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.

Materials:
Computer with Logger Pro software, lab pro, motion detector, horizontal track, two carts, 500 g masses(2), triple beam balance, bubble level.

Introduction:
The experiment uses the carts on the track with some mass to analyze the conservation of momentum.
Momentum is calculate by p=mv (m-mass, v-velocity). Consider the system of the two carts as an isolated system, and the momentum of this system will be conserved. If the two carts have a perfectly inelastic collision the law of conservation of momentum applies.
 

Pi = Pf
m1v1 + m2v2 = (m1 + m2)V

* The above formula will be use to calculate the conservation of momentum. v1 and v2 are the initial velocities of the carts and V is the combined velocities after the collision.

Procedure:
1. Set up the apparatus as shown in Figure 1. Use the bubble level to verify that the track is as level as possible. Record each mass of the carts. Connect the lab pro to the computer and the motion detector to the lab pro. Make sure the motion detector is working properly before collecting data. Also the track should be leveled correctly.

     m1 = 0.4998 kg,  m2 = 0.5041 kg 

3. With the second cart (m2) at rest give the first cart (m1) a moderate push away from the motion detector and towards m2. Observe the position vs time graph before and after the collision. Draw an example.

                            Graph 1

The slope of the line should look constant at first and then the slope should decrease since there was a collision between the two objects.

 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?

* Yes, this is a good approximation because the speed before the impact is needed in order to   calculate the impulse after the two carts collide. 

4. Repeat for two more collisions. Calculate the momentum of the system the instant before and after the collision for each trial and find the percent difference. Put your results in a data table.

       After collecting the data for the first run, we selected a small area of the graph right before the collision and made a linear fit. This is the initial velocity for the initial momentum. We did the same for right after the collision. This is the final velocity (V) for the final momentum. Then we performed 2 more trials.

  

Table 1

The table shows 3 trials with the calculated momentum and the percent difference. 

sample:

p1 = m1v1                                                      p2 = m2v1

     = (0.4998 kg)(0.5762 m/s)                               = (0.5041 km)(0.5762 m/s)

     = 0.2689 kgm/s                                               = 0.2700 kgm/s

5. Place an extra 500 g on the second cart and repeat steps 3 and 4. Print out one representative graph showing the position vs time for a typical collision.What should the velocity vs time graph look like? acceleration?
 

                             Graph 2

Before and after collision position vs time graph for the two carts

                           Graph3

The initial velocities are constant and decrease after the collision but still move together at a slower rate.

                             Graph 4

Acceleration vs time graph before and after collision.

Table 2

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

Table 3

 * The table shows the momentum before and after the collision. This time the first cart is the one that has the 500 g on top.

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?

The average for all 9 trial was 21.7% because in the second experiment we obtained a 138% difference. We agreed to keep the result as an example for sources of error in this lab. If we take out the value and average the 8 trials we got a 7.19% difference. This is more supportive data to to verifying the law of conservation of linear momentum

8. For each of the nine trials above calculate the kinetic energy of the system before and after the collision. Find the percent kinetic energy lost during each collision. Put this information in a separate data table.


Table 4

Percent difference for all 9 trials of the experiment.

Conclusion:

    The lab helps to clearly understand that momentum is conserved in a perfect inelastic collision. We demonstrated this by calculating the initial and final momentum of 3 different scenarios involving 2 carts. Overall the nine trials for the two carts support that momentum is conversed for linear inelastic collisions.

    On Table 2 we can see that we came across a source of error. This is because we did not get a perfect inelastic collision during that trial. Another source that some one could come across is that the track is not leveled correctly creating an incline on the track.

Balanced Torque and Center of Gravity

Purpose:
To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses.

Equipment:
Meter stick, meter stick clamps (knife edge clamp), balance support, mass set, weight hangers, unknown masses, balance.

Introduction:
The condition for rotational equilibrium is that the net torque on an object about some point in the body, O, is zero. Torque is defined as the force times the lever arm of the force with respect to the chosen point O. The lever arm is the perpendicular distance from zero to the line of action of the force.


Diagram 1.1

The two masses are perpendicular to the lever arm.

Procedure:
1. Balance the meter stick on clamp and record the position of the balance point. What point in the meter stick does this correspond to?


Diagram 1.2


the balance point was rounded to 49 cm





2. Select two different masses and using the meter stick clamps and weight hangers, suspend one on each side of the meter stick support at different distances from the support. Adjust the positions so the system is balanced. Record the masses and positions. Is it necessary to include the mass of the clamps in your caculation? Sum the torques about your pivot point O and compare with the expected value.
     Yes the mass of the clamps should be included because they contribute to the total weight of the system. 


 Diagram 1.3

Torque1 - Torque2 = 0 N m

length 1 =   40 cm = 0.40 m                                             length 2 = 31.16 cm = 0.3116 m 
wight 1 =   (0.1745 kg)(9.8 m/s^2)                                   weight 2 = (0.2206 kg)(9.8 m/s^2)
            =   1.7 N                                                                         =  2.2 N  

Sum of Torques
T1 - T2 = 0 N m
(length 1)(weight 1) - (length 2)(weight 2) = 0 N m
(0.40 m)(1.7 N) - (0.3116 m)(2.2 N) = 0 N m
0.001 N = 0 N

Percent Error:

solve for one of the lengths (L2)
(length 1)(weight 1) - (length 2)(weight 2) = 0 N m
(0.40 m)(1.7 N) - (L2)(2.2 N) = 0 N m
L2 = 0.312 m

((0.3116 m - 0.312  m) / 0.312) * 100 = 0.1 % error

3. Place the same two masses used above at different locations on the same side of the support and balance the system with a third mass on the opposite side. Record the masses and positions. Calculate the net torque on this system about the point support and compare with the expected value.

Diagram 1.4

Diagram shows the distance between the center to the weights. The net force of torque equals zero.

      mass1 = 174.5 g =  0.1745 kg                                      mass2 = 270.8 g =  0.2708 kg     
      length1 = 15.0 cm = 0.15 m                                         length2 = 34.0 cm = 0.34 m
                                        
                                                 mass3 = 220.6 g =  0.2206 kg                          
                                                 length3 = 30.0 cm = 0.30 m 

net torque: 

T2 - T1 -T3 = 0
(length 2)(weight 2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(0.3116 m)(2.2 N)   - (0.40 m)(1.7 N)     -  (0.30 m)(2.16 N)   = 0 N m 
0.0016 N m = 0 N m

Percent Error:

solve for one length (L2)
T2 - T1 -T3 = 0
(length 2)(weight 2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(L2)(2.2 N)   - (0.40 m)(1.7 N)     -  (0.30 m)(2.16 N)   = 0 N m 
L2 = 0.3394 m

((0.3394 - 0.34) / 0.3394) * 100 = 0.2 % error 


4. Replace one of the above masses with an unknown mass. Readjust the positions of the masses until equilibrium is achieved, recording all values. Using the equilibrium condition for rotational motion, calculate the unknown mass. Measure the mass of the unknown on a balance and compare the two masses by finding the percent difference.

 Diagram 1.4

In this step we are solving for the unknown mass of weight 2.

    mass3 = 220.6 g =  0.2206 kg      mass1 = 174.5 g =  0.1745 kg            mass2 = ???
              length3 = 30.0 cm = 0.30 m         length1 = 15.0 cm = 0.15 m     length2 = 30.2 cm = 0.302 m

solve for the unknown weight (w2)

T2 - T1 -T3 = 0
(length 2)(w2) - (length 1)(weight 1) - (length 3)(weight 3) = 0 N m
(0.302 m)(9.8 m/s^2)w2  - (0.40 m)(1.7 N)  -  (0.30 m)(2.16 N)   = 0 N m
w2 = 0.3043 kg = 304.3 g

percent error: 

((304.6 - 304.3) / 304.6) * 100 = 0.01 % error
 


5. Place about 200 grams at 90 cm on the meter stick and balance the system by changing the balance point of the meter stick. From this information, calculate the mass of the meter stick. Compare this with the meter stick mass obtained from the balance. Should the clamp holding the meter stick be included as part of the mass of the meter stick?


Diagram 1.5

w1 = 199.9 grams at the 90 cm mark

length1 = 12.0 cm = 0.12 m
length2 = 29 cm = 0.29 m 

mass of meter stick????
remember that the torques in the system are equal.
T1 = T2
w1*L1 = w2*L2
m1*g*L1 = m2*g*L2
(199.9 kg)(9.8 m/s^2)(0.12 m) = m2(9.8m/s^2)(0.29 m)
m2 = 82.717 g

percent error:

((82.7 g - 81.8 g) / 81.8 g) * 100 = 1.1 % error




6. With the 200 grams still at 90 cm mark, imagine that you now position an additional 100 grams mass at the 30 cm mark on the meter stick. Calculate the position of the center of gravity of this combination (two masses and meter stick). Where should the point of support on the meter stick be to balance this system? Check your result by actually placing the 100 g at the 30 cm mark and balancing this system. Compare the calculated and experimental result.

Diagram 1.6

Pictorial representation to solve for x

m1 = 0.0818 kg             L1 = 49 cm = 0.49 m
m2 = 0.100 kg               L2 = 30 cm = 0.30 m
m3 = 0.200 kg               L3 = 90 cm = 0.90 m

center of mass = 1/M Σ mi*xi = (m1x1 + m2x2 + m3x3) / (m1 +m2 + m3)

cm = 0.655 m = 65.5 m

percent error:
actual center = 65.1 cm

((65.1 cm - 65.5 cm) / 65.1 cm) * 100 = 0.6% error 

Conclusion:

    The lab is a helpful way in learning how to calculate torque using the center of balace of the meter stick as an example.There were different cases in calculating the net torque. The first part we just had to balance the meter stick in the center. After we had to place two weights on opposite ends until we found the new balance point. This point was located at 0.4m and 0.3 m from the center. The percent error that we obtained was only a 0.1%. Then we placed a third wight on one end and we did the same. Calculated the torques and set then up to 0 since the system is at equilibrium. 
    All the percent errors that we calculated were really small meaning that our calculations were accurate. Some sources of error that may occur during the experiment is that the center of balance may be off and it will ruin the data collected. 

Human Power

Introduction:

The main purpose of the lab is to calculate the power output one can generate walking up a set of stairs. Power is defined as the rate at which energy is converted from one form to another. To find the gravitational potential energy we measured the vertical height of the stairs and recorded our own mass in kilograms. With the formula △PE = mgh, (m - mass, g - gravity, h - height) we can calculate the potential energy. In order to find the power out put we used divided our potential energy by the average time it took to climb the stairs.

Materials:

• two meter sticks
• stopwatch 
• kilogram scale

Procedure:

• Determine your own mass in kilograms. 
• Measure the vertical height of the stairs.****Make a sketch for the method the height was found.
• After having a person recording the time, walk or run up the stairs to the second floor. 
• Everyone will do two trials to climb up the second floor. Record both times.
• calculate the total power output by using the formulas from the introduction. Obtain the average power output from both trials.
• Put the average output power on the board and calculate the average power of the entire class.
• Determine your power output in terms of horsepower. 

Data:

Sketch of the Stairs

Picture 1.1  The stair set leading to the second floor that has a height of 4.29 meters.

Calculations
Table1

*Note: power is in watts

                           
                                                     Gravitational potential energy 
F = mg                                           △PE = mgh
F= (63.16 kg)(9.8m/s^2)                  △PE = (63.16kg)(9.8m/s^2)(4.29m)
F= 619 N                                        △PE = 2655.37 J


Power:
Power1 = △PE / △t1                Power2 = △PE / △t2                
P1 = 2655.37 J/ 4.25 s              P2 = 2655.37 J / 5.07 s 
P1 = 624.79 w                         P2 = 523.74 w

Horse Power:
1 HP = 745.699872 watts

HP1 = 624.79 w (1 HP/ 745.699872 w)
HP1 = 0.84 hp

HP2 = 523.74 w (1 HP/ 745.699872 w)
HP2 = 0.7 hp

Class Averages:
634.98  watts
.82 hp

Percent difference: 

((measured class value - measured value) / class value) * 100

△PE
((634.98 w - 574.3 w) / 634.98 w) *100 = 9.55 %

HP
((0.82 hp - 0.77 hp) / 0.82 hp) *100 =  6.09 %



Questions:

1. Is it okay to use your hands and arms on the hand-railing to assist you in your climb up the stairs? Explain why or why not.
 If one is trying to be consistent it would be more beneficial to not use the railing or if it was used, use it in both trials. This is because as you swing around the corner, you can feel a greater acceleration than a normal turn. The assist will have a greater influence on the time since the mass, height, and gravity are used as constants.Overall the gravitational potential energy will be the same for both scenarios.


2. Discuss some of the problems with the accuracy of this experiment.
  Some of the problems that can occur during the lab are that the measurement of the height was inaccurate or the time could've been off by a few milliseconds due to different reaction times from the person walking and the record individual. 

Follow up Questions:

1. Two people of the same mass climb the same flight of stairs. Hinrik climbs the stairs in 25 seconds. Valdis takes 35 seconds. Which person does the most work? Which person expands the most power? Explain your answers.

   They both do the same amount of work. To calculate this we use △PE = mgh and the mass in this case are the same, giving the same value for the amount of work done. Hinrik is the one who expands more power since he had the ten seconds less than Valdis. The lab helps demonstrate a clearer picture. Table 1 shows that when my time climbing up the stairs was less, the more power I used and my slower time had less power.

2. A box that weights 1000 Newtons is lifted a distance of 20.0 meters straight up by a rope and pulley system. The work is done in 10.0 seconds. What is the power developed in watts and kilowatts.

h = 20.0 m                             Power = △PE / △t
t = 10 s                                            = (1000 J * 20 m) / 10 s
w = 1000 N                                      = 2000 watts or 2 kilowatts

3. Brynhildur climbs up a ladder to a height of 5.0 meters. If she is 64 kg:

    a) What work does she do?
             The work that she does is lift the 64 kg against gravity to the height of 5 meters.

    b) What is the increase in the gravitational potential energy of the person at this height?
             Work = mgh
                      = 64 kg * 9.8 m/s^2 * 5 m
                      = 3136 J

    c) Where does the energy come from to cause this increase in P.E.?
              The energy comes from her as she puts in work to climb up the ladder while the constant force of gravity pulls her down.

4. Which requires more work: lifting a 50 kg box vertically for distance of 2m , or lifting a 25kg box vertically for a distance of 4 meters?

50 kg box:                                                   25 kg box:
  △PE = mgh                                              △PE = mgh         
          = 50 kg * 9.8 m/s^2 * 2 m                         = 25 kg * 9.8 m/s^2 * 4 m
          = 980 J                                                    = 980 J                                                 
Both require the same amount of work.

Conclusion:


    The lab helps give a greater understanding of gravitational potential energy. In order to climb the set of stairs, there's a need for work since the individual needs to fight against the downward force of gravity. In both trials the same mass and height are equal and g remains a constant 9.8 m/s^2 so the same amount of potential energy is the same but the faster one goes the more power is generated since the the work is cramped into a smaller time. The method to find power is to use the formula power = mgh / t. 
    The percent difference between the gravitational potential energy from the class and mine was a 9.55 %. While the horsepower difference was about 6.1%. This makes clear sense since when i observed the class do their trials i noticed that mostly everyone was rushing while i maintained a steady pace.
    The most obvious sources of error are that we had a person telling us when to go and then another person running the time. This could be a conflict because the reaction times among the individuals might not be the same. Another factor is the way we measured the height of the stairs.

Centripetal Force

Introduction:
The purpose of this lab is to experiment with Newton's second law of motion for uniform circular motion. Using a centripetal force apparatus to rotate a known mass around a known radius. The velocity can be calculated by timing the number of revolutions the object travels and the total distance the mass has traveled. To determine the centripetal force to maintain the object in the circular path can be calculated by F = mv^2/r.

Materials:
Centripetal force apparatus, metric scale, vernier caliper, stop watch, slotted weight set, triple beam balance.

Procedure:
1. Assemble the centripetal force apparatus so that when the weights rotate around the mass hangs right over the vertical cross arm that indicates the radius.

2.Spin the apparatus 50 times and time how long it takes while maintaining a constant speed. Use the same mass and radius for all the the first four trials.

3. Obtain the average time to calculate the velocity of the weights. Then you can calculate the centripetal force exerted on the mass during the rotation.

4. Determine the centripetal force by attaching a hanging weight to the mass until it is position over the radius indicator post. The force should be about the same.

5. Add 100 g and repeat steps 2, 3, and 4.

centripetal force apparatus

Formulas:
f = 50/t
v = 2pi*r*f
F = mv^2/r 


Data:

Table1.2

*top 5 trials are with the first mass and the bottom was added 100 g. Note: radius is in meters.

 Known mass = 474.5g (475g)

revolutions (t) = 50 

*convert units

f = 50/t   so, f = 50/32.06 s    (revolutions over time)

                     = 1.56 Hz

linear speed

v = 2pi*r*f so, v = 2pi(0.18542)(1.56)

                         = 1.817 m/s

calculated cf

F = mv^2/r so, F= (0.4745mg)(1.817 m/s)^2/0.18542m

                        = 8.444 N

Measured cf 8.428 N

Percent difference

(8.444-8.428/8.428)* 100 = 0.19%

Force Diagram 

Tension force

  *net force = 0

gravity

Conclusion: 

The lab shows that while keeping the radius constant, the centripetal force becomes lower as the mass of the object increases. The results that we obtained were closed to the measured centripetal force. The percent difference for most of them was between 0.19% and -7.1%. There was one trial where we got a percent difference that was high (-23.7%) but we kept it as an example of source of error. The reason for such a difference is because of confusion between the number of rotations and time, and the revolutions not being constant. Some of the sources of error which we came across were that at moments the revolutions didn't seem as if they were going at a constant velocity. When the revolutions are too fast the object overpasses the radius generating some different results.Another source of error is not remembering to use unit conversions if you used something else than meters..

Drag Force on a Coffee Filter

Introduction:
This lab studies the relationship between air drag force and the velocity of a falling body. It also looks into the velocity dependence of the drag force. The drag is created in the opposite direction of the objects motion. Using a motion sensor and the graphing software we can gather the terminal velocities from the graphs generated by Logger Pro.

Procedure:
1.  Gather nine coffee filters, and make sure the shape stays the same. If the shape is altered for some it will create more/less surface area than others, compromising the results. Place the motion detector on the floor facing upward and hold the nine filters 1.5 m above the sensor. Drop the filters to collect the data and a graph will be shown with a constant decreasing slope. As the time increases the position decreases because the filters are falling from 1.5 m to 0 m. On the graph select the portion where the graph shows the constant speed.

2.  The portion of the graph that was selected can be analyzed by selecting curve fit. Linear curve is the shape of the graph and once selected it gives you the slope. Record the slope because it represents the average velocity of the falling filters. Repeat the steps but every 4 trials remove one filter. Table 1.1 displays the slopes and average slope of each coffee filter trial.

3. Curve fit the linear graph. The slope of this line represents the terminal velocity.

4. How does your value of n compare with the value given in the text? what does the other fit parameter represent? Our n value equal 1.9, the book's value is 2. This value represents the cross section of the air filters. 


Table 1.1

Note: unit mesurement of m/s for the trials and average.

5. Make a graph that plots packet weight vs terminal speed. Perform a power fit to the data and record the power n given by the computer. Check the percent error between your n and the theoretical n. Table 1.2 shows the graph of number of filters vs terminal speed.


 Table 1.2

                     

               filters vs terminal speed

Conclusion:

This lab helped explain that the drag force on a falling object eventually equals the magnitude of the gravity force with an Fnet of 0. At this point there is no further acceleration and the object falls at a constant speed. Once this object reaches the terminal speed it will continue to fall constantly until it hits the ground or another external force is applied.The most common sources of error that we came across was that if there's and object or movement near the motion detector it can compromise the results.

Working with Spreadsheet

Introduction:
The lab's main purpose is to familiaize ourselves and learn new techniques for using the electronic spreadsheet.

Procedure:

• Using Microsoft Exel create a spreadsheet that calculates the values of the function f(x)=Asin(Bx+C) using the following values: A= 5, B = 3, C =  π/3. These values are the constants and should be labeled with amplitude, frequency, and phase. Make a colum for 'x' and one for 'f(x)'; for the first x value enter a zero and for f(x) enter the given formula. The x colum should run from 0 to 10 with an incriment of 0.1 radians. Using the formula have the spreadsheet calculate the all x values by dragging the select box from the cell with the formula to the last cell (Figure 1).
• Copy the calculated values and after loading the graphical analysis application paste the data on to the program. This should show the graph for the value (name the axis).
• Select a portion of the graph to be analyze and find the best curve fit that fits the function (sin function). After the functions has been fit a box displays a value for A, B, and C that fit the plotted graph. These values that are displayed on the box should match the constant values that were assigned to the function (Figure 2).  
• Repeat the above process to calculate the position vs time values of a free falling object. The constants for the function are g = 9.8 m/s^2, Vo = 50 m/s, and Xo = 1000m and change in time t = 0.2. Analyze the graph and curve fit it to a quadratic function (Figure 3). 

 Figure 1

On the left, the table for the sine function is shown. The right side shows the position vs time values for a free-falling particle. 

 Figure 2

The selected portion of the graph was analyzed and the curve fit was a sine function. The values given by the curve fit are the same as the constant values of the original function.

   Figure 3

Position vs Time graph for a particle with an initial position of 1000m

Conclusion:
The lab was excellent practice for using spreadsheets to calculate values for a given function. Using the spreadsheet is a great method to calculate since one can plug the formula, constant values, and solve for x. For example, on the first portion of the lab we have function, the x values (0-10 incrementing by 0.1), and by saving the constants on the formula and dragging the select box down all the values are calculated by Excel. An experimental error which we came across was that our free falling particle graph was showing a different graph because on the process of calculating we got stuck on a loop that would only calculate 3 values repeatably. We solved the issue by re-assigning the constants and incrementing our time once more.After we curve fit our data using the quadratic equation since it was the closest to the graph we obtained. The value A represents the initial position of the object, B represents velocity, and C is half of gravity.

Vector Addition of Forces

Introduction:

This lab concentrates on the study of vector addition by graphing and by using the components. After all the calculations have been done a circular force table can be used to check the results.

Materials:
Circular force table, masses plus holders, string, protractor, and four pulleys.

Procedure:
1.) Gather the given magnitudes and angles and use the scale of 1 cm = 20 g. Then make a vector diagram showing the forces and find their resultant. Determine the magnitude and the direction of the resultant force using a ruler and protractor.  

2.) Make a second vector diagram and show the same three forces.Find the resulting vector but this time using components. On the diagram show both the resultant and the components plus the force that is required to cancel out the resultant.


3.) Mount the three pulleys on the edge of the force table at the given angles. Attach the strings to the center ring so that each run over the pulley and attach the mass holder as well. Hang the appropriate masses on the strings. Set up the fourth pulley at 180 degrees opposite from the angle calculated for the resultant of the first three vectors (record the mass and the angles). After adding the weights to the fourth pulley you should notice that the ring is at equilibrium if the resultants are correct.


4.) Check the results on the simulation. Add the vectors and obtain the resultant.







Vector Diagram

The diagram shows vectors A, B, C and D as the sum. (1 cm = 20 g)

Second Vector Diagram






































Vector Calculations:

Rx = (100Cos(0)) + (200Cos(71)) + (160Cos(144)) = 36.1

Ry = (100Sin(0))  + (200Sin(71))  + (160Sin(144))  = 283

ø = 82.7°              R = sqrt(283^2 + 36.1^2) = 285.3  

The simulation confirmed the results as correct. *Note: scale to 1=10.

 

Conclusion:
The lab sole purpose of the lab was to add the vectors and to find the counter balance so that the ring in the middle of the circular force table would be at equilibrium. The vector diagram shows the given vectors graphed on quadrant I with D as the resultant; the resultant D is the sum of the three given vectors.
In order to find the magnitude of the D component we had to find the Rx and Ry values, then we solved to find the magnitude of D by squaring the Rx and Ry values and taking the square root of both. To find the angle of the D component we just had to take the inverse tangent of Rx and Ry, giving us 82.7 degrees. Some of the errors that could’ve happed are that when trying to simulate the program with the given link, one could’ve miscalled the magnitude of the vectors resulting with a sum that is incorrect.