Question

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

a) 0.45 Ns
b) 0.0495 Ns
c) 4.9 Ns
d) 0.00045 Ns

Answer 1

To calculate the impulse exerted by the floor on the egg, we use the impulse-momentum principle. We calculate the initial and final velocities of the egg and then use the equation J = m(vf - vi) to calculate the impulse.

Step-by-step explanation:

To calculate the impulse exerted by the floor on the egg, we need to use the impulse-momentum principle. The impulse (J) is equal to the change in momentum (Δp) of the egg.

First, we need to calculate the initial velocity (vi) of the egg before it falls. We can use the kinematic equation: (vi)² = (vf)² - 2gΔy, where g is the acceleration due to gravity and Δy is the height. Plugging in the values, we get vi = √(2gΔy).

Next, we calculate the final velocity (vf) when the egg hits the floor. The final velocity is given by vf = √(vi)² + 2gΔy, where Δy is the height and g is the acceleration due to gravity. Plugging in the values, we get vf = √(2gΔy).

Finally, we can calculate the impulse (J) using the equation J = m(vf - vi), where m is the mass of the egg. Plugging in the values, we get J = 5.0g(√(2gΔy) - √(2gΔy)).