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
2 answers
correct.
Force exerted by floor on egg
1 answer