An egg rolls off a kitchen counter and breaks as it hits the floor. The counter is 1.0 m high, the mass of the egg is about 50 g, and the time interval during the collision is about 0.010 s.
How large is the impulse that the floor exerts on the egg?
use equations of motion to find the velocity just before it hits the floor:
Vf^2 = Vi^2 + 2gx
Final velocity = 4.42m/s
Impulse is change in momentum so:
m(Vf - Vi) = 0.05(0 - 4.42)
= - 0.221 kg.m/s
correct.
Force exerted by floor on egg
The impulse that the floor exerts on the egg is approximately -0.221 kg·m/s.
To find the impulse that the floor exerts on the egg, we need to calculate the change in momentum of the egg during the collision.
The equation to calculate impulse is:
Impulse = change in momentum
First, let's find the final velocity (Vf) of the egg just before it hits the floor using the equations of motion. The equation is:
Vf^2 = Vi^2 + 2gx
where Vf is the final velocity, Vi is the initial velocity (which we assume to be zero because the egg is at rest on the counter), g is the acceleration due to gravity (approximately 9.8 m/s^2), and x is the vertical distance the egg falls (1.0 m).
Substituting the values into the equation, we get:
Vf^2 = 0^2 + 2 * 9.8 * 1.0
Vf^2 = 19.6
Vf ≈ 4.42 m/s
Next, we can calculate the change in momentum of the egg during the collision using the equation:
Impulse = m * (Vf - Vi)
where m is the mass of the egg (about 50 g or 0.05 kg), Vf is the final velocity, and Vi is the initial velocity (which is zero).
Substituting the values into the equation, we get:
Impulse = 0.05 * (0 - 4.42)
Impulse ≈ -0.221 kg.m/s
Therefore, the impulse that the floor exerts on the egg is approximately -0.221 kg.m/s. The negative sign indicates that the impulse is in the opposite direction of the initial velocity.