THE MATH BEHIND THE MACHINE ANSWER KEY
The catapult is our culminating first-semester physics project. With it, we are able to apply many of the skills and
concepts learned throughout the semester. EX: our catapult uses a spring to provide it with energy to fire a
projectile. This spring contains elastic potential energy (EPE) which is converted into the translational kinetic
energy (TKE) and gravitational potential energy (GPE) of a marble projectile when the catapult fires, both forms
of energy we studied in Chapter 8. The marble the catapult fires behaves according to the projectile motion
principles that we studied in Chapter 3. However, we also learned that projectile motion is a combination of two
simpler motions which are vertical motion and horizontal motion that we studied in Chapter 2. Finally, to design,
build, and fire the catapult effectively we will use the skills of science which we learned in Chapter 1.
There are some quantities that we will measure in our catapult and there are others that we will calculate.
Each group of students will measure the following using rulers, protractors, scales:
 l, the length of their catapult throwing arm (in m)
 , the launch angle of the catapult (in degrees),
 m, the mass of the marble projectile (in kg)
 k, the spring constant of their spring (in N/m)
 h, the launch height of your marble (in m)
These measurements, when used in conjunction with some formulas we have studied this
semester will allow us to calculate







the initial velocity, vo (in m/s), at which the catapult fires,
the distance, x (in m), that the marble will travel before hitting the ground
the height (in m)of the marble projectile y(x) at any moment in the marble’s flight,
the total elastic potential energy (EPE = ½kx2) the catapult contains when it is ready to fire,
the kinetic energy (TKE = ½mvo2) of the marble when fired,
its gravitational potential energy (GPE = mgh) of the marble when fired,
the remaining elastic potential energy (EPE = ½kx2) of the spring after the catapult is fired.
MEASURING.
To measure the length of the catapult throwing arm, l, and the launch height, h, of the marble, you will
use a simple meter stick and convert your units to meters. The launch angle, , is found with the
protractor built into the catapult. The mass of the marble projectile, m, is 0.005 kg. The spring constant,
k, of your braided rope catapult is 2000 N/m. All these values are quantities that are measured.
spring constant, k,
in Newtons/meter
length, l, in meters
mass of marble, m,
in kgs

angle, , in degrees
launch height, h, in meters
1
CALCULATING.
ENERGY.
In Chapter 8 we looked at energy considerations. Consider that there are three
important energy states that the catapult is in:
DRAW:
DRAW:
DRAW:
1st STATE.
2nd STATE.
3rd STATE.
Energy = 0.
Catapult is in a relaxed
position.
E in = _EPE_.
Eout = __EPE+ GPE + TKE_.
Catapult is pulled back by the Work
done by the hand which contained CPE
(Chemical Potential Energy). The
work done has given the catapult
__EPE_ (form of energy).
Catapult fires. The chain stops it before the spring
completely releases so the catapult still has some
EPE (form of energy). The marble has gained
height so it has GPE_ (form of energy). The
marble is moving so the catapult also has _TKE_
(form of energy).
The energy formula for the catapult is summarized as follows:
Ein = Eout
__ EPE _ = _ (EPE) + GPE + TKE
With the formulas plugged in:
__½kx2
= __mgh_+ __ ½mvo2
Solve the formula above for vo, your “hediondo” (without plugging in numbers):
½kx2 = mgh + ½mvo2
−mgh
−mgh_______
2
½kx −mgh = ½mvo2
½kx2 −mgh = ½mvo2
½m
½m
2
½kx −mgh = vo2
½m
1 kx2  mgh
2
 vo
1 m
2
Now, given the initial conditions from before that you measured, find the initial velocity of the marble that
is projected from your catapult. (Remember: the mass of the marble projectile, m, is 0.005 kg. The
spring constant, k, of your braided rope catapult is 2000 N/m). Assume the braided rope is stretched x =
0.005 m and the firing height is 0.2 m.
1 (2000 N / m)(0.005 m) 2  (0.005 kg)(9.8 m / s 2 )(0.2 m)
2
 vo = __2.47 m/s_m/s
1 (0.005 kg)
2
2
--------------------------------------------------------------------------SIMPLE MACHINES. In Chapter 8 we looked at simple machines. A catapult is a
not-so-simple machine that involves several simple machines.
MULTIPLE CHOICE QUESTIONS. Choose the letter of the best answer.
1.
___d.____ The catapult is mainly what kind of simple machine?
a. wheel & axle
d. lever
b. screw
e. pulley
c. wedge
f. inclined plane
2. ___a.____ When we built the base of the catapult, to ensure that it was strong we used_____
instead of ______.
a. screws, nails
d. wedges, screws
b. nails, screws
e. both a. and c. are correct
c. screws, wedges
f. both b. and d. are correct
3. ____a.___ Screws are simple machines themselves but at the top end of every screw there is
almost always a _____ and at the tip of every screw there is almost always a _____.
a. wheel & axle, wedge
d. lever, inclined plane
b. wheel & axle, inclined plane
e. wheel & axle, inclined plane
c. inclined plane, wedge
f. inclined plane, lever
4. ____a.____ To twist and tighten the rope that would become our spring we used a ______.
a. wheel & axle
d. lever
b. screw
e. pulley
c. wedge
f. inclined plane
5. ____b.___ The fulcrum of the lever that is in the catapult is where ______.
a. the protractor is attached to the base of the catapult
b. the rope braid wraps around the throwing arm
c. the bottle cap is attached to the throwing arm
d. the rope braid is attached to the base of the catapult
3
-----------------------------------------------------------------------------MOTION FORMULAS. In Chapter 2 & 3 we looked at motion formulas.
The energy formulas from above permit you to get the initial velocity, vo (in m/s), at which the
marble projectile will be fired. Once the marble fires, a different formula is used to find how far (x)
in meters, the marble will go. The formula you will use to find this is related to the formulas we
used in Quarter 1 for vertical and horizontal motion.
The catapult formula is as follows:

 2
4.9

x
y  yo  (tan(90   ) x  2
 v o (cos(90   )) 2  .


MATCHING. Although the catapult formula looks complicated initially, we will find that it is
an appearance only. Let’s take apart this formula and conquer it. Match each variable from
the formula on the left with its definition on the right:
___e.__ 1.
___a.__ 2.
___d.__ 3.
___c.__ 4.
___b.__ 5.
y
yo
x

vo
a.
b.
c.
d.
e.
initial height of the marble (projectile), in m
initial velocity of the marble (projectile), in m/s
launch angle, in degrees
horizontal distance the projectile moves, in m.
height of the projectile at any moment, in m
MULTIPLE CHOICE QUESTIONS. Choose the letter of the best answer.
1.
___d.___ The catapult formula looks most like which of the formulas we worked with in Quarter
1?
a. x = vxot
d. y = yo + vyot – ½gt2
b. x = vxot + ½at2
e. y = yo – ½gt2
2
c. x = vxot – ½at
2. ___e.____ One BIG difference between the catapult formula and the one we worked with in
Quarter 1 is that one particular quantity is missing from the CATAPULT formula
which is:
a. initial velocity
d. gravity
b. initial height
e. time elapsed
c. height thrown
3. ____b.____ Another BIG difference between the catapult formula and the one we worked with in
Quarter 1 is that the quantity from Quest. Nr. 2 has been replaced by what quantity?
a. initial height
d. initial velocity
b. horizontal distance the projectile moves
e. gravity
c. time elapsed
4.
___d.____ The number “4.9” that appears in the catapult formula comes from:
a. the spring constant
d. ½ × gravity (½g)
b. the initial height
e. it is not clear just by looking
c. the initial velocity
4

4.9
Look closely at this catapult formula: y  y o  (tan(90   ) x   2
 v o (cos(90   )) 2

Simplified, the catapult formula actually looks like this:
y = yo +
 2
x


a x – bx
2

4.9
where a = tan (90 − ) and b =  2
 v o (cos(90   )) 2


.


5. ___b.____ At first the catapult formula looks like a lot of mumbo-jumbo but if we look at it
carefully, we can see that it is simply the formula of a __________.
a. line
b. parabola
c. circle
d. ellipse
e. hyperbola
-------------------------------------------------------------------------------------------------------------Realize that you will measure your launch angle  with the protractor and you will have found
the initial velocity vo from the energy formula as shown above so that a and b are just numbers
that you will put in front of x and x2 in the formula.
EXAMPLE: Suppose that the launch angle of your catapult  = 40 and the initial height, yo,
of the launch arm is the same as before 0.2 m, and your initial velocity,vo, that you got from your
energy formulas was vo = 2.47 m/s. Plug in your values into your catapult formula and you get:

 2

 2
4.9
4.9
 x  y  0.2  (tan(50) x  
x
y  0.2  (tan(90  40) x  
2
2
2
2
 2.47 (cos(90  40)) 
 2.47 (cos(50)) 





 2
4.9
x
y  0.2  (1.191753593) x  
2 
 6.08  (0.6427876097) ) 

 2
4.9
 x
y  0.2  (1.191753593) x  
 6.08  (0.4131759112)) 
4.9

 2
y  0.2  (1.191753593) x  
x
 2.51210954 
y  0.2  (1.191753593) x  1.950551885x 2
y = 0.2 + 1.191753593x – 1.950551885x2
Graph the formula on your graphing calculator. Draw what it
looks like here at right.
a. Looking at the formula, how do you know that the parabola
opens downward? The negative sign in front of the “x2”
term.
b. Label the y-intercept on your graph (x, y). The y-intercept
represents the _initial height_ the marble is thrown at.
c. What does the vertex of the parabola represent in real life in
terms of our catapult? ___the maximum height of the marble____
d. What does the positive x-intercept of the parabola represent in real life in terms of our
catapult? ___the place the marble lands, the horizontal throwing distance___.
5
FINDING THE MAXIMUM HEIGHT OF THE MARBLE. To find the
maximum height, graph the catapult formula in your graphing calculator, select CALC, and find the
maximum.
EX: Using the same values as before, we could graph the following.
Plot 1 Plot 2 Plot 3
2
2
2
\Y1 = 0.2 + xtan (50) – 4.9x /((2.47 *(cos(50)) )
\Y2 =
\Y3 =
\Y4 =
\Y5 =
…and we would get the following
graph:
\Y1 =
2nd
TRACE
To find the maximum, hit CALC which is
and choose 3: maximum, which, in this case is y = 0.38 m_.
Draw the coordinates of the maximum height (x, y) as
they appear on your calculator screen on the graph at
left…
X=.30549047 Y=0.38203523
FINDING HOW FAR THE MARBLE WILL GO. To find the distance the marble
should go, graph the catapult formula in your graphing calculator, select CALC, and find the zero.
EX: Using the same values as before, we would graph the following.
Plot 1 Plot 2 Plot 3
\Y1 = 0.2 + xtan (50) – 4.9x2/((2.472*(cos(50))2)
\Y2 =
\Y3 =
\Y4 =
\Y5 =
\Y1 =
TRACE
To find the maximum, hit CALC which is 2nd
and choose 2: zero. Find the zero of your parabola
which, in this case is x = _0.75 m__ .
Draw the coordinates of the horizontal distance
travelled (x, y) as they appear on your calculator
screen on the graph at left…
X=.74805219
Y=0
This is theoretically how far (in meters) your catapult
should throw when you shoot it the first time.
EN
THE
6
D