How Does Physics Explain a Fly Sticking into Cake Icing?

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AI Thread Summary
The discussion revolves around calculating the forces and displacement involved when a fly lands on cake icing. The average force exerted on the fly during the collision is calculated to be -0.005 N, with the cake experiencing an equal and opposite force of 0.005 N due to Newton's third law. The acceleration of the fly is determined to be -5 m/s², leading to a calculated dent in the icing of 0.00025 m. The calculations are confirmed to be correct, and an alternative method using the work-energy theorem is mentioned for part C. Overall, the physics principles applied effectively explain the interaction between the fly and the cake icing.
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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
 
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All looking good. FYI, part C could also have been attacked with the work-energy theorem.
 
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