What is the Average Pressure on an Apple During Impact?

[nhmllr]

Homework Statement

1. A 0.3 kg apple falls from rest through a height of 40 cm onto a flat surface. Upon impact, the apple comes to rest in 0.1 s, and 4 cm² of the apple comes into contact with the surface during the impact. What is the average pressure exerted on the apple during the impact? Ignore air resistance.
(A) 67,000 Pa
(B) 21,000 Pa
(C) 6,700 Pa
(D) 210 Pa
(E) 67 Pa
(the answer is b)

Homework Equations

p = mv
f = Δp/Δt
v = √(2gh)
pressure = force / surface area

The Attempt at a Solution

The velocity the apple is traveling at when it hits the ground is
v = √(2gh) = √(2* 10 * 0.4) = 2.83 m/s
The average force on the apple is
Δp/Δt = (0.3 * 2.83) / (0.1) = 8.48 N
So the average pressure on the apple should be
force/surface area = 8.48/(0.04²) = 5300 Pa

But that is not an answer. What did I do wrong?

 

[Dick]

Homework Statement

1. A 0.3 kg apple falls from rest through a height of 40 cm onto a flat surface. Upon impact, the apple comes to rest in 0.1 s, and 4 cm² of the apple comes into contact with the surface during the impact. What is the average pressure exerted on the apple during the impact? Ignore air resistance.
(A) 67,000 Pa
(B) 21,000 Pa
(C) 6,700 Pa
(D) 210 Pa
(E) 67 Pa
(the answer is b)

Homework Equations

p = mv
f = Δp/Δt
v = √(2gh)
pressure = force / surface area

The Attempt at a Solution

The velocity the apple is traveling at when it hits the ground is
v = √(2gh) = √(2* 10 * 0.4) = 2.83 m/s
The average force on the apple is
Δp/Δt = (0.3 * 2.83) / (0.1) = 8.48 N
So the average pressure on the apple should be
force/surface area = 8.48/(0.04²) = 5300 Pa

But that is not an answer. What did I do wrong?

4 cm^2 is not the same as (4 cm)^2.

 

[nhmllr]

4 cm^2 is not the same as (4 cm)^2.

...Whoops

Thanks