# Question 5fb0a

May 29, 2017

The maximum speed at the bottom of the drop is $26.4 \frac{m}{s}$

3-The speed at the bottom of the drop is independent of the mass of the car.

#### Explanation:

To find the speed of the car we will need to find how long (seconds) it will fall.

Frictionless fall gravity time is: $t = \sqrt{\frac{2 d}{g}}$

$t = \sqrt{\frac{2 \cdot 35.4 \cancel{m}}{9.81 \frac{\cancel{m}}{s} ^ 2}}$

$t = \sqrt{\frac{70.8 {s}^{2}}{9.81}}$

$t = \sqrt{7.2 {s}^{2}} = 2.7 s$

Then we can determine the falling velocity by: $v = g \cdot t$

$v = 9.81 \frac{m}{s} ^ 2 \cdot 2.7 s = 26.4 \frac{m}{s}$

The gravity equations are here:
https://www.mansfieldct.org/Schools/MMS/staff/hand/lawsgravprac.htm

To calculate the speed of the car at the bottom of the drop if its mass were doubled, we would find that there is no reference to mass in gravity equations. That immediately indicates: "The speed at the bottom of the drop is independent of the mass of the car."
Information on free-falling objects is here:
http://www.physicsclassroom.com/class/1DKin/Lesson-5/The-Big-Misconception

Important Note: If the mass were doubled, there would be a $4 \times K E$ increase in the kinetic energy of the car which would mean it would result in a deeper hole on impact, while still maintaining the same speed of decent.
https://socratic.org/questions/how-does-kinetic-energy-change-when-velocity-is-doubled102211

Jun 2, 2017

Part A

Using Law of Conservation of Energy, kinetic energy at the bottom of drop equals change in potential energy. Ignoring frictional losses
$\frac{1}{2} m {v}^{2} = m g h$
$\implies {v}^{2} = 2 g h$
$\implies v = \sqrt{2 g h}$ ........(1)
Inserting given values and taking $g = 9.81 m {s}^{-} 2$, we get
$v = \sqrt{2 \times 9.81 \times 35.4}$
$v = 26.4 m {s}^{-} 1$, rounded to one decimal place

Part B
3. Evident from equation (1)