Lab 13: The Ballistic Pendulum

Purpose: To use the ballistic pendulum to determine the innitial velocity of a projectile using conservation of momentum and conservation of energy.

Equipment: Ballistic pendulum, carbon paper, meter stick, clamp box, triple beam balance, plumb.

Introduction: In this experiment a steel ball will be shot into the bob of a pendulum and the height, h, to which the pendulum bob moves, as shown in Figure 1, will determine the initial velocity, V, of the bob after it receives the moving ball.
                
                Figure 1:   

     

If we equate the kinetic energy of the bob and  ball at the bottom to the potential energy of the bob and ball at the height, h, that they are raised to, we get:

                                                       (K.E)bottom = (P.E)top
                                                       1/2( M+m) V^2 = ( M+m) gh

      Where M is the mass of the pendulum and m is the mass of the ball. Solving for V we get:

                                                        V = √(2gh) ----------(1)

Using conservation of momentum we know the momentum before impact (collision) should be the same as the momentum after impact. Therefore:
                                                                     
                                                                   pf = pi 
                                                     or
                                                              mvo= (M+m)V -----------(2)

       Where vo is the initial velocity of the ball before impact. By using equations (1) and (2) we can therefore find the initial velocity, vo, of the ball.
       We can also determine the initial velocity f the ball by shooting the ball as above but this time allowing the ball to miss the pendulum bob and travel horizontally under the influence of gravity. In this case we simply have a projectile problem where we cam measure the distance traveled horizontally and vertically (see Figure 2) and then determine the initial velocity, vo, of the ball.

       Figure 2:

       Starting with equations:
         
                                                       △x = voxt + 1/2axt^2 -------------(3)
                                                       △y = voyt + 1/2ayt^2 -------------(4)

       You should be able to derive the initial velocity of the ball in the horizontal direction (assuming that and known).


Part I: Determination of initial velocity from conservation of energy

1. Set the apparatus near one edge of the table as shown in figure 2. Make sure that the base is accurately horizontal, as shown by a level. Clamp the frame to the table. To make the gun ready for shooting, rest the pendulum on the rack, put the ball in position on the end of the rod and, holding the base with one hand, pull back on the ball with the other until the collar on the rod engages the trigger. This compresses the spring a definite amount, and the ball is given the same initial velocity every time the gun is shot.

2. Release the pendulum from the rack and allow it to hang freely. When the pendulum is at rest, pull the trigger, thereby propelling the ball into the pendulum bob with a definite velocity. This causes the pendulum to swing from a vertical position to an inclined position with the pawl engaged in some particular tooth of the rack.

3. Shoot the ball into the cylinder about nine times, recording each point on the rack at which the pendulum comes to rest. This in general will not be exactly the same for all cases but may vary by several teeth of the rack. The mean of these observation gives the mean highest position of the pendulum. Raise the pendulum until its pawl is engaged in the tooth corresponding most closely to the mean value and measure h1, the elevation above the surface of the base of the base of the index point for the center of gravity. Next release the pendulum and allow it to hang in its lower most position and measure h2. The difference between these two values gives h, the vertical distance through which the center of gravity of the system is raised after shooting the ball.

Record h:

Our average point on the rack is 12.9, which corresponds to 12.3cm
So,  h1= 12.3cm
We measured h2 = 3.8cm

h= h2 - h1= 12.3cm- 3.8cm= 8.5cm


4. Carefully remove the pendulum from its support. Weigh and record the masses of the pendulum and of the    ball. Replace the pendulum and carefully adjust the thumb screw.

M (mass of pendulum) = 194g
m ( mass of the ball) = 56.8g


5. From the data calculate the initial velocity v using equations (1) and (2).

   Because:
               V = √(2gh)
               mvo= (M+m)V
   So:
              vo= (M+m) ×√(2gh) / m
                  = (194+56.8) ×√(2×9.8×0.085) / 56.8
                  = 5.7 m/s



Part II: Determination of initial velocity from measurements of range and fall

1. To obtain the data for this part of the experiment the pendulum is positioned up on the rack so that it will not interfere with the free flight of the ball. One observer should watch carefully to determine the point at which the ball strikes the floor. The measurements in this part of the experiment are made with reference to this point and the point of departure of the ball. Clamp the frame to the table. as it is important that the apparatus not be moved until the measurements have been completed. A piece of paper taped to the floor at the proper place and cover with carbon paper will help in the exact determination of the spot at which the ball strikes the floor.

2. Shoot the ball a number of times, nothing each time the point at which it strikes the floor. Determine, relative to the edge of the paper, the average position of impact of the ball. Determine the distances △x and △y calculate vo by the use of equations (3) and (4). Make careful stretches in your lab report show all of the  distances involved.

      
      The distance from the ball to the paper: 258.4cm
      The distance of the ball on the paper: 17.4cm

 The distance of the ball on the paper 

       △x = 258.4+17.6= 276cm
       △y(height) = 99.7cm

          △y = voyt + 1/2ayt^2
       0.997 = 0 + (1/2) × 9.8 × t^2 
               t = 0.45s

          △x = voxt + 1/2axt^2
        2.76 = vo × 0.45 +0 
           vo = 6.1 m/s

     
      Percent of difference between part I and part II:
     
      (6.1-5.7) / [(6.1+5.7) /2] = 6.8%


Conclusion:

What you learned
    
     After doing this Ballistic Pendulum lab, I learned that the momentum and energy of the of the ball and pendulum's system are both conserved if there is no external forces. We used the the law of conservation of momentum and conservation of energy to find the initial velocity of the ball, and then compare to the value we got from measurement of range and fall. We found these two values are really close, which proves the law of conservation of momentum and conservation of energy.   
   

Source of error:

    1. The table is not exactly horizontal, so our h is not accurate.
    2. The friction of the system and the air resistance are external forces that can't be avoided, which will lose the energy of the system.


Which method did you think was more accurate? Explain.
 
    The method 1 will be more accurate.
    Although in both method 1 and method 2, we assume that there is no external forces, the friction and air resistance still can't be avoided. But in method 2, the ball will go through the air for a long time that will be influenced by air resistance more than method 1.

Lab 12: Inelastic Collisions

Purpose: To analyze the motion of two low friction carts during an inelastic collision and verify that the law of conservation of linear momentum is obeyed

 Equipment: Computer with Logger Pro software, lab pro, motion detector, horizontal track,two carts, 500 g masses(3), triple beam balance, bubble level

 Introduction: This experiment uses the carts and track as shown in the figure. If we regard thesystem of the two carts as an isolated system, the momentum of this system will beconserved. If the two carts have a perfectly inelastic collision, that is, stick together after the collision, the law of conservation of momentum says

Pi = Pf

m₁v_{1 +} m₂v₂ = (m₁ + m₂) * V

where v₁ and v₂ are the velocities before the collision and V is the velocity of thecombined mass after the collision.

Procedure:

1.    Set up the apparatus as shown in Figure 1. Use the bubble level to verify that the track is aslevel as possible. Record the mass of each cart. Connect the lab pro to the computer and themotion detector to the lab pro. On the computer, start the Logger Pro software; open theMechanics folder and the Graphlab file.


2.     First, check to see that the motion detector is working properly by clicking the Collect buttonto start collecting data. Move the cart nearest the detector back and forth a few times whileobserving the position vs time graph being drawn by the computer. Does it provide a reasonablegraph of the motion of the cart? Remember to be aware of unwanted reflections caused byobjects in between the motion detector and the cart. Also, position the carts so that their velcropads are facing each other. This will insure that they will stick together after the collision. 


 3.     With the second cart (m2) at rest give the first cart (m1) a moderate push away from themotion detector and towards m2. Observe the position vs time graph before and after thecollision. What should these graphs look like? Draw an example: 

 

        m1= 501.5g      m2= 516.1g   

        the graph look like two parts of continuous increasing liner curves. The slope of first part should be lower than the second part. 

The slope of the position vs. time graph directly before and directly after the collision gives thevelocity directly before and directly after the collision. To avoid the problem of dealing with friction forces (Remember, we are assuming the system is isolated.), we will find the velocity of the carts at the instant before and after the collision.

Is this a good approximation? Why or why not?

        

        It's a good approximation, because we focus on the change of the velocity after the collision, so we      should only compare the instant velocity before and after the collision. Due to the friction and air resistance, the velocity will have big error if we use the velocity in a large range of  time. 

For the velocity before the collision, select a very small range of data points just before the collision. Avoid the portion of the curve which represents the collision. Choose Analyze/LinearFit. Record the slope (velocity) of this line. Repeat for a very small range of data points just afterthe collision. Record this slope (velocity) as well. 

p-t graph when m1= 501.5g and m2= 516.1g

     Slope before the collision: 0.444
     Slope after the collision: 0.150

4.     Repeat for two more collisions. Calculate the momentum of the system the instant before andafter the collision for each trial and find the percent difference. Put your results in an Excel datatable. Show sample calculations here:

      

      m1= 501.5g     

      m2= 516.1g 

      vbefore  is equal to the slope of a very small range of data points before the collision.
      vafter  is equal to the slope of a very small range of data points after the collision.
   
      pi = m1× vbefore +  m2 × 0

      pf = (m1+ m2) × vafter

      perror = (pf - pi ) / pf

5.     Place an extra 500 g on the second cart and repeat steps 3 and 4. Sketch one representativegraph showing the position vs time for a typical collision. (What do velocity vs. time and acceleration vs. time look like? 

      m1= 501.5g     
      m2= 516.1g +495g= 1011.1g

p-t graph when m1= 501.5g and m2= 1011.1g

a-t graph when m1= 501.5g and m2= 1011.1g

             The acceleration is almost zero before and after the collision.

v-t graph when m1= 501.5g and m2= 1011.1g

             The velocity is almost constant  before and after the collision, because according to the a-t graph,  the acceleration is almost 0. The velocity after the collision is lower than the velocity before the collision.

        pi = m1× vbefore +  m2 × 0

        pf = (m1+ m2) × vafter
     
        perror = (pf - pi ) / pf

6. Remove the 500g  from the second cart and place it on the first cart. Repeat steps 3 and 4.

        m1= 501.5g +495g= 996.5g   
        m2= 516.1g

      

p-t graph when m1= 996.5g and m2= 516.1g

7.     Find the average of all of the percent differences found above. This average represents your verification of the law of conservation of linear momentum. How well is the law obeyed based on the results of your experiment? Explain.

     

      Based on our result, we found our average value of momentum has less than 10 percent difference between pf and pi. We can say in the acceptable error range, the momentum before and after the collision is almost the same.

8.     For each of the nine trials above calculate the kinetic energy of the system before and afterthe collision. Find the percent kinetic energy lost during each collision. Put this information in aseparate data table. Show sample calculations here:

the percent kinetic energy lost for each trial 

9.     Do a theoretical calculation for ΔK/K in a perfectly inelastic collision for the three situations:

       
     1. a mass, m, colliding with an identical mass, m, initially at rest.
             
          mvb = 2mva ------ vb = 2va 
          K = (1/2)×m×(vb)^2
          ΔK = (1/2)×2m×(va)^2- (1/2)×m×(vb)^2

          
          ΔK/K= -50%       
     
    
     2.  a mass, m, colliding with a mass, 2m, initially at rest.

          mvb = 3mva ------ vb = 3va 

          K = (1/2)×m×(vb)^2

          ΔK = (1/2)×3m×(va)^2- (1/2)×m×(vb)^2

          

          ΔK/K= -66.7%

     3.  a mass, 2m, colliding with a mass, m, initially at rest.

     

          2mvb = 3mva ------ vb = (3/2)va 

          K = (1/2)×2m×(vb)^2

          ΔK = (1/2)×3m×(va)^2- (1/2)×2m×(vb)^2

          

          ΔK/K= -33.3% 

Conclusions:

•     What you learned

      I learned that momentum is equal to mass times velocity. If there is no external forces in a system, then the momentum of this system is conserved.
     An impulse delivered to a particle changes the particle's momentum. The momentum after the impulse is equal to the momentum before the interaction plus the impulse that arises from the interaction.
     Inelastic collisions is a collision in which kinetic energy is not conserved, even though the momentum of the system can be conserved.

• Sources of error

     1. The table is not exactly horizontal, so the component of the gravity force will influence the result.

     2. The friction and the air resistance are external forces of the system that can't be avoided, so our graphs are not perfect.

• Compare your experimental numbers calculated in part 8 above with the results of your theoretical calculations in part 9.

     We find in part 8, the calculated values of percent energy lost of trials 4 to 9 are close to the theoretical values in part 9, but the values of trail 1,2 and 3 are little bit biased. The reason is may be that in trail 1,2 and 3, we didn't put the bar on either car. So, the mass of the system is the smallest, which is easy to be influenced by external forces, like friction and air resistance.

• If momentum is conserved, is kinetic energy also conserved?

    No. Inelastic collisions is a collision in which kinetic energy is not conserved. But if there is no external forces in the system, the momentum of the system can also be conserved.