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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

Respuesta :

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

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abiolataiwo2015

How large is the impulse that the floor exerts on the egg is 22.15 N.

h =Height of the counter=1.0m

m=mass of the egg=50g=0.05kg

t=time interval of collision=0.010s

g=acceleration due to gravity=9.81m/s²

First step is to calculate the velocity of the egg just before hitting the floor

v²=u²+2gh, where u=0

Let plug in the formula

v²=0²+2×9.81×1.0

v²=4.43/s

Second step is to calculate the change in momentum of the egg

Change in momentum=0.05×4.43/s

Change in momentum=0.2215 kg.m/s

Third step is to calculate the force exerted on the egg by the floor using this formula

Force exerted on the egg=Change in momentum/Time

Let plug in the formula

Force exerted on the egg=0.2215/0.010

Force exerted on the egg=22.15 N

Inconclusion how large is the impulse that the floor exerts on the egg is 22.15 N.

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https://brainly.com/question/14425219

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