# Egg- 50g Cotton Ball- .408g Straws- 1.26g Rubber Bands- 2.18g Time- 2.79 seconds 9.82 meter drop...

## Question:

Egg- 50g

Cotton Ball- .408g

Straws- 1.26g

Rubber Bands- 2.18g

Time- 2.79 seconds

9.82 meter drop

Find the potential energy, kinetic energy, velocity before impact, and velocity half way down. (dropped egg from 9.82 meter hight, straws and rubber bands and cotton ball were used to protect egg)

## Potential and Kinetic Energy:

A body of mass "m" at a height "h" possesses potential energy while it is falling it gains a velocity "v" and has kinetic energy associated with it.

The expressions for the two energies are as follows:

{eq}Potential \ Energy =m\ g\ h\\ Kinetic \ Energy = \dfrac{1}{2}mv^2 {/eq}

Thus we have from the expression of conservation of energy

{eq}\begin{align*} \text{Potential Energy} & = \text{Kinetic Energy} \\ mgh& = \frac{1}{2}mv^2 \\ v & = \sqrt{2gh} \\ \end{align*} {/eq}

We have the following items:

i) Egg- 50g

ii) Cotton Ball- 0.408 g

iii) Straws- 1.26 g

iv) Rubber Bands- 2.18 g

We have a total mass of the items as

{eq}m = (50+0.408+1.26+2.18) \ g = 53.85 \ g = 53.85 \times 10^{-3} \ kg {/eq}

Time of fall is t = 2.79 sec.

The fall in height is h = 9.82 m.

• Before impact we have

1) Potential Energy

{eq}\begin{align*} E_p & =0 &\text{Height = 0} \end{align*} {/eq}

2) Velocity

{eq}\begin{align*} v & = \sqrt{2gh} &\text{See context section}\\ &= \sqrt{2gh'}\\ & = \sqrt{2\times 9.8 \times 9.82} \ m/s\\ & =13.9 \ m/s \end{align*} {/eq}

3) Kinetic Energy

{eq}\begin{align*} E_k & =\frac{1}{2}mv^2 \\ & = \frac{1}{2}\times 53.85 \times 10^{-3} \times 13.9^2 \ J \\ & = 5.18 \ J \end{align*} {/eq}

• At halfway ie at {eq}h' = \dfrac{h}{2} {/eq} we have the

1) Potential Energy

{eq}\begin{align*} E_p' & =mgh' \\ & = 53.85 \times 10^{-3} \ kg \times 9.8 \ m/s^2 \times \frac{9.82}{2} \ J \\ & = 2.59 \ J \end{align*} {/eq}

2) Velocity

{eq}\begin{align*} v' & = \sqrt{2gh'} &\text{See context section}\\ &= \sqrt{2gh'}\\ & = \sqrt{2\times 9.8 \times \frac{9.82}{2}} \ m/s\\ & =9.81 \ m/s \end{align*} {/eq}

3) Kinetic Energy

{eq}\begin{align*} E_k' & =\frac{1}{2}mv'^2 \\ & = \frac{1}{2}\times 53.85 \times 10^{-3} \times 9.81^2 \ J \\ & = 2.59 \ J \end{align*} {/eq}