Repeat the above procedure two more times so that you will have three trials altogether.

    Change the mass on the gliders for these two trials by adding masses symmetrically to the

    sides of one of the gliders.

    5. Calculate the momentum for each trial before and after the collision. Is linear momentum

    conserved? Calculate the total kinetic energy both before and after collision. Is kinetic

    energy conserved?

    Photogate 1 Photogate 2 Glider 1 Glider 2

     mm 12

     Fig. 1: Experimental Setup for Elastic Collision


    1. Repeat the above procedure for a completely inelastic collision. Use the wax plug and needle on the gliders so that they stick together, and place a counter mass at the back of glider 2. After the collision, stop the gliders after one rectangle of cardboard has passed through a photogate. Only one timing is necessary, as v = v. 1f2f

    2. Calculate the momenta both before and after collision. Is momentum conserved for the inelastic case? Calculate the total kinetic energy both before and after collision. Is kinetic energy conserved?

    Photogate 1 Photogate 2 Glider 1 Glider 2

    Wax Plug Needle

     mm 12

     Fig. 2: Experimental Setup for Inelastic Collision

    - 34 -


    Table A. Elastic Collision L = ____________________ m , L = ____________________ m 12

     Before Collision

     ;;Trial 1 2 Total Total Glider Glider

    m p m p vv Momentum Energy tt1 1i1i2 2i2i1i2i (J) (kg?m/s) (kg) (m/s) (kg) (m/s) (kg?m/s) (kg?m/s) (s) (s)




     After Collision

     ;;Trial 1 2 Total Total Glider Glider

    m p m p vv Momentum Energy tt1 1f1f2 2f2f1f2f (J) (kg?m/s) (m/s) (m/s) (kg) (kg) (kg?m/s) (kg?m/s) (s) (s)




    Table B. Inelastic Collision

     Before Collision

     ;;Trial 1 2 Total Total Glider Glider

    m p m p vv Momentum Energy tt1 1i1i2 2i2i1i2i (J) (kg?m/s) (kg) (m/s) (kg) (m/s) (kg?m/s) (kg?m/s) (s) (s)




     After Collision

     ;;Trial 1 2 Total Total Glider Glider

    m p m p vv Momentum Energy tt1 1f1f2 2f2f1f2f (J) (kg?m/s) (m/s) (m/s) (kg) (kg) (kg?m/s) (kg?m/s) (s) (s)




    - 35 -

Experiment 11: CENTER OF MASS


     To determine the center of mass (CM) of objects of uniform density and thickness.


     One way to locate the center of mass of an object of uniform thickness is to balance the object on a pivot. Another way is to drop a plumb line from one corner and trace out the plumb line, then repeat this for another corner. The point of intersection of the two lines is the center of mass.

     A third way to locate the center of mass is to divide an object into rectangles, and imagine the mass of each rectangle to be concentrated at the rectangle’s center. Each rectangle can be weighted by area instead of mass, as the object has uniform surface density and thickness. Then the center of mass of the object is the average of the positions of these centers, weighted according to area.

    2 For example, suppose that the upper bar of Object A in Fig. 2 has an area of A = 10 cm 1

    extending from y = 10 cm to y = 14 cm, so y = 12 cm, and suppose the vertical stem has an area 1

    2of A = 15 cm and extends from y = 0 cm to y = 10 cm, so y = 5 cm. Then, 22

    10?1215?5;?;?195Y= = = 7.8 cm. CM 101525


    (Ax(AyiiiiX = Y = cmcm(A(Aii


     Cardboard sheet Ruler String

    Scissors Plumb line Nail


     1. Cut out the following geometric shapes from a sheet of cardboard of uniform thickness.

     Object A Object B

    Object C

    Fig. 1

    - 36 -

    2. Puncture one corner of Object A with a nail, let it hang freely, and trace the path of the

    plumb line across the object, starting at the nail. Repeat this from another corner, and find

    the point where these two lines intersect. Mark this point, and label it as the “experimental”

    center of mass.

     3. Verify that this is the center of mass by balancing the object with your index finger.

     4. Calculate the position of the center of mass for object A, by imagining object A as two

    rectangles of unequal mass, each replaced by point masses at their geometric centers.

     5. Mark and label the "calculated" position of the center of mass on the object by placing it

    against a sheet of graph paper. See Fig. 2.

     y y y

    x x x O O O

     Object A Object B

    Object C

    Fig. 2.

     6. Compare the "experimental" and the "calculated" center of mass positions.

     7. Repeat procedure for object B.

     8. With object C, place it against the coordinates and locate the calculated X position. See CM

    Fig. 3.


    x O


    Fig. 3.

     9. Draw a line parallel to the y-axis through the calculated X. Tape a string along this line. CM

10. Experimentally locate Y by sliding your index finger along the mounted string to find the CM

    point of balance. Mark this point "exp" Y. Compare with "calculated" Y. CMCM

    - 37 -

     by using the plumb lines from two different corners. 11. Verify YCM


    Object Calculated Experimental

     _______ cm Y _______ cm X _______ cm Y _______ cm XCMCMCMCM

     _______ cm Y _______ cm X _______ cm Y _______ cm XCMCMCMCM

     _______ cm Y _______ cm X _______ cm Y _______ cm XCMCMCMCM

    - 38 -



     The object of this experiment is to use the method of balancing torques to determine the center of mass of a non-homogeneous meter stick, and to determine the unknown mass of an object.


     If a rigid object is in rotational equilibrium, the net torque acting on it, about an axis, is zero. This equilibrium condition can be stated as:

     ( ~ = 0

    where ~;= Fd, F is the applied force, and d is the distance from the horizontal axis of rotation to the point where the downward force is applied. The plus sign {+} corresponds to a counter-clockwise torque and the negative sign {-} corresponds to a clockwise torque.

     The center of mass, denoted here by CM, is the point at which the mass of the object can be considered to be concentrated. The position x of the CM of a non-homogeneous meter stick can be determined by balancing the torque of the stick on one side of the fulcrum with the torque of a known mass on the other side of the fulcrum.

    Fulcrum at midpoint of stick

    d d2x 1



     F = mg 11 F = mg 22~ ~; clockwisecounter-clockwise

    is a negative torque is a positive torque

    d + (Fd) = 0 (~ = F1122

    Fig. 1

     Having established the position of the CM and knowing the mass of the stick, the same procedure can be used to determine the unknown mass of another object.


     Weighted meter stick Electronic balance Scissors & light string Metal cube

    - 39 -

     1 knife-edge clamp Knife-edge stand Hooked masses



     indicated on the electronic balance. 1. Record the mass of the non-uniform meter stick m1

    2. Set up the apparatus as shown in Fig. 1, making sure that the fulcrum is at the midpoint of

    the stick, and using one of the hooked standard masses suspended from a loop of string

    as m. Slide m in along the stick until the stick is in equilibrium. Record m and d. 2222

    Be sure to include the mass of the string in the mass of m. 2

    3. Use Eq. 1 to obtain your first estimate of x, the distance of the CM from the weighted end

    of the stick. This equation was obtained from the equilibrium condition:

;~+ ~ = Fd - Fd = 0 counter-clockwiseclockwise1122

     Fd = Fd1122

    Fmg22 d= d = d 122 Fmg11

    m2 d = d= distance of CM from fulcrum 12 m1

     Eq. 1

     x = position of CM from weighted end

     = fulcrum position minus d 1

    4. Move the fulcrum 5.0 cm away from the midpoint, toward the weighted end of the stick as

    shown in Fig. 2. Slide m to establish equilibrium. Record m and m and the new 212

    value of d. Use Eq. 1 to calculate the new value of d and subtract this from the position 21

    of the fulcrum to obtain your second estimate of x, the position of the CM.

    Midpoint of stick Fulcrum

     d2d x 1


    m 2

     F = mg 22 F = mg 11

    - 40 -

    Fig. 2

     and balance the stick on the knife-edge 5. To obtain your third estimate of x, remove m2

    clamp. Record x as the position of the fulcrum.


1. Having calculated the position of the center of mass on the previous page, set up the

    apparatus as shown in Fig. 1, only this time m, a metal cube, will be an unknown mass. 2

    Using a loop of string, hang this unknown mass on the stick and slide it along the stick to


2. Record the new value of d. Use Eq. 2 to obtain your estimate of the unknown mass, m. 22

    F1 ( ~ = 0 , so Fd + (Fd) = 0 and F = d, giving mg = 112221 2d2

    mg1d. 1d2

    d1m = m 21d2

    Eq. 2

3. Weigh the metal cube on the electronic balance and find the percent difference between

    the two measurements of m. 2


1. The equilibrium condition can be used even when there are several torques involved.

    Set up the apparatus as shown below, using the same m as in Part B, and using hooked 2

    standard masses suspended from loops of string as m and m: 34

    Fulcrum at midpoint of stick

     d2d 4

     dd 3x 1

     mmm 423

     F F F F 2143

     Fig. 3

2. Use Eq. 3 to obtain another estimate of the unknown mass m. 2

    - 41 -

    (;~ = 0 , so Fd + Fd Fd - Fd= 0 giving F = 11223344 2

    FdFdFd334411 d2

    mdmdmd443311 m = 2d2

    Eq. 3

    3. Find the percent difference between this measurement and the value obtained directly

    from the electronic balance.


Data Table A: Determination of the Center of Gravity

    m (stick) m dFulcrum Position x d* 1221 (m) (kg) (kg)(m) (m)

At midpoint

    (Steps 1 3)

    Fig. 1

At 5.0 cm from midpoint

    (Step 4)

    Fig. 2


    (Step 5)

Data Table B: Unknown Mass m2

    m (stick) m dd* 1221(kg)(m)(kg)(m) (~ = 0

    (Steps 1-2)

    Fig. 1

Unknown mass

    from weighing

    (Step 3)

    Percent Difference _________________

Data Table C: Multiple Torque System

    - 42 -

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