A 5.0-g egg falls from a counter 90 cm high and breaks upon hitting the floor. What is the impulse exerted on the egg by the floor?

To find the impulse exerted on the egg by the floor, we need to determine the change in momentum as the egg strikes the floor.

Firstly, let's find the velocity of the egg just before impact. Assuming the counter is the point of reference, the egg is initially at rest, and is then accelerated by gravity, so we can use the following kinematic equation:

v^2 = u^2 + 2as

where: - v is the final velocity - u is the initial velocity (0 m/s, since the egg is initially at rest) - a is the acceleration due to gravity (9.81 m/s^2) - s is the distance fallen (90 cm = 0.9 m)

Plugging in the values, we get:

v^2 = 0 + 2(9.81 m/s^2)(0.9 m)

v = sqrt(2 * 9.81 * 0.9) v = √(17.658) v ≈ 4.2 m/s (approximate value)

Next, we calculate the momentum p of the egg just before impact using the formula:

p = mv

where: - m is the mass of the egg (5.0 g = 0.005 kg) - v is the final velocity (≈4.2 m/s)

p = (0.005 kg) * (4.2 m/s) p ≈ 0.021 kg*m/s

When the egg hits the floor, its velocity changes from 4.2 m/s to 0 m/s almost instantaneously. Thus, the impulse, which is the change in momentum, is:

impulse = change in momentum = final momentum - initial momentum

Since the egg comes to a stop, the final momentum is 0 kg*m/s. The impulse is equal to the negative of the initial momentum because it’s acting in the opposite direction (upwards, as the floor pushes back on the egg).

impulse = 0 - 0.021 kg*m/s impulse ≈ -0.021 kg*m/s

The negative sign indicates that the impulse is in the direction opposite to the initial momentum. However, when we speak of the magnitude of impulse, we can report it as:

|impulse| ≈ 0.021 kg*m/s

So, the impulse exerted on the egg by the floor is approximately 0.021 kg*m/s.