Send the link below via email or IMCopy
Present to your audienceStart remote presentation
- Invited audience members will follow you as you navigate and present
- People invited to a presentation do not need a Prezi account
- This link expires 10 minutes after you close the presentation
- A maximum of 30 users can follow your presentation
- Learn more about this feature in our knowledge base article
Egg Drop Physics Project
Transcript of Egg Drop Physics Project
Egg Drop Project
Unfortunately, our drop was unsuccessful. We disregarded air resistance and did not factor in the effects of the center of mass being at the top of the device where we put the egg. Because the center of mass was so high, the gravitational force caused the device to turn in midair and land on its side. Therefore, the levers did not come into play because the device did not land right side up.
Description - The Physics Behind It
The Physics Behind the Drop
The egg-holding device has momentum while falling, because
In order to protect the egg, we created a cage to
prolong the time of impact, thus reducing the
force. This is because impulse is equal to force
times time, so the longer the time, the lesser the
force has to be. In order to achieve this, we
created small "levers" out of bent - but not
broken - toothpicks. This created a spring effect.
The "levers" would bend upon impact and increase
the time of the impulse.
it has mass and velocity. Its velocity increases due to an
acceleration from gravity. Thus the device's momentum
increases as it falls. But then, when it hits the ground, its
momentum becomes zero. This is because the ground exerts
an impulse on the device. Impulse is equal and opposite to the
momentum right before it hits the ground. The impulse causes
the velocity to become zero. Only the velocity affects the change
in momentum, because the mass stays constant.
2. Find t using the formula y = 1/2 * g * t^2 when Vyi is 0 m/s.
y = 1/2 * g * t^2
2.95 m = 1/2 * 9.8 m/s^2 * t^2
2.95 = 4.9 * t^2
0.602 = t^2
t = 0.77 s
3. Find Vyf using the formula Vyf = gt + Vyi
Vyf = gt + Vyi
Vyf = 9.8 m/s * 0.77 s + 0 m/s
Vyf = 6.47 m/s
4. Find momentum of device upon impact with the floor and final momentum using P = m * v
P upon impact = m * V upon impact P final = m * V final
P upon impact = .8654 kg * 6.47 m/s P final = .8654 kg * o m/s
P upon impact = P initial = 5.59 kg * m/s P final = o kg * m/s
1. Find the impulse exerted by the floor using the formula P initial + Impulse = P final
Pi + I = Pf
5.59 kg * m/s + I = 0 kg * m/s
I = -5.59 kg * m/s or N
1. Measure change in
y: 2.95 m.
By our group
Pieces of the Whole
Physics is Phun!
However, we did achieve having our egg be securely in the device. It fit snugly in the holder and did not bounce out upon impact.
We would choose to redesign our own device with a few improvements. Firstly, we would make a larger base of support. This is because we realized that even if the device landed right side up, it may have tipped over because its area of support was not large enough. We would also try to balance out the top-heaviness caused by the egg by adding more toothpicks to the bottom simply to concentrate mass there. This would hopefully counteract its tendency to turn in mid-air
We would add more toothpicks here (along the bottom).
We would create a wider base, or area of support, for the device to land on.
Egg goes in here!