• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22

Crows dropping nuts

Extracts from this document...

Introduction

Crows dropping nuts

Task:

  1. Introduction:

Crows love dropping nuts but their beaks are not strong enough to break the nuts open. To crack open the shell they will repeatedly drop the nut on a hard surface until it opens. In this investigation we will determine a possible function that would model the behaviour of the data given below.

  1. Data table given:

Height of drops (m)

1.7

2.0

2.9

4.1

5.6

6.3

7.0

8.0

10.0

13.9

Number of drops

42.0

21.0

10.3

6.8

5.1

4.8

4.4

4.1

3.7

3.2

  1. Definition of variables as well as constraints:
  • Variables: Representing the Number of Drops and Height of Drops,

Let ‘h’ be the height above ground (in metres) and ‘n’ be the number of drops favourable outcome (nut crack). Whereby ‘h’ is the independent variable and ‘n’ is the dependent variable. Therefore all graphs will be plotted as ‘h’ (height of drops) and ‘n’ (number of drops).

Another variable that should also be accounted for is the size of the nut ‘Sn’. It was stated that the average number of drops will also be investigated using medium and small nuts. Therefore ‘Sn’ will be used to illustrate the size of the nuts.

  • Constraints: Representing the boundary values and types of numbers for h and n,

h is a positive integer such that: 0 < h. Height is a displacement measure, it tells you the vertical displacement of an object from a ‘ground’ position. For this data it is assumed that h = 0 is the ‘hard surface’ whereby the nut impacts.

 n ≥ 1 because it must take at least one drop for a nut to successfully crack from a very high height. For one particular trial the number of drops it takes for a nut to break must take an integer value. For example, the nut cannot break on 1.4 drops; it would either break on the first or second attempt.

...read more.

Middle

image07.png

Thus after having calculated the motion of the nut under constant acceleration, therefore it was also important to calculate the momentum of the nut, as this is the quantity that is conserved under an inelastic collision.  (Citation http://en.wikipedia.org/wiki/Momentum)

image08.pngimage08.png

p = momentum      

m = mass

v = velocity

p = (kgms-1)

m = (kg)

v = (ms-1)

Hence knowing the velocity of the nut from the previous equation, the following was deduced;

image09.png

Thus it was assumed that the impact of the nut will be proportional to the momentum,

image11.png

However as the nut would bounce off the surface not all the force of the impact with the ground go into deforming the nut (breaking) (Citation http://en.wikipedia.org/wiki/Deformation).

Therefore another force is introduced,

image12.pngimage12.png  where image13.pngimage13.png represents the percentage of force that is absorbed by the nut on impact, and is why the notation image13.pngimage13.png is chosen as it is the constant of deformation in this model. Hence the range is:image14.pngimage14.png.

The following is a representation of the formula used to calculate the impact of the nut, whereby, image15.pngimage15.png reflects the force of impact. Therefore by substituting the previous expression for momentum ( image17.pngimage17.png ) into the formula for ‘deformation of the nut’. The next step was to calculate the required number of drops for the nut to crack.

image18.pngimage18.pngBut, image19.pngimage19.png

Therefore the image20.pngimage20.png = image21.pngimage21.png will be substituted for image23.pngimage23.pngin the equation below,

image24.png

It is evident that a nut will break when it is exposed to enough impact force. There will be a threshold were after sufficient impact force the shell cannot contain the energy caused by impact and will fracture to conserve energy.

image25.pngimage25.png where by image26.pngimage26.png

For each drop the nut will get weaker (i.e. more deformation to its atomic/molecular lattice). Hence the following equation is a representation of the number of impacts,

image27.png

Therefore when,  image28.pngimage28.png

...read more.

Conclusion

It can be assumed through the results above that so long as the data being collected maintains an accurate curve with minimal variation; the reciprocal power function will fit the data quiet accurately. However if there was variation within the data as found in the small nut, hence the equation will not correctly fit the data.  However in the first equation (reciprocal function) it was proven that the model was not an accurate fit to the data. Whereby the curve of the data resulted in having a negative value. Hence the equation needed to be modified for a better fitEven though after modifying the data and forcing a horizontal asymptote to better fit the data points it appeared that the model still did not fit the data accurately, and hence the equation needed to be changed.However if all the parameters were modified as shown in the graphs above it appeared that the equation fits the data quiet well. However this equation has many limitations. If the curve of a particular data (other than the ones investigated) had a steeper slope thus the equation would not obtain an accurate fit. Moreso if there was more data calculated, thus it would be safe to assume that the equation after a certain amount of time would completely miss the data, because it apparent that the curve is constantly descending, thus after a specific period of time the curve will cross into the negative value. Whereby the equation has reached its utmost limit.  
  • However as for this investigation it can be concluded that a reciprocal function is not as accurate as a reciprocal power function.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. A logistic model

    2.9 ? (?3.8 ? 10?5 )(1? 104 ) ? 3.28 Hence the equation of the linear growth factor is: r ? ?3.8 ?10?5 u ? 3.28 {10} n n Using equations {1} and {2}, one can find the equation for un+1: 11 IB Mathematics HL Type II Portfolio: Creating a logistic model International School of Helsingborg - Christian Jorgensen un?1 ?

  2. Maths IA Type 2 Modelling a Functional Building. The independent variable in ...

    40 100 50 72 2.5 9000 3600 41 102.5 44.721 72 2.5 8049.845 3219.938 42 105 38.730 72 2.5 6971.370 2788.548 43 107.5 31.623 72 2.5 5692.100 2276.840 44 110 22.361 72 2.5 4024.922 1609.969 45 112.5 0 72 2.5 0 0 Total: 795688.274 318275.310 Now, after determining the total

  1. Crows Dropping Nuts

    The first 5 points are: Height of drop 1.7 2 2.9 4.1 5.6 Number of drops 42 21 10.3 6.8 5.1 [A]= and [B]= so, [C] = [B]-1 x [A] [C]= = 4.46781392a4+ -66.20812432b3+353.927455c2+-812.134389d+687.7474501. I graphed the equation using a GDC: This function modeled all of the first 5 data

  2. matrix power

    You then multiply the pairs together and the sum of the products will give a single number which is the first digit of the new matrix. Therefore the matrix equation being solved will look like: Another easier way of solving matrices powers is raise the power of the digits inside

  1. Creating a logistic model

    y = Putting this data into a graph, we have: Model for Growth Rate = 2.5 When we have a new initial growth rate, the ordered pairs i.e. (un, rn) have now changed to: (60000, 1) (10000, 2.5) To find the linear growth factor, we form the two equations: 2.5 = m(10000)

  2. Population trends. The aim of this investigation is to find out more about different ...

    The data given shows, as a starting point the year 1950, it then continues for another 45 years of data which has been given in multiples of 5. Year Population in millions 1950 554.8 1955 609.0 1960 657.5 1965 729.2 1970 830.7 1975 927.8 1980 998.9 1985 1070.0 1990 1155.3

  1. The investigation given asks for the attempt in finding a rule which allows us ...

    Based on the Wolfram alpha online integrator, 134x3-23x2+40x-18=4.66667≈4.667. Therefore, the different between the gained answer and the actual value is 0.021. Comparison of Results Calculated Value Actual Value % Error Function 1 1.980 1.981 0.05 Function 2 8.246 8.260 0.169 Function 3 4.688 4.667 0.450 Percentage Error = Calculated-ActualActual × 100% Based on the obtained percentage errors, the

  2. The aim of the assignment is to examine the gold medalist heights for the ...

    and at (1948, 198). A cubic function is effective for this set of data specifically because of the concave up and down shape, although if it were for Gold Medal High Jump results in the future, a logarithmic model as it illustrates humans infinitely increasing ability.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work