ASSIGNMENT
FAKULTI TEKNOLOGI KEJURUTERAAN
MEKANIKAL DAN PEMBUATAN
UNIVERSITI TEKNIKAL MALAYSIA MELAKA
BMMM 2313 DYNAMICS AND MECHANICS OF MACHINE
ASSIGNMENT
Ping-pong Launcher Design Challenge
LECTURER’S NAME MOHD AFDHAL BIN SHAMSUDIN
STUDENT’S NAME MATRIX NO:
S1. PRAVIN KUMAR A/L JEYAKUMAR B091910362
S2. MUHAMMAD SYUKRI BIN AHMAD B091910112
S3. MUHAMMAD LUQMAN HAZIQ BIN ZULKIFLEE B091910433
S4. NUR HAFIZAN BIN ABDUL RAZAK B091910137
S5. MUHAMMAD HAZIQ ZAFRIN BIN ALWI B091910316
INTRODUCTION
If a particle is thrown obliquely near to the surface of the earth, it travels under relentless
acceleration in a circular course toward the middle of the earth (we assume that the particle
remains close to the surface of the earth). A projectile is called the direction of such a particle
and the acceleration is called projectile motion. Air resistance to the body's motion must be
presumed to be lacking in projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
Along the x-axis: uniform velocity, responsible for
the horizontal (forward) motion of the particle.
Along y-axis: uniform acceleration, responsible for
the vertical (downwards) motion of the particle.
Acceleration of horizontal projectile motion and vertical projectile motion of a particle: When
a particle is projected at any speed in the air, acceleration due to gravity is the only force that
acts on it during it is time in the air (g). This acceleration works backward vertically. In the
horizontal direction, there is no acceleration, which implies that the particle's momentum in
the horizontal direction remains unchanged.
MECHANISM
For this assignment project, we must make a ping-pong launcher for do the projectile motion
experiment in real situation. First of all, we will have equipment to proceed the ping-pong
launcher. What we use is the block of wood, rubber band, paper roll pipe, and etc. what we
do is first sketch up the design for ping-pong launcher.
Sketch
After done on sketch product, we start to measure the work piece to cut out. The requirement
on this project is 6.1cm height above ground, so we decide to make all of that into one
complete product. In the measurement, we need to do precision and accurately to prevent
error on experiment start. Every single number on the measurement we take out to calculate
the smoothness while experiment begin.
Final Product
This is final product that has been build using all listed work piece and apparatus. When the
final product has been finished, we must do some testing on this product to determine the
smoothness of ping-pong to launch and the strength of launcher. So far, the test was run with
good condition and run smoothly. It time to do the real task that has been given by our group
to launch the ping-pong ball to the target based on the requirement has been given in the
assignment task.
Set-up
First thing to do is by setting up the work place for do a launching. What we need is the
measure tape, target marker as a marking the ping-pong to hit it and the ping-pong
launcher.by following the task has been given, we set up the distance of 1.5 m for a target
from the ping-pong launcher.
Target Set-up
We set up the target at 1.5m from the ping-pong launcher. By using the measuring tape,we
can set the distance accurately and it more easier to calculate when to launching the ping-
pong boll to hit the target.
Angle of Launcher
The launcher must have angle to launch the ping-pong ball thru the target. So, by using mark
on the paper roll and stand, we can find out what angle is suitable for perfect launch of the
ping-pong ball to hit the target perfectly. What we can see bottom of the paper roll pipe the
stand holding the pipe is the angle stand. Function of angle stand is used for set the angle of
pipe launcher and hold at that selected angle position by using angle protector ruler.
Ready for launch
When all set-up was ready in all position, it time for launch the ping-pong ball to the target.
The rubber band pulled and hook to the trigger and anytime the trigger will released the
rubber band and start to push the ping-pong ball.rubber band we set-up with specific length
and strength on it.the length of rubber band is important in this project.
Relation with dynamic & mechanic machine
In the subject we learn about dynamic and mechanic have a topic that called as a projectile
motion. In this topic, Projectile motion is the motion of an object thrown or projected into the
air, subject to only the acceleration of gravity. The object is called a projectile, and its path is
called its trajectory. By doing this project assignment, we can find out the projectile motion is
really happened while doing this experiment. We using acceleration, velocity, angle to hit the
target perfectly. The shape of ball travel we can see it became curve from upward to
downward because of gravity.
Rubber strength
Mass (kg) Force (N) Displacement, x (m)
0 0 0
0.1 0.981 0.008
0.2 1.962 0.009
0.3 2.943 0.010
0.4 3.924 0.011
0.5 4.905 0.013
Graph of F against x
5.5
4.905
5
4.5
3.924
4
3.5
2.943
3
Force N
2.5
1.962
2
1.5
0.981
1
0.5
0
0
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014
Dispalcement x, m
Rubber length
This is how to determine the length of rubber by using weight and hanging to determine the
length off rubber. The equipment used is bottle as a container of load, rubber as a determiner
on length and water is a load to force downward of rubber. We must scale the load by using
weight meter or indicator. Hanging the bottle with rubber and hang on table. The load will
start pull rubber downward depend on load we put in. take re length reading and load reading.
Calculation
Initial requirement of the project.
I. First, we use this formula to find value Vi (initial velocity) and t (time of flight).
Horizontal Projectile θ=25 °
h = 0 . 061 m
S = So + Vot
d/R = 1.5 m
R = 0 + ( Vo cos ϴ ) t
m = 0.0027 kg
1.5 m = 0 + ( Vo cos θ ) t
2V sin θ
1.5 m = 0 + ( Vo cos θ ) x 2V sin θ
g t= , thus substitute to t
g
2sin θ cos θ to find vo .
1.5 m = 0 + V2 ( g )
2
V sin (2θ) 2 sin θ cos θ=sin(2 θ)
1.5 m = 0 +
g
2
V sin (2(25 °))
1.5 m =
9.81
V2 = 19.209
V = 4.383 ms-1 .
II. Then substitute value of V into 1st equation to get t.
R = 0 + ( Vo cos ϴ ) t
1.5 m = 0 + ( 4.383 cos 25 ° ) t
1.5 m = (3.972) t
t = 0.377 s .
III. Apply spring elongation formula to find x (rubber elongation).
m = kx2 / v2 K = slope of the graph
4.905−0.981
√
m v2 =
x= ( ¿)¿ 0.013−0.008
k
= 784.8 N/m
x=√ ( 0.0027 ) ¿ ¿ ¿
3
x=8.129 . 10 .
Discussion
By definition, a projectile has a single force that acts upon it which is the
force of gravity. From the calculation, we use the horizontal projectile motion
equation because the horizontal distance can be expressed as x=V(t). Vertical
1
distance from the ground is described by the formula y=Vyot- gt2.
2
Conclusion
In order to create the ping pong ball launcher, we first needed to determine
the drag coefficient that would then help us in calculating the angle to launch the ball.
We started off by calculating the theoretical value for the drag parameter and then
performed an experiment determine our experimental value for the drag parameter.
The experimental value for our drag coefficient was 5% away from the theoretical
value that we calculated. The values of the drag coefficient then enabled us to
determine the kinematic equations that enabled us to create a ping pong ball
launcher. It is necessary to account for air resistance as the drag coefficient affects
the trajectories of the ball. The graphs illustrate that the drag coefficient reduces the
distance that the ball travels by about 25% which is a huge margin of error if it is not
accounted for. Therefore, we had to take this into account in order to build a
launcher that will accurately hit the target.
