Question

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. Part AHow large is the impulse that the floor exerts on the egg?Express your answer to two significant figures and include the appropriate units.Part BHow large is the force exerted on the egg by the floor when stopping it?Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

A. 0.22 kg m/s

B. 22.36 N

Explanation:

50 g = 0.05 kg

As the egg is falling from 1m high, its potential energy is converted to kinetic energy at the floor. So if we use the floor as a reference point:

$E_p = E_k$

$mgh = mv^2/2$

where m is the egg mass and h is the vertical distance traveled, v is the egg velocity when it hits the floor, which is what we are looking for

We can start by divide both sides by m

$gh = v^2/2$

Let g = 10m/s2

$v^2 = 2gh = 2*10*1 = 20$

$v = \sqrt{20} = 4.47 m/s$

A. The momentum, and also impulse of the egg when it hits the floor is

$\Delta M = m\Delta v = 0.05*4.47 = 0.22 kgm/s$

B. The impulse caused an impact force which lasted for 0.01s

$\Delta M = F\Delta t = F*0.01$

$F = \Delta M/0.01 = 0.22 / 0.01 = 22.36 N$