Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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 $17.0 m/s$

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

c) the speed of the car at point $C$ is $33.4 m/s$

d) Final height $56.7 m$

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

i) Mass of car $m = 825 kg$

ii) The height $h = 42 m$

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

iv) Initial velocity $v_{0} = 17 m/s$

Step 2: To understand the concept

The mechanical energy $E_{m e c}$ of a system is the sum of its kinetic energy $K$ and potential energy $U$ .

Formula:

$K_{2} + U_{2} = K_{1} + U_{1}$

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

$U = m g h$

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

At point $A, U_{A} = m g h$ and at initial point $U_{0} = m g h .$

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" $\begin{matrix} v_{0} = v_{A} \\ = 17.0 m/s \end{matrix}$

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

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

∴ $U_{B} = \frac{m g h}{2}$

From principle of conservation of mechanical energy,

$K_{0} + U_{0} = K_{B} + U_{B}$

$\Rightarrow \frac{1}{2} m v_{0}^{2} + m g h = \frac{1}{2} m v_{B}^{2} + \frac{m g h}{2}$

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

$v_{B}^{2} = 2 \times \frac{(\frac{m g h}{2} + \frac{1}{2} m v_{0}^{2})}{m}$

$\Rightarrow v_{B} = \sqrt{2 \times \frac{(\frac{m g h}{2} + \frac{1}{2} m v_{0}^{2})}{m}}$

$\Rightarrow v_{B} = \sqrt{(g h + v_{0}^{2})}$

$\Rightarrow v_{B} = \sqrt{(9.80 \times 42.0 + {17.0}^{2})}$

$\Rightarrow v_{B} = \sqrt{700.6}$

$\Rightarrow v_{B} = 26.5 m/s$

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,

$K_{0} + U_{0} = K_{C} + U_{C}$

$\Rightarrow \frac{1}{2} m v_{0}^{2} + m g h = \frac{1}{2} m v_{C}^{2} + 0$

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

$v_{C}^{2} = 2 \times \frac{(m g h + \frac{1}{2} m v_{0}^{2})}{m}$

$\Rightarrow v_{C} = \sqrt{2 \times (g h + \frac{1}{2} v_{0}^{2})}$

$\Rightarrow v_{C} = \sqrt{(2 g h + v_{0}^{2})}$

$\Rightarrow v_{C} = \sqrt{(2 \times 9.80 \times 42.0 + {17.0}^{2})}$

$\Rightarrow v_{C} = \sqrt{1112.2}$

$\Rightarrow v_{C} = 33.4 m / s$

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

$K_{0} + U_{0} = K_{f} + U_{f}$

$\Rightarrow \frac{1}{2} m v_{0}^{2} + m g h = \frac{1}{2} m v_{B}^{2} + m g h_{f}$

$\Rightarrow h_{f} = h + \frac{v_{0}^{2}}{2} g$

$\Rightarrow h_{f} = 42.0 + \frac{17^{2}}{2} \times 9.80$

$\Rightarrow h_{f} = 56.7 m$

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.

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Fundamentals Of Physics

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Short Answer

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)?

Step by Step Solution

Step 1: Given

Step 2: To understand the concept

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

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

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

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

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

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Fundamentals Of Physics

Answers without the blur.

Short Answer

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)?

Step by Step Solution

Step 1: Given

Step 2: To understand the concept

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

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

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

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

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

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