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Expert-verified Found in: Page 203 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Problem 2, what is the speed of the car at (a) point A, (b) point B(c) point C? (d) How high will the car go on the last hill, which is too high for it to cross? (e) If we substitute a second car with twice the mass, what then are the answers to (a) through and (d)? a) The speed of the car at point $A$ is $17.0\text{m/s}$

b) The speed of the car at point $B$ is $26.5\text{m/s}$

c) the speed of the car at point $C$ is $33.4\text{m/s}$

d) Final height $56.7\text{m}$

e) Even if a second car has twice the mass, it is evident that the above results do not depend on mass. Thus, a different mass for the coaster must lead to the same results.

See the step by step solution

## Step 1: Given

i) Mass of car $m=825\text{​\hspace{0.17em}kg}$

ii) The height $h=42\text{\hspace{0.17em}m}$

iii) Gravitational acceleration $g=9.8{\text{m/s}}^{\text{2}}$

iv) Initial velocity ${v}_{0}=17\text{m/s}$

## Step 2: To understand the concept

The mechanical energy ${{E}}_{mec}$ of a system is the sum of its kinetic energy ${}{K}$ and potential energy ${}{U}$.

Formula:

${K}_{2}+{U}_{2}={K}_{1}+{U}_{1}$

$K=\frac{1}{2}m{v}_{0}^{2}$

$U=mgh$

## Step 3: (a) Calculate the speed of the car at point A

At point $A,{U}_{A}=mgh$ and at initial point ${U}_{0}=mgh.$

That means initial position of the car and point $A$ is on same height, so,

${U}_{A}={U}_{0}$

${K}_{0}+{U}_{0}={K}_{A}+{U}_{A}$

${K}_{0}={K}_{A}$

∴role="math" localid="1661168162281" $\begin{array}{c}{v}_{0}={v}_{A}\\ =17.0\text{m/s}\end{array}$

## Step 4: (b) Calculate the speed of the car at point B

The height of point $B$ from bottom $=h/2$

${U}_{B}=\frac{mgh}{2}$

From principle of conservation of mechanical energy,

${K}_{0}+{U}_{0}={K}_{B}+{U}_{B}$

$⇒\frac{1}{2}m{v}_{0}^{2}+mgh=\frac{1}{2}m{v}_{B}^{2}+\frac{mgh}{2}$

We have to find ${V}_{B}$, so rearrange the equation; we can get

${v}_{B}^{2}=2×\frac{\left(\frac{mgh}{2}+\frac{1}{2}m{v}_{0}^{2}\right)}{m}$

$⇒{v}_{B}=\sqrt{2×\frac{\left(\frac{mgh}{2}+\frac{1}{2}m{v}_{0}^{2}\right)}{m}}$

$⇒{v}_{B}=\sqrt{\left(gh+{v}_{0}^{2}\right)}$

$⇒{v}_{B}=\sqrt{\left(9.80×42.0+{17.0}^{2}\right)}$

$⇒{v}_{B}=\sqrt{700.6}$

$⇒{v}_{B}=26.5\text{m/s}$

## Step 5: (c) Calculate the speed of the car at point C

The height of point from bottom $=0$

${U}_{C}=0$

From principle of conservation of mechanical energy,

${K}_{0}+{U}_{0}={K}_{C}+{U}_{C}$

$⇒\frac{1}{2}m{v}_{0}^{2}+mgh=\frac{1}{2}m{v}_{C}^{2}+0$

We have to find ${v}_{B}$ , so rearrange the equation; we can get

${v}_{C}^{2}=2×\frac{\left(mgh+\frac{1}{2}m{v}_{0}^{2}\right)}{m}$

$⇒{v}_{C}=\sqrt{2×\left(gh+\frac{1}{2}{v}_{0}^{2}\right)}$

$⇒{v}_{C}=\sqrt{\left(2gh+{v}_{0}^{2}\right)}$

$⇒{v}_{C}=\sqrt{\left(2×9.80×42.0+{17.0}^{2}\right)}$

$⇒{v}_{C}=\sqrt{1112.2}$

$⇒{v}_{C}=33.4m/s$

## Step 6: (d) Calculate how high will the car go on the last hill, which is too high for it to cross

To find the “final” height, we set ${K}_{f}=0.$ In this case, we have

${K}_{0}+{U}_{0}={K}_{f}+{U}_{f}$

$⇒\frac{1}{2}m{v}_{0}^{2}+mgh=\frac{1}{2}m{v}_{B}^{2}+mg{h}_{f}$

$⇒{h}_{f}=h+\frac{{v}_{0}^{2}}{2}g$

$⇒{h}_{f}=42.0+\frac{{17}^{2}}{2}×9.80$

$⇒{h}_{f}=56.7\text{m}$

## Step 7: (e) Calculate the answers to (a) through and (d) if we substitute a second car with twice the mass

From the formula, we can see that the results derived above are independent of mass. So change in mass would not change the values found above. ### Want to see more solutions like these? 