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Q9P

Expert-verifiedFound in: Page 203

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.

** **

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 ${v}_{0}=17\text{m/s}$

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

**Formula:**

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

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

$U=mgh$

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" $\begin{array}{c}{v}_{0}={v}_{A}\\ =17.0\text{m/s}\end{array}$

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

∴${U}_{B}=\frac{mgh}{2}$

From principle of conservation of mechanical energy,

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

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

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

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

$\Rightarrow {v}_{B}=\sqrt{2\times \frac{(\frac{mgh}{2}+\frac{1}{2}m{v}_{0}^{2})}{m}}$

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

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

$\Rightarrow {v}_{B}=\sqrt{700.6}$

$\Rightarrow {v}_{B}=26.5\text{m/s}$

The height of point from bottom $=0$

∴${U}_{C}=0$

From principle of conservation of mechanical energy,

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

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

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

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

$\Rightarrow {v}_{C}=\sqrt{2\times (gh+\frac{1}{2}{v}_{0}^{2})}$

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

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

$\Rightarrow {v}_{C}=\sqrt{1112.2}$

$\Rightarrow {v}_{C}=33.4m/s$

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

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

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

$\Rightarrow {h}_{f}=h+\frac{{v}_{0}^{2}}{2}g$

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

$\Rightarrow {h}_{f}=56.7\text{m}$

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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