How Does Physics Explain a Fly Sticking into Cake Icing?

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SUMMARY

The discussion focuses on the physics of a fly colliding with cake icing, specifically calculating the average force on the fly and the cake, as well as the depth of the icing dent. The average force on the fly during the collision is calculated to be -0.005 N, while the force exerted on the cake is 0.005 N, in accordance with Newton's third law. The depth of the dent in the icing is determined to be 0.00025 m using kinematic equations. The calculations utilize principles of momentum, impulse, and acceleration.

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  • Understanding of Newton's laws of motion
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Homework Statement


A fly traveling horizontally in the positive x direction lands on the side of your slice of birthday cake and gets stuck in the icing. Just before striking the cake the fly is traveling at 0.05ms-1 and the fly has a mass of 1 gram.

a) If the collision lasts 10ms, what is the average force on the fly during the collision?
b) What is the average force on the cake?
c) How far is the icing dented, assuming the force during the collision is constant?

Homework Equations



FΔt=Δp=Δmv
F=ma
Xf=Xi+Vit+0.5at2

The Attempt at a Solution



Hey guys :D

I was wondering if my answers are right, it seems correct to me but I'm not too sure...

A)
So i just find the change of momentum which is Δmv = 0 - (0.001*0.005) = -0.00005kgms-1

Next, I substitute the change in momentum into the impulse formula and find the force exerted on the fly by the cake

F = -0.00005/0.01 = -0.005N

B)

From Newton's third law, the average force on the cake exerted by the fly is just the negative of the force on the fly exerted by the cake: 0.005N

C)

This one i had some trouble, but i pretty sure it's correct...

From the calculated force on the fly from question A, we can use the formula F=ma to determine the acceleration

a=F/m=-0.005/0.001= -5ms-2

From this acceleration, we can determine the how far the fly went inside the cake by using the kinematic equation Xf=Xi+Vit+0.5at2

Xf=0+0.05(0.01)-0.5(5)(0.01)2=0.0005-0.00025=0.00025m
 
Last edited:
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All looking good. FYI, part C could also have been attacked with the work-energy theorem.
 

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