A surprising demonstration involves dropping egg from a third-floor window so that the egg lands on a foam-rubber pad without breaking. If a 56-g egg falls (from rest) 12m, and the 5cm thick foam pad stops it in 6.25ms, by how much is the pad compressed? (NOte: assume constant upward acceleration as the egg compresses the foam-rubber pad)

Calculate the speed V = sqrt (2gH) of the egg when it hits the pad. I get 15.3 m/s. Multiply that by the mass M to get the momentum, MV = 0.856 kg m/s. The average upward force on the egg, Fav, multiplied by the time to bring it to a stop (6.25*10^-3 s) equals the momentum. Therefore Fav = 137 Newtons.

The compression equals the average velocity during deceleration, 7.65 m/s multiplied by the time of contact,6.25*10^-3 s. I get 4.8 cm, just a bit less than the thickness of the pad. A thinner pad would not have worked.

Well, this is quite an "egg"citing question! Let's go ahead and "crack" it.

Given:
Mass of the egg (m) = 56 g = 0.056 kg
Initial velocity (u) = 0 m/s (since it falls from rest)
Distance (s) = 12 m
Time taken (t) = 6.25 ms = 0.00625 s

Now we know that the downward force (weight) acting on the egg is given by the equation F = m * g, where g is the acceleration due to gravity.

Using Newton's second law (F = m * a), we can also say that the force exerted by the foam pad (which is equal to the force required to stop the egg) is given by F = m * a.

Since the force exerted by the pad is equal to the weight of the egg, we can equate the two equations:

m * a = m * g

Canceling out the mass (m) on both sides, we get:

a = g

This means that the acceleration experienced by the egg is equal to the acceleration due to gravity (9.8 m/s^2).

Now, let's calculate the compression of the pad. To do this, we need to find the final velocity (v) of the egg when it hits the pad.

Using the equation v = u + a * t:

v = 0 + 9.8 * 0.00625
v = 0.06125 m/s

Since the foam pad brings the egg to a stop, its final velocity (v) is 0 m/s. Hence, we can use the equation v^2 = u^2 + 2 * a * s to find the distance the pad compresses.

0 = 0.06125^2 + 2 * 9.8 * s

Solving for s:

s = - (0.06125^2) / (2 * 9.8)
s = -0.000197 m

But wait a minute, negative compression doesn't make sense here! Since compression is a "squishy" thing, let's take the absolute value:

s = 0.000197 m

So, the pad compresses by approximately 0.000197 meters or 0.197 mm.

That's one resilient foam pad, able to handle an egg falling from such "elevated" heights without cracking under the pressure!

To solve this problem, we can use the equations of motion and the concept of acceleration.

Given:
Mass of the egg (m) = 56 g = 0.056 kg
Distance fallen (d) = 12 m
Stopping time (t) = 6.25 ms = 0.00625 s
Thickness of the foam pad (h) = 5 cm = 0.05 m

Let's solve it step-by-step:

Step 1: Find the initial velocity (u) of the egg when it hits the foam pad.
We can use the equation of motion:
s = ut + (1/2)at^2

Since the egg starts from rest and falls downwards, the initial velocity (u) is 0 m/s.
Thus, the equation simplifies to:
s = (1/2)at^2

Plugging in the values:
12 m = (1/2) * a * (0.00625 s)^2

Step 2: Solve for acceleration (a).
Rearranging the equation:
a = (2 * 12 m) / (0.00625 s)^2

Step 3: Find the final velocity (v) of the egg when it is stopped by the foam pad.
Using the equation of motion:
v = u + at

Since the initial velocity (u) is 0 m/s, the equation simplifies to:
v = at

Plugging in the values:
v = a * 0.00625 s

Step 4: Calculate the average velocity (v_avg) during the 6.25 ms of stopping time.
v_avg = 0.5 * (0 + v)

Step 5: Calculate the compression distance (x) of the foam pad.
x = v_avg * t

Step 6: Determine the actual compression distance considering the thickness of the foam pad.
Actual compression distance = x - h

Substituting the values and calculating each step will give you the final answer.

To find out how much the foam pad is compressed when the egg lands on it, we need to determine the distance the pad is compressed.

We can use the kinematic equation that relates distance, initial velocity, time, and acceleration:

d = vi * t + 0.5 * a * t^2

In this case, we know the initial velocity of the egg is zero (since it starts from rest), and the time it takes for the pad to stop the egg is 6.25 ms or 6.25 × 10^-3 s.

Let's start by finding the acceleration of the egg when it lands on the foam pad. We can use the equation of motion to calculate it:

vf = vi + a * t

Since the final velocity (vf) is zero (egg comes to rest on the pad), the equation becomes:

0 = 0 + a * 6.25 × 10^-3

Simplifying the equation, we find that the acceleration is zero. This means the egg experiences no acceleration after it lands on the foam pad since it comes to rest immediately.

Now, we can use the distance formula to find the compression of the foam pad. Plugging in the known values:

d = 0 * 6.25 × 10^-3 + 0.5 * 0 * (6.25 × 10^-3)^2

Simplifying further, we see that the term "0 * 6.25 × 10^-3" becomes zero, so we are left with:

d = 0

Therefore, the foam pad is not compressed at all when the egg lands on it.

It's important to note that in reality, foam pads compress to some extent when an object is dropped on them. However, based on the given information and assumptions (constant upward acceleration as the egg compresses the foam pad), the answer indicates that there is no visible compression in this scenario.