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$$
\beta=2 \pi c \tilde{\omega}_{\mathrm{obs}}\left(\frac{\mu}{2 D}\right)^{1 / 2}
$$
Given that $\tilde{\omega}_{\mathrm{obs}}=2886 \mathrm{~cm}^{-1}$ and $D=440.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$ for $\mathrm{H}^{35} \mathrm{Cl}$, calculate $\beta$. | 1.81 | 1.81 | 5-10 | $10^{10} \mathrm{~m}^{-1}$ | chemmc |
||
Two narrow slits separated by $0.10 \mathrm{~mm}$ are illuminated by light of wavelength $600 \mathrm{~nm}$. If a detector is located $2.00 \mathrm{~m}$ beyond the slits, what is the distance between the central maximum and the first maximum? | 12 | 12 | 1-44 | mm | chemmc |
||
$$
\text { If we locate an electron to within } 20 \mathrm{pm} \text {, then what is the uncertainty in its speed? }
$$ | 3.7 | 3.7 | 1-46 | $10^7 \mathrm{~m} \cdot \mathrm{s}^{-1}$ | chemmc |
||
The mean temperature of the earth's surface is $288 \mathrm{~K}$. What is the maximum wavelength of the earth's blackbody radiation? | 1.01 | 1.01 | 1-14 | 10^{-5} \mathrm{~m} | chemmc |
||
The power output of a laser is measured in units of watts (W), where one watt is equal to one joule per second. $\left(1 \mathrm{~W}=1 \mathrm{~J} \cdot \mathrm{s}^{-1}\right.$.) What is the number of photons emitted per second by a $1.00 \mathrm{~mW}$ nitrogen laser? The wavelength emitted by a nitrogen laser is $337 \mathrm{~nm}$. | 1.70 | 1.70 | 1-16 | $
10^{15} \text { photon } \cdot \mathrm{s}^{-1}
$ | chemmc |
||
Sirius, one of the hottest known stars, has approximately a blackbody spectrum with $\lambda_{\max }=260 \mathrm{~nm}$. Estimate the surface temperature of Sirius.
| 11000 | 11000 | 1-7 | $\mathrm{~K}$
| chemmc |
||
A ground-state hydrogen atom absorbs a photon of light that has a wavelength of $97.2 \mathrm{~nm}$. It then gives off a photon that has a wavelength of $486 \mathrm{~nm}$. What is the final state of the hydrogen atom? | 2 | no units | 2 | 1-26 | chemmc |
||
It turns out that the solution of the Schrödinger equation for the Morse potential can be expressed as
$$
G(v)=\tilde{\omega}_{\mathrm{e}}\left(v+\frac{1}{2}\right)-\tilde{\omega}_{\mathrm{e}} \tilde{x}_{\mathrm{e}}\left(v+\frac{1}{2}\right)^2
$$
The Harmonic Oscillator and Vibrational Spectroscopy
where
$$
\tilde{x}_{\mathrm{e}}=\frac{h c \tilde{\omega}_{\mathrm{e}}}{4 D}
$$
Given that $\tilde{\omega}_{\mathrm{e}}=2886 \mathrm{~cm}^{-1}$ and $D=440.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$ for $\mathrm{H}^{35} \mathrm{Cl}$, calculate $\tilde{x}_{\mathrm{e}}$. | 0.01961 | only first part taken of the question | 0.01961 | 5-12 | chemmc |
||
In the infrared spectrum of $\mathrm{H}^{127} \mathrm{I}$, there is an intense line at $2309 \mathrm{~cm}^{-1}$. Calculate the force constant of $\mathrm{H}^{127} \mathrm{I}$. | 313 | 313 | 5-13 | $ \mathrm{~N} \cdot \mathrm{m}^{-1}$ | chemmc |
||
Calculate the percentage difference between $e^x$ and $1+x$ for $x=0.0050$ | 1.25 | Math Part D (after chapter 4) | 1.25 | D-1 | $10^{-3} \%$ | chemmc |
|
Calculate the kinetic energy of an electron in a beam of electrons accelerated by a voltage increment of $100 \mathrm{~V}$ | 1.602 | 1.602 | 1-39 | $10^{-17} \mathrm{~J} \cdot$ electron ${ }^{-1}$ | chemmc |
||
We must first convert $\phi$ from electron volts to joules.
$$
\begin{aligned}
\phi & =2.28 \mathrm{eV}=(2.28 \mathrm{eV})\left(1.602 \times 10^{-19} \mathrm{~J} \cdot \mathrm{eV}^{-1}\right) \\
& =3.65 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
Using Equation $h v_0=\phi$, we have
$$
v_0=\frac{3.65 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}=5.51 \times 10^{14} \mathrm{~Hz}$$
| Given that the work function for sodium metal is $2.28 \mathrm{eV}$, what is the threshold frequency $v_0$ for sodium? | 5.51 | 5.51 | 1.1_3 | $10^{14}\mathrm{~Hz}$ | chemmc |
|
The mass of an electron is $9.109 \times 10^{-31} \mathrm{~kg}$. One percent of the speed of light is
$$
v=(0.0100)\left(2.998 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)=2.998 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}
$$
The momentum of the electron is given by
$$
\begin{aligned}
p=m_{\mathrm{e}} v & =\left(9.109 \times 10^{-31} \mathrm{~kg}\right)\left(2.998 \times 10^6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) \\
& =2.73 \times 10^{-24} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}
\end{aligned}
$$
The de Broglie wavelength of this electron is
$$
\begin{aligned}
\lambda=\frac{h}{p} & =\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{2.73 \times 10^{-24} \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}}=2.43 \times 10^{-10} \mathrm{~m} \\
& =243 \mathrm{pm}
\end{aligned}
$$
This wavelength is of atomic dimensions.
| Calculate the de Broglie wavelength of an electron traveling at $1.00 \%$ of the speed of light. | 243 | 243 | 1.1_11 | $\mathrm{pm}$ | chemmc |
|
Because $u(\theta, \phi)$ is independent of $r$, we start with
$$
\nabla^2 u=\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial u}{\partial \theta}\right)+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 u}{\partial \phi^2}
$$
Substituting
$$
u(\theta, \phi)=-\left(\frac{3}{8 \pi}\right)^{1 / 2} e^{i \phi} \sin \theta
$$
into $\nabla^2 u$ gives
$$
\begin{aligned}
\nabla^2 u & =-\left(\frac{3}{8 \pi}\right)^{1 / 2}\left[\frac{e^{i \phi}}{r^2 \sin \theta}\left(\cos ^2 \theta-\sin ^2 \theta\right)-\frac{\sin \theta}{r^2 \sin ^2 \theta} e^{i \phi}\right] \\
& =-\left(\frac{3}{8 \pi}\right)^{1 / 2} \frac{e^{i \phi}}{r^2}\left(\frac{1-2 \sin ^2 \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\
& =2\left(\frac{3}{8 \pi}\right)^{1 / 2} \frac{e^{i \phi} \sin \theta}{r^2}
\end{aligned}
$$
or $c=-2$. | Show that $u(\theta, \phi)=Y_1^1(\theta, \phi)$ given in Example $$
\begin{aligned}
&Y_1^1(\theta, \phi)=-\left(\frac{3}{8 \pi}\right)^{1 / 2} e^{i \phi} \sin \theta\\
&Y_1^{-1}(\theta, \phi)=\left(\frac{3}{8 \pi}\right)^{1 / 2} e^{-i \phi} \sin \theta
\end{aligned}
$$ satisfies the equation $\nabla^2 u=\frac{c}{r^2} u$, where $c$ is a constant. What is the value of $c$ ? | -2 | -2 | E.E_4 | chemmc |
||
We want to find the constant $c$ such that
$$
I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=1
$$
First notice that $\Psi_2(1,2)$ can be factored into the product of a spatial part and a spin part:
$$
\begin{aligned}
\Psi_2(1,2) & =1 s(1) 1 s(2)[\alpha(1) \beta(2)-\alpha(2) \beta(1)] \\
& =1 s\left(\mathbf{r}_1\right) 1 s\left(\mathbf{r}_2\right)\left[\alpha\left(\sigma_1\right) \beta\left(\sigma_2\right)-\alpha\left(\sigma_2\right) \beta\left(\sigma_1\right)\right]
\end{aligned}
$$
The normalization integral becomes the product of three integrals:
$$
I=c^2\langle 1 s(1) \mid 1 s(1)\rangle\langle 1 s(2) \mid 1 s(2)\rangle\langle\alpha(1) \beta(1)-\alpha(2) \beta(1) \mid \alpha(1) \beta(2)-\alpha(2) \beta(1)\rangle
$$
The spatial integrals are equal to 1 because we have taken the $1 s$ orbitals to be normalized. Now let's look at the spin integrals. When the two terms in the integrand of the spin integral are multiplied, we get four integrals. One of them is
$$
\begin{aligned}
\iint \alpha^*\left(\sigma_1\right) \beta^*\left(\sigma_2\right) \alpha\left(\sigma_1\right) \beta\left(\sigma_2\right) d \sigma_1 d \sigma_2 & =\langle\alpha(1) \beta(2) \mid \alpha(1) \beta(2)\rangle \\
& =\langle\alpha(1) \mid \alpha(1)\rangle\langle\beta(2) \mid \beta(2)\rangle=1
\end{aligned}
$$
where once again we point out that integrating over $\sigma_1$ and $\sigma_2$ is purely symbolic; $\sigma_1$ and $\sigma_2$ are discrete variables. Another is
$$
\langle\alpha(1) \beta(2) \mid \alpha(2) \beta(1)\rangle=\langle\alpha(1) \mid \beta(1)\rangle\langle\beta(2) \mid \alpha(2)\rangle=0
$$
The other two are equal to 1 and 0 , and so
$$
I=c^2\left\langle\Psi_2(1,2) \mid \Psi_2(1,2)\right\rangle=2 c^2=1
$$
or $c=1 / \sqrt{2}$. | The wave function $\Psi_2(1,2)$ given by Equation 9.39 is not normalized as it stands. Determine the normalization constant of $\Psi_2(1,2)$ given that the "1s" parts are normalized. | $1 / \sqrt{2}$ | 0.70710678 | 9.9_4 | chemmc |
||
The equations for $c_1$ and $c_2$ associated with Equation $$
\left|\begin{array}{ll}
H_{11}-E S_{11} & H_{12}-E S_{12} \\
H_{12}-E S_{12} & H_{22}-E S_{22}
\end{array}\right|=0
$$ are
$$
c_1(\alpha-E)+c_2 \beta=0 \quad \text { and } \quad c_1 \beta+c_2(\alpha-E)=0
$$
For $E=\alpha+\beta$, either equation yields $c_1=c_2$. Thus,
$$
\psi_{\mathrm{b}}=c_1\left(2 p_{z 1}+2 p_{z 2}\right)
$$
The value of $c_1$ can be found by requiring that the wave function be normalized. The normalization condition on $\psi_\pi$ gives $c_1^2(1+2 S+1)=1$. Using the Hückel assumption that $S=0$, we find that $c_1=1 / \sqrt{2}$.
Substituting $E=\alpha-\beta$ into either of the equations for $c_1$ and $c_2$ yields $c_1=-c_2$, or
$$
\psi_{\mathrm{a}}=c_1\left(2 p_{z 1}-2 p_{z 2}\right)
$$
The normalization condition gives $c^2(1-2 S+1)=1$, or $c_1=1 / \sqrt{2}$.
| Find the bonding and antibonding Hückel molecular orbitals for ethene. | $1 / \sqrt{2}$ | 0.70710678 | 11.11_11 | chemmc |
||
Letting $\xi=\alpha^{1 / 2} x$ in Table 5.3 , we have
$$
\begin{aligned}
& \psi_0(\xi)=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\xi^2 / 2} \\
& \psi_1(\xi)=\sqrt{2}\left(\frac{\alpha}{\pi}\right)^{1 / 4} \xi e^{-\xi^2 / 2} \\
& \psi_2(\xi)=\frac{1}{\sqrt{2}}\left(\frac{\alpha}{\pi}\right)^{1 / 4}\left(2 \xi^2-1\right) e^{-\xi^2 / 2}
\end{aligned}
$$
The dipole transition moment is given by the integral
$$
I_{0 \rightarrow v} \propto \int_{-\infty}^{\infty} \psi_v(\xi) \xi \psi_0(\xi) d \xi
$$
The transition is allowed if $I_{0 \rightarrow v} \neq 0$ and is forbidden if $I_{0 \rightarrow v}=0$. For $v=1$, we have
$$
I_{0 \rightarrow 1} \propto\left(\frac{2 \alpha}{\pi}\right)^{1 / 2} \int_{-\infty}^{\infty} \xi^2 e^{-\xi^2} d \xi \neq 0
$$
because the integrand is everywhere positive. For $v=2$,
$$
I_{0 \rightarrow 2} \propto\left(\frac{\alpha}{2 \pi}\right)^{1 / 2} \int_{-\infty}^{\infty}\left(2 \xi^3-\xi\right) e^{-\xi^2} d \xi=0
$$
because the integrand is an odd function and the limits go from $-\infty$ to $+\infty$.
| Using the explicit formulas for the Hermite polynomials given in Table 5.3 as below $$
\begin{array}{ll}
H_0(\xi)=1 & H_1(\xi)=2 \xi \\
H_2(\xi)=4 \xi^2-2 & H_3(\xi)=8 \xi^3-12 \xi \\
H_4(\xi)=16 \xi^4-48 \xi^2+12 & H_5(\xi)=32 \xi^5-160 \xi^3+120 \xi
\end{array}
$$, show that a $0 \rightarrow 1$ vibrational transition is allowed and that a $0 \rightarrow 2$ transition is forbidden. | 0 | 0 | 5.5_12 | chemmc |
||
According to Equation $$
\begin{aligned}
&v_{\mathrm{obs}}=2 B(J+1) \quad J=0,1,2, \ldots\\
&B=\frac{h}{8 \pi^2 I}
\end{aligned}
$$, the spacing of the lines in the microwave spectrum of $\mathrm{H}^{35} \mathrm{Cl}$ is given by
$$
2 B=\frac{h}{4 \pi^2 I}
$$
$$
\frac{h}{4 \pi^2 I}=6.26 \times 10^{11} \mathrm{~Hz}
$$
Solving this equation for $I$, we have
$$
I=\frac{6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}}{4 \pi^2\left(6.26 \times 10^{11} \mathrm{~s}^{-1}\right)}=2.68 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{m}^2
$$
The reduced mass of $\mathrm{H}^{35} \mathrm{Cl}$ is
$$
\mu=\frac{(1.00 \mathrm{amu})(35.0 \mathrm{amu})}{36.0 \mathrm{amu}}\left(1.661 \times 10^{-27} \mathrm{~kg} \cdot \mathrm{amu}^{-1}\right)=1.66 \times 10^{-27} \mathrm{~kg}
$$
Using the fact that $I=\mu l^2$, we obtain
$$
l=\left(\frac{2.68 \times 10^{-47} \mathrm{~kg} \cdot \mathrm{m}^2}{1.661 \times 10^{-27} \mathrm{~kg}}\right)^{1 / 2}=1.29 \times 10^{-10} \mathrm{~m}=129 \mathrm{pm}
$$
| To a good approximation, the microwave spectrum of $\mathrm{H}^{35} \mathrm{Cl}$ consists of a series of equally spaced lines, separated by $6.26 \times 10^{11} \mathrm{~Hz}$. Calculate the bond length of $\mathrm{H}^{35} \mathrm{Cl}$. | 129 | 129 | 6.6_2 | $\mathrm{pm}$ | chemmc |
|
To find $1 E_{\mathrm{h}}$ expressed in joules, we substitute the SI values of $m_{\mathrm{e}}, e$, $4 \pi \epsilon_0$, and $\hbar$ into the above equation. Using these values from Table $$$
\begin{array}{lll}
\hline \text { Property } & \text { Atomic unit } & \text { SI equivalent } \\
\hline \text { Mass } & \text { Mass of an electron, } m_{\mathrm{e}} & 9.1094 \times 10^{-31} \mathrm{~kg} \\
\text { Charge } & \text { Charge on a proton, } e & 1.6022 \times 10^{-19} \mathrm{C} \\
\text { Angular momentum } & \text { Planck constant divided by } 2 \pi, \hbar & 1.0546 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} \\
\text { Length } & \text { Bohr radius, } a_0=\frac{4 \pi \epsilon_0 \hbar^2}{m_{\mathrm{e}} e^2} & 5.2918 \times 10^{-11} \mathrm{~m} \\
\text { Energy } & \frac{m_{\mathrm{e}} e^4}{16 \pi^2 \epsilon_0^2 \hbar^2}=\frac{e^2}{4 \pi \epsilon_0 a_0}=E_{\mathrm{h}} & 4.3597 \times 10^{-18} \mathrm{~J} \\
\text { Permittivity } & \kappa_0=4 \pi \epsilon_0 & 1.1127 \times 10^{-10} \mathrm{C}^2 \cdot \mathrm{J}^{-1} \cdot \mathrm{m}^{-1} \\
\hline
\end{array}
$$, we find
Atomic and Molecular Calculations Are Expressed in Atomic Units
$$
\begin{aligned}
1 E_{\mathrm{h}} & =\frac{\left(9.1094 \times 10^{-31} \mathrm{~kg}\right)\left(1.6022 \times 10^{-19} \mathrm{C}\right)^4}{\left(1.1127 \times 10^{-10} \mathrm{C}^2 \cdot \mathrm{J}^{-1} \cdot \mathrm{m}^{-1}\right)^2\left(1.0546 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)^2} \\
& =4.3597 \times 10^{-18} \mathrm{~J}
\end{aligned}
$$
If we multiply this result by the Avogadro constant, we obtain
$$
1 E_{\mathrm{h}}=2625.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}
$$
To express $1 E_{\mathrm{h}}$ in wave numbers $\left(\mathrm{cm}^{-1}\right)$, we use the fact that $1 E_{\mathrm{h}}=4.3597 \times 10^{-18} \mathrm{~J}$ along with the equation
$$
\begin{aligned}
\tilde{v} & =\frac{1}{\lambda}=\frac{h v}{h c}=\frac{E}{c h}=\frac{4.3597 \times 10^{-18} \mathrm{~J}}{\left(2.9979 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\right)\left(6.6261 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)} \\
& =2.1947 \times 10^7 \mathrm{~m}^{-1}=2.1947 \times 10^5 \mathrm{~cm}^{-1}
\end{aligned}
$$
so that we can write
$$
1 E_{\mathrm{h}}=2.1947 \times 10^5 \mathrm{~cm}^{-1}
$$
Last, to express $1 E_{\mathrm{h}}$ in terms of electron volts, we use the conversion factor
$$
1 \mathrm{eV}=1.6022 \times 10^{-19} \mathrm{~J}
$$
Using the value of $1 E_{\mathrm{h}}$ in joules obtained previously, we have
$$
\begin{aligned}
1 E_{\mathrm{h}} & =\left(4.3597 \times 10^{-18} \mathrm{~J}\right)\left(\frac{1 \mathrm{eV}}{1.6022 \times 10^{-19} \mathrm{~J}}\right) \\
& =27.211 \mathrm{eV}
\end{aligned}
$$
| The unit of energy in atomic units is given by
$$
1 E_{\mathrm{h}}=\frac{m_{\mathrm{e}} e^4}{16 \pi^2 \epsilon_0^2 \hbar^2}
$$
Express $1 E_{\mathrm{h}}$ in electron volts $(\mathrm{eV})$. | 27.211 | 27.211 | 9.9_1 | $\mathrm{eV}$ | chemmc |
|
The probability that the particle will be found between 0 and $a / 2$ is
$$
\operatorname{Prob}(0 \leq x \leq a / 2)=\int_0^{a / 2} \psi^*(x) \psi(x) d x=\frac{2}{a} \int_0^{a / 2} \sin ^2 \frac{n \pi x}{a} d x
$$
If we let $n \pi x / a$ be $z$, then we find
$$
\begin{aligned}
\operatorname{Prob}(0 \leq x \leq a / 2) & =\frac{2}{n \pi} \int_0^{n \pi / 2} \sin ^2 z d z=\frac{2}{n \pi}\left|\frac{z}{2}-\frac{\sin 2 z}{4}\right|_0^{n \pi / 2} \\
& =\frac{2}{n \pi}\left(\frac{n \pi}{4}-\frac{\sin n \pi}{4}\right)=\frac{1}{2} \quad \text { (for all } n \text { ) }
\end{aligned}
$$
Thus, the probability that the particle lies in one-half of the interval $0 \leq x \leq a$ is $\frac{1}{2}$. | Calculate the probability that a particle in a one-dimensional box of length $a$ is found between 0 and $a / 2$. | $\frac{1}{2}$ | 0.5 | 3.3_6 | chemmc |
||
An automobile with a mass of $1000 \mathrm{~kg}$, including passengers, settles $1.0 \mathrm{~cm}$ closer to the road for every additional $100 \mathrm{~kg}$ of passengers. It is driven with a constant horizontal component of speed $20 \mathrm{~km} / \mathrm{h}$ over a washboard road with sinusoidal bumps. The amplitude and wavelength of the sine curve are $5.0 \mathrm{~cm}$ and $20 \mathrm{~cm}$, respectively. The distance between the front and back wheels is $2.4 \mathrm{~m}$. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of the wheels and springs and assume that the wheels are always in contact with the road.
| -0.16 | -0.16 | Problem 3.40 | $ \mathrm{~mm}$ | class |
||
Find the shortest path between the $(x, y, z)$ points $(0,-1,0)$ and $(0,1,0)$ on the conical surface $z=1-\sqrt{x^2+y^2}$. What is the length of the path? Note: this is the shortest mountain path around a volcano. | $2 \sqrt{2} \sin \frac{\pi}{2 \sqrt{2}}$ | 2.534324263 | Problem 6.14 | class |
|||
In the blizzard of ' 88 , a rancher was forced to drop hay bales from an airplane to feed her cattle. The plane flew horizontally at $160 \mathrm{~km} / \mathrm{hr}$ and dropped the bales from a height of $80 \mathrm{~m}$ above the flat range. She wanted the bales of hay to land $30 \mathrm{~m}$ behind the cattle so as to not hit them. How far behind the cattle should she push the bales out of the airplane? | 210 | 210 | Problem 2.6 | $\mathrm{~m}$ | class |
||
Consider a damped harmonic oscillator. After four cycles the amplitude of the oscillator has dropped to $1 / e$ of its initial value. Find the ratio of the frequency of the damped oscillator to its natural frequency.
| $\frac{8 \pi}{\sqrt{64 \pi^2+1}}$ | 0.9992093669 | Problem 3.44 | class |
|||
What is the minimum escape velocity of a spacecraft from the moon? | 2380 | 2380 | Problem 8.28 | $\mathrm{~m} / \mathrm{s}$ | class |
||
A rocket has an initial mass of $7 \times 10^4 \mathrm{~kg}$ and on firing burns its fuel at a rate of 250 $\mathrm{kg} / \mathrm{s}$. The exhaust velocity is $2500 \mathrm{~m} / \mathrm{s}$. If the rocket has a vertical ascent from resting on the earth, how long after the rocket engines fire will the rocket lift off? | 25 | 25 | Problem 9.60 | $\mathrm{~s}$ | class |
||
A spacecraft of mass $10,000 \mathrm{~kg}$ is parked in a circular orbit $200 \mathrm{~km}$ above Earth's surface. What is the minimum energy required (neglect the fuel mass burned) to place the satellite in a synchronous orbit (i.e., $\tau=24 \mathrm{hr}$ )? | 2.57 | 2.57 | Problem 8.42 | $10^{11} \mathrm{~J}$ | class |
||
A clown is juggling four balls simultaneously. Students use a video tape to determine that it takes the clown $0.9 \mathrm{~s}$ to cycle each ball through his hands (including catching, transferring, and throwing) and to be ready to catch the next ball. What is the minimum vertical speed the clown must throw up each ball?
| 13.2 | 13.2 | Problem 2.4 | $\mathrm{~m} \cdot \mathrm{s}^{-1}$ | class |
||
A deuteron (nucleus of deuterium atom consisting of a proton and a neutron) with speed $14.9 \mathrm{~km} / \mathrm{s}$ collides elastically with a neutron at rest. Use the approximation that the deuteron is twice the mass of the neutron. If the deuteron is scattered through a $\mathrm{LAB}$ angle $\psi=10^{\circ}$, what is the final speed of the deuteron? | 14.44 | 14.44 | Problem 9.22 | $\mathrm{~km} / \mathrm{s}$ | class |
||
A student drops a water-filled balloon from the roof of the tallest building in town trying to hit her roommate on the ground (who is too quick). The first student ducks back but hears the water splash $4.021 \mathrm{~s}$ after dropping the balloon. If the speed of sound is $331 \mathrm{~m} / \mathrm{s}$, find the height of the building, neglecting air resistance. | 71 | 71 | Problem 2.30 | $\mathrm{~m}$ | class |
||
A steel ball of velocity $5 \mathrm{~m} / \mathrm{s}$ strikes a smooth, heavy steel plate at an angle of $30^{\circ}$ from the normal. If the coefficient of restitution is 0.8 , at what velocity does the steel ball bounce off the plate? | $4.3$ | 4.3 | Problem 9.42 | $\mathrm{~m} / \mathrm{s}$ | class |
||
Include air resistance proportional to the square of the ball's speed in the previous problem. Let the drag coefficient be $c_W=0.5$, the softball radius be $5 \mathrm{~cm}$ and the mass be $200 \mathrm{~g}$. Find the initial speed of the softball needed now to clear the fence. | 35.2 | 35.2 | Problem 2.18 | $\mathrm{~m} \cdot \mathrm{s}^{-1}$ | class |
||
A child slides a block of mass $2 \mathrm{~kg}$ along a slick kitchen floor. If the initial speed is 4 $\mathrm{m} / \mathrm{s}$ and the block hits a spring with spring constant $6 \mathrm{~N} / \mathrm{m}$, what is the maximum compression of the spring? | 2.3 | 2.3 | Problem 2.26 | $\mathrm{~m}$ | class |
||
An Earth satellite has a perigee of $300 \mathrm{~km}$ and an apogee of $3,500 \mathrm{~km}$ above Earth's surface. How far is the satellite above Earth when it has rotated $90^{\circ}$ around Earth from perigee? | 1590 | 1590 | Problem 8.24 | $\mathrm{~km}$ | class |
||
Two masses $m_1=100 \mathrm{~g}$ and $m_2=200 \mathrm{~g}$ slide freely in a horizontal frictionless track and are connected by a spring whose force constant is $k=0.5 \mathrm{~N} / \mathrm{m}$. Find the frequency of oscillatory motion for this system. | 2.74 | 2.74 | Problem 3.6 | $\mathrm{rad} \cdot \mathrm{s}^{-1}$ | class |
||
Calculate the minimum $\Delta v$ required to place a satellite already in Earth's heliocentric orbit (assumed circular) into the orbit of Venus (also assumed circular and coplanar with Earth). Consider only the gravitational attraction of the Sun. | 5275 | 5275 | Problem 8.38 | $\mathrm{~m} / \mathrm{s}$ | class |
||
A potato of mass $0.5 \mathrm{~kg}$ moves under Earth's gravity with an air resistive force of $-k m v$. Find the terminal velocity if the potato is released from rest and $k=$ $0.01 \mathrm{~s}^{-1}$. | 1000 | 1000 | Problem 2.54 | $\mathrm{~m} / \mathrm{s}$ | class |
||
The height of a hill in meters is given by $z=2 x y-3 x^2-4 y^2-18 x+28 y+12$, where $x$ is the distance east and $y$ is the distance north of the origin. What is the $x$ distance of the top of the hill? | -2 | -2 | Problem 1.40 | m | class |
||
Shot towers were popular in the eighteenth and nineteenth centuries for dropping melted lead down tall towers to form spheres for bullets. The lead solidified while falling and often landed in water to cool the lead bullets. Many such shot towers were built in New York State. Assume a shot tower was constructed at latitude $42^{\circ} \mathrm{N}$, and the lead fell a distance of $27 \mathrm{~m}$. How far did the lead bullets land from the direct vertical? | 2.26 | 2.26 | Problem 10.22 | $\mathrm{~mm}$ | class |
||
A simple harmonic oscillator consists of a 100-g mass attached to a spring whose force constant is $10^4 \mathrm{dyne} / \mathrm{cm}$. The mass is displaced $3 \mathrm{~cm}$ and released from rest. Calculate the natural frequency $\nu_0$. | 6.9 | 6.9 | Problem 3.2 | $10^{-2} \mathrm{~s}^{-1}$ | class |
||
Use the function described in Example 4.3, $x_{n+1}=\alpha x_n\left(1-x_n^2\right)$ where $\alpha=2.5$. Consider two starting values of $x_1$ that are similar, 0.9000000 and 0.9000001 . Determine the lowest value of $n$ for which the two values diverge by more than $30 \%$. | 30 | 30 | Problem 4.14 | class |
|||
A gun fires a projectile of mass $10 \mathrm{~kg}$ of the type to which the curves of Figure 2-3 apply. The muzzle velocity is $140 \mathrm{~m} / \mathrm{s}$. Through what angle must the barrel be elevated to hit a target on the same horizontal plane as the gun and $1000 \mathrm{~m}$ away? Compare the results with those for the case of no retardation. | 17.4 | 17.4 | Problem 2.20 | $^{\circ}$ | class |
||
A spacecraft is placed in orbit $200 \mathrm{~km}$ above Earth in a circular orbit. Calculate the minimum escape speed from Earth. | 3.23 | 3.23 | Problem 8.30 | $ \mathrm{~km} / \mathrm{s}$ | class |
||
Find the value of the integral $\int_S(\nabla \times \mathbf{A}) \cdot d \mathbf{a}$ if the vector $\mathbf{A}=y \mathbf{i}+z \mathbf{j}+x \mathbf{k}$ and $S$ is the surface defined by the paraboloid $z=1-x^2-y^2$, where $z \geq 0$. | $-\pi$ | -3.141592 | Problem 1.38 | class |
|||
A skier weighing $90 \mathrm{~kg}$ starts from rest down a hill inclined at $17^{\circ}$. He skis $100 \mathrm{~m}$ down the hill and then coasts for $70 \mathrm{~m}$ along level snow until he stops. Find the coefficient of kinetic friction between the skis and the snow. | 0.18 | 0.18 | Problem 2.24 | class |
|||
Consider a comet moving in a parabolic orbit in the plane of Earth's orbit. If the distance of closest approach of the comet to the $\operatorname{Sun}$ is $\beta r_E$, where $r_E$ is the radius of Earth's (assumed) circular orbit and where $\beta<1$, show that the time the comet spends within the orbit of Earth is given by
$$
\sqrt{2(1-\beta)} \cdot(1+2 \beta) / 3 \pi \times 1 \text { year }
$$
If the comet approaches the Sun to the distance of the perihelion of Mercury, how many days is it within Earth's orbit? | 76 | 76 | Problem 8.12 | $ \text { days }$ | class |
||
A British warship fires a projectile due south near the Falkland Islands during World War I at latitude $50^{\circ} \mathrm{S}$. If the shells are fired at $37^{\circ}$ elevation with a speed of $800 \mathrm{~m} / \mathrm{s}$, by how much do the shells miss their target? | 260 | 260 | Problem 10.18 | $\mathrm{~m}$ | class |
||
Two double stars of the same mass as the sun rotate about their common center of mass. Their separation is 4 light years. What is their period of revolution?
| 9 | 9 | Problem 8.46 | $10^7 \mathrm{yr}$ | class |
||
To perform a rescue, a lunar landing craft needs to hover just above the surface of the moon, which has a gravitational acceleration of $g / 6$. The exhaust velocity is $2000 \mathrm{~m} / \mathrm{s}$, but fuel amounting to only 20 percent of the total mass may be used. How long can the landing craft hover? | 273 | 273 | Problem 9.62 | $\mathrm{~s}$ | class |
||
In an elastic collision of two particles with masses $m_1$ and $m_2$, the initial velocities are $\mathbf{u}_1$ and $\mathbf{u}_2=\alpha \mathbf{u}_1$. If the initial kinetic energies of the two particles are equal, find the conditions on $u_1 / u_2$ such that $m_1$ is at rest after the collision and $\alpha$ is positive. | $3 \pm 2 \sqrt{2}$ | 5.828427125 | Problem 9.36 | class |
|||
Astronaut Stumblebum wanders too far away from the space shuttle orbiter while repairing a broken communications satellite. Stumblebum realizes that the orbiter is moving away from him at $3 \mathrm{~m} / \mathrm{s}$. Stumblebum and his maneuvering unit have a mass of $100 \mathrm{~kg}$, including a pressurized tank of mass $10 \mathrm{~kg}$. The tank includes only $2 \mathrm{~kg}$ of gas that is used to propel him in space. The gas escapes with a constant velocity of $100 \mathrm{~m} / \mathrm{s}$. With what velocity will Stumblebum have to throw the empty tank away to reach the orbiter? | 11 | 11 | Problem 9.12 | $ \mathrm{~m} / \mathrm{s}$ | class |
||
null | A deuteron (nucleus of deuterium atom consisting of a proton and a neutron) with speed $14.9$ km / s collides elastically with a neutron at rest. Use the approximation that the deuteron is twice the mass of the neutron. If the deuteron is scattered through a LAB angle $\psi = 10^\circ$, the final speed of the deuteron is $v_d = 14.44$ km / s and the final speed of the neutron is $v_n = 5.18$ km / s. Another set of solutions for the final speed is $v_d = 5.12$ km / s for the deuteron and $v_n = 19.79$ km / s for the neutron. What is the maximum possible scattering angle of the deuteron? | $74.8^\circ$, $5.2^\circ$ | 30 | 9.22 B. | $^\circ$ | class |
|
null | A steel ball of velocity $5$ m/s strikes a smooth, heavy steel plate at an angle of $30^\circ$ from the normal. If the coefficient of restitution is 0.8, at what angle from the normal does the steel ball bounce off the plate? | $36^\circ$ | 36 | 9.42 B. | $^\circ$ | class |
|
null | A string is set into motion by being struck at a point $L/4$ from one end by a triangular hammer. The initial velocity is greatest at $x = L/4$ and decreases linearly to zero at $x = 0$ and $x = L/2$. The region $L/2 \leq x \leq L$ is initially undisturbed. Determine the subsequent motion of the string. How many decibels down from the fundamental are the second harmonics?' | 4.4, 13.3 | 4.4 | 13.6 | dB | class |
|
null | In the blizzard of ' 88 , a rancher was forced to drop hay bales from an airplane to feed her cattle. The plane flew horizontally at $160 \mathrm{~km} / \mathrm{hr}$ and dropped the bales from a height of $80 \mathrm{~m}$ above the flat range. To not hit the cattle, what is the largest time error she could make while pushing the bales out of the airplane? Ignore air resistance. | 0.68 | 0.68 | 2.6 B. | seconds | class |
|
null | A free neutron is unstable and decays into a proton and an electron. How much energy other than the rest energies of the proton and electron is available if a neutron at rest decays? (This is an example of nuclear beta decay. Another particle, called a neutrino-- actually an antineutrino $\bar v$ is also produced.) | 0.8 | 0.8 | 14.30 | $MeV$ | class |
|
null | A new single-stage rocket is developed in the year 2023, having a gas exhaust velocity of $4000$ m/s. The total mass of the rocket is $10^5$ kg, with $90$% of its mass being fuel. The fuel burns quickly in $100$ s at a constant rate. For testing purposes, the rocket is launched vertically at rest from Earth's surface. Neglecting air resistance and assuming that the acceleration of gravity is constant, the launched object can reach 3700 km above the surface of Earth. If the object has a radius of $20$ cm and the air resistance is proportional to the square of the object's speed with $c_w = 0.2$, assuming the density of air is constant, the maximum height reached is 890 km. Now also include the fact that the acceleration of gravity decreases as the object soars above Earth. Find the height reached. | 950 | 950 | 9.64 C. | km | class |
|
null | Calculate the effective gravitational field vector $g$ at Earth's surface at the poles. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result $g = 9.780356[1 + 0.0052885sin^2\lambda - 0.0000059 sin^2 (2\lambda )]$ $m/s^2$ where $\lambda$ is the latitude? | 9.832 | 9.832 | 10.20 | $m/s^2$ | class |
|
null | In nuclear and particle physics, momentum is usually quoted in $MeV / c$ to facilitate calculations. Calculate the kinetic energy of an electron if it has a momentum of $1000$ $MeV/c$ | $T_{electron} = 999.5$, $T_{proton} = 433$ | 999.5 | 14.32 | $MeV$ | class |
|
null | A skier weighing $90$ kg starts from rest down a hill inclined at $17^\circ$. He skis 100 m down the hill and then coasts for 70 m along level snow until he stops. Given a coefficient of kinetic friction of $\mu_k = 0.18$, what velocity does the skier have at the bottom of the hill? | 15.6 | Uses answer from part A. (coefficient of kinetic friction) | 15.6 | 2.24 B. | $m/s$ | class |
null | A rocket starts from rest in free space by emitting mass. At what fraction of the initial mass is the momentum a maximum? | $e^{-1}$ | 0.367879 | 9.54 | class |
||
null | A particle moves in a plane elliptical orbit described by the position vector $r = 2b \sin \omega ti + b \cos \omega tj$
What is the angle between $v$ and $a$ at time $t = \frac{\pi}{2\omega}$ ? | $90^\circ$ | 90 | 1.10 B. | $^\circ$ | class |
|
null | An Earth satellite has a perigee of $300$ km and an apogee of $3,500$ km above Earth's surface. How far is the satellite above Earth when it has moved halfway from perigee to apogee? | 1900 | 1900 | 8.24 (b) | $km$ | class |
|
null | In a typical model rocket (Estes Alpha III) the Estes C6 solid rocket engine provides a total impulse of $8.5$ N-s. Assume the total rocket mass at launch is $54$ g and that it has a rocket engine of mass $20$ g that burns evenly for $1.5$ s. The rocket diameter is $24$ mm. Assume a constant burn rate of the propellent mass ($11$ g), a rocket exhaust speed $800$ m/s, vertical ascent, and drag coefficient $c_w = 0.75$. Take into account the change of rocket mass with time and omit the effect of gravity. The rocket's speed at burn out is 131 m/s. How far has the rocket traveled at that moment? | 108 | 108 | 9.66 B. | m | class |
|
null | A deuteron (nucleus of deuterium atom consisting of a proton and a neutron) with speed $14.9$ km / s collides elastically with a neutron at rest. Use the approximation that the deuteron is twice the mass of the neutron. If the deuteron is scattered through a LAB angle $\psi = 10^\circ$, what is the final speed of the neutron? | 5.18 | 9.22 A, but only the neutron (not deuteron, which has already been annotated) | 5.18 | 9.22 A. | km / s | class |
null | A new single-stage rocket is developed in the year 2023, having a gas exhaust velocity of $4000$ m/s. The total mass of the rocket is $10^5$ kg, with $90$% of its mass being fuel. The fuel burns quickly in $100$ s at a constant rate. For testing purposes, the rocket is launched vertically at rest from Earth's surface. Neglecting air resistance and assuming that the acceleration of gravity is constant, the launched object can reach 3700 km above the surface of Earth. If the object has a radius of $20$ cm and the air resistance is proportional to the square of the object's speed with $c_w = 0.2$, determine the maximum height reached. Assume the density of air is constant. | 890 | 890 | 9.64 B. | km | class |
|
null | A new single-stage rocket is developed in the year 2023, having a gas exhaust velocity of $4000$ m/s. The total mass of the rocket is $10^5$ kg, with $90$% of its mass being fuel. The fuel burns quickly in $100$ s at a constant rate. For testing purposes, the rocket is launched vertically at rest from Earth's surface. Neglecting air resistance and assuming that the acceleration of gravity is constant, the launched object can reach 3700 km above the surface of Earth. If the object has a radius of $20$ cm and the air resistance is proportional to the square of the object's speed with $c_w = 0.2$, assuming the density of air is constant, the maximum height reached is 890 km. Including the fact that the acceleration of gravity decreases as the object soars above Earth, the height reached is 950 km. Now add the effects of the decrease in air density with altitude to the calculation. We can very roughly represent the air density by $log_{10}(\rho) = -0.05h + 0.11$ where $\rho$ is the air density in $kg/m^3$ and $h$ is the altitude above Earth in km. Determine how high the object now goes. | 8900 | 8900 | 9.64 D. | km | class |
|
null | A racer attempting to break the land speed record rockets by two markers spaced $100$ m apart on the ground in a time of $0.4$ $\mu s$ as measured by an observer on the ground. How far apart do the two markers appear to the racer? | 55.3 | Omitted the last two questions / answers | 55.3 | 14.12 | $m$ | class |
null | A billiard ball of intial velocity $u_1$ collides with another billard ball (same mass) initially at rest. The first ball moves off at $\psi = 45^\circ$. For an elastic collision, say the velocities of both balls after the collision is $v_1 = v_2 = \frac{u_1}{\sqrt(2)}$. At what LAB angle does the second ball emerge? | 45 | 45 | 9.34 B. | $^\circ$ | class |
|
null | Calculate the effective gravitational field vector $\textbf{g}$ at Earth's surface at the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. | 9.780 | Only calculating the equator, not the polar. | 9.780 | 10.20 B. | $m/s^2$ | class |
null | An astronaut travels to the nearest star system, 4 light years away, and returns at a speed $0.3c$. How much has the astronaut aged relative to those people remaining on Earth? | 25.4 | Astronaut ages 25.4, people on Earth age 26.7 | 25.4 | 14.20 | years | class |
null | In a typical model rocket (Estes Alpha III) the Estes C6 solid rocket engine provides a total impulse of $8.5$ N-s. Assume the total rocket mass at launch is $54$ g and that it has a rocket engine of mass $20$ g that burns evenly for $1.5$ s. The rocket diameter is $24$ mm. Assume a constant burn rate of the propellent mass ($11$ g), a rocket exhaust speed $800$ m/s, vertical ascent, and drag coefficient $c_w = 0.75$. Take into account the change of rocket mass with time and omit the effect of gravity. Find the rocket's speed at burn out. | 131 | 131 | 9.66 A. | m/s | class |
|
null | A new single-stage rocket is developed in the year 2023, having a gas exhaust velocity of $4000$ m/s. The total mass of the rocket is $10^5$ kg, with $90$% of its mass being fuel. The fuel burns quickly in $100$ s at a constant rate. For testing purposes, the rocket is launched vertically at rest from Earth's surface. Neglect air resistance and assume that the acceleration of gravity is constant. Determine how high the launched object can reach above the surface of Earth. | 3700 | 3700 | 9.64 A. | km | class |
|
null | Include air resistance proportional to the square of the ball's speed in the previous problem. Let the drag coefficient be $c_w = 0.5$, the softball radius be $5$ cm and the mass be $200$ g. Given a speed of 35.2 m/s, find the initial elevation angle that allows the ball to most easily clear the fence. | $40.7^\circ$ | Used answer given in part A. for speed | 40.7 | 2.18 B. | $^\circ$ | class |
null | Show that the small angular deviation of $\epsilon$ of a plumb line from the true vertical (i.e., toward the center of Earth) at a point on Earth's surface at a latitude $\lambda$ is $\epsilon = \frac{R\omega^2sin\lambda cos\lambda}{g_0 - R\omega^2 cos^2\lambda}$ where R is the radius of Earth. What is the value (in seconds of arc) of the maximum deviation? Note that the entire denominator in the answer is actually the effective $g$, and $g_0$ denotes the pure gravitational component. | 6 | 6 | 10.12 | min | class |
|
null | A potato of mass 0.5 kg moves under Earth's gravity with an air resistive force of -$kmv$. The terminal velocity of the potato when released from rest is $v = 1000$ m/s, with $k=0.01s^{-1}$. Find the maximum height of the potato if it has the same value of k, but it is initially shot directly upward with a student-made potato gun with an initial velocity of $120$ m/s. | 680 | Used terminal velocity answer of 1000 m/s from part A. | 680 | 2.54 B. | $m$ | class |
The static frictional force has the approximate maximum value$$f_{\max }=\mu_s N$$ Then we have, $y$-direction
$$
-F_g \cos \theta+N=0
$$
$x$-direction
$$
-f_s+F_g \sin \theta=m \ddot{x}
$$
The static frictional force $f_s$ will be some value $f_s \leq f_{\max }$ required to keep $\ddot{x}=0$ -that is, to keep the block at rest. However, as the angle $\theta$ of the plane increases, eventually the static frictional force will be unable to keep the block at rest. At that angle $\theta^{\prime}, f_s$ becomes
$$
f_s\left(\theta=\theta^{\prime}\right)=f_{\max }=\mu_s N=\mu_s F_g \cos \theta
$$
and
$$
\begin{aligned}
m \ddot{x} & =F_g \sin \theta-f_{\max } \\
m \ddot{x} & =F_g \sin \theta-\mu_s F_g \cos \theta \\
\ddot{x} & =g\left(\sin \theta-\mu_s \cos \theta\right)
\end{aligned}
$$
Just before the block starts to slide, the acceleration $\ddot{x}=0$, so
$$
\begin{aligned}
\sin \theta-\mu_s \cos \theta & =0 \\
\tan \theta=\mu_s & =0.4 \\
\theta=\tan ^{-1}(0.4) & =22^{\circ}
\end{aligned}
$$
| If the coefficient of static friction between the block and plane in the previous example is $\mu_s=0.4$, at what angle $\theta$ will the block start sliding if it is initially at rest? | 22 | 22 | 2.2 | $^{\circ}$ | class |
|
The Equation $\tau^2=\frac{4 \pi^2 a^3}{G(m_1+m_2)} \cong \frac{4 \pi^2 a^3}{G m_2}$ relates the period of motion with the semimajor axes. Because $m$ (Halley's comet) $\ll m_{\text {Sun }}$
$$
\begin{aligned}
a & =\left(\frac{G m_{\text {Sun }} \tau^2}{4 \pi^2}\right)^{1 / 3} \\
& =\left[\frac{\left.\left(6.67 \times 10^{-11} \frac{\mathrm{Nm}^2}{\mathrm{~kg}^2}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)\left(76 \mathrm{yr} \frac{365 \mathrm{day}}{\mathrm{yr}} \frac{24 \mathrm{hr}}{\mathrm{day}} \frac{3600 \mathrm{~s}}{\mathrm{hr}}\right)^2\right]}{4 \pi^2}\right]^{1 / 3} \\
a & =2.68 \times 10^{12} \mathrm{m}
\end{aligned}
$$
Using Equation $\left.\begin{array}{l}r_{\min }=a(1-\varepsilon)=\frac{\alpha}{1+\varepsilon} \ r_{\max }=a(1+\varepsilon)=\frac{\alpha}{1-\varepsilon}\end{array}\right\}$ , we can determine $r_{\min }$ and $r_{\max }$
$$
\begin{aligned}
& r_{\min }=2.68 \times 10^{12} \mathrm{~m}(1-0.967)=8.8 \times 10^{10} \mathrm{~m} \\
\end{aligned}
$$
| Halley's comet, which passed around the sun early in 1986, moves in a highly elliptical orbit with an eccentricity of 0.967 and a period of 76 years. Calculate its minimum distances from the Sun. | 8.8 | 8.8 | 8.4 | $10^{10} \mathrm{m}$ | class |
|
Using $\mathbf{F}=m \mathrm{~g}$, the force components become
$x$-direction
$$
0=m \ddot{x}
$$
y-direction
$-m g=m \ddot{y}$ Neglect the height of the gun, and assume $x=y=0$ at $t=0$. Then
$$
\begin{aligned}
& \ddot{x}=0 \\
& \dot{x}=v_0 \cos \theta \\
& x=v_0 t \cos \theta \\
& y=-\frac{-g t^2}{2}+v_0 t \sin \theta \\
&
\end{aligned}
$$
and
$$
\begin{aligned}
& \ddot{y}=-g \\
& \dot{y}=-g t+v_0 \sin \theta \\
& y=\frac{-g t^2}{2}+v_0 t \sin \theta
\end{aligned}
$$
We can find the range by determining the value of $x$ when the projectile falls back to ground, that is, when $y=0$.
$$
y=t\left(\frac{-g t}{2}+v_0 \sin \theta\right)=0
$$
One value of $y=0$ occurs for $t=0$ and the other one for $t=T$.
$$
\begin{aligned}
\frac{-g T}{2}+v_0 \sin \theta & =0 \\
T & =\frac{2 v_0 \sin \theta}{g}
\end{aligned}
$$The range $R$ is found from
$$
\begin{aligned}
x(t=T) & =\text { range }=\frac{2 v_0^2}{g} \sin \theta \cos \theta \\
R & =\text { range }=\frac{v_0^2}{g} \sin 2 \theta
\end{aligned}$$We have $v_0=1450 \mathrm{~m} / \mathrm{s}$ and $\theta=55^{\circ}$, so the range becomes
$$
R=\frac{(1450 \mathrm{~m} / \mathrm{s})^2}{9.8 \mathrm{~m} / \mathrm{s}^2}\left[\sin \left(110^{\circ}\right)\right]=202 \mathrm{~km}$$.
| We treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$. The Germans used a long-range gun named Big Bertha in World War I to bombard Paris. Its muzzle velocity was $1,450 \mathrm{~m} / \mathrm{s}$. Find its predicted range of flight if $\theta=55^{\circ}$. | 72 | 202 | 2.6 | $\mathrm{km}$ | class |
|
$$
\begin{aligned}
\frac{m}{k} & =\frac{m}{G m M_{\text {Sun }}}=\frac{1}{G M_{\text {Sun }}} \\
& =\frac{1}{\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{s}^2 \cdot \mathrm{kg}\right)\left(1.99 \times 10^{30} \mathrm{~kg}\right)} \\
& =7.53 \times 10^{-21} \mathrm{~s}^2 / \mathrm{m}^3
\end{aligned}
$$
Because $k / m$ occurs so often in solar system calculations, we write it as well.
$$
\begin{aligned}
\frac{k}{m} & =1.33 \times 10^{20} \mathrm{~m}^3 / \mathrm{s}^2 \\
a_t & =\frac{1}{2}\left(r_{\text {Earth-Sun }}+r_{\text {Mars-Sun }}\right) \\
& =\frac{1}{2}\left(1.50 \times 10^{11} \mathrm{~m}+2.28 \times 10^{11} \mathrm{~m}\right) \\
& =1.89 \times 10^{11} \mathrm{~m} \\
T_t & =\pi\left(7.53 \times 10^{-21} \mathrm{~s}^2 / \mathrm{m}^3\right)^{1 / 2}\left(1.89 \times 10^{11} \mathrm{~m}\right)^{3 / 2} \\
& =2.24 \times 10^7 \mathrm{~s} \\
\end{aligned}
$$
| Calculate the time needed for a spacecraft to make a Hohmann transfer from Earth to Mars | 2.24 | 2.24 | 8.5 | $10^7 \mathrm{~s}$ | class |
|
We continue to use our simple model of the ocean surrounding Earth. Newton proposed a solution to this calculation by imagining that two wells be dug, one along the direction of high tide (our $x$-axis) and one along the direction of low tide (our $y$-axis). If the tidal height change we want to determine is $h$, then the difference in potential energy of mass $m$ due to the height difference is $m g h$. Let's calculate the difference in work if we move the mass $m$ from point $c$ to the center of Earth and then to point $a$. This work $W$ done by gravity must equal the potential energy change $m g h$. The work $W$ is
$$
W=\int_{r+\delta_1}^0 F_{T_y} d y+\int_0^{r+\delta_2} F_{T_x} d x
$$. The small distances $\delta_1$ and $\delta_2$ are to account for the small variations from a spherical Earth, but these values are so small they can be henceforth neglected. The value for $W$ becomes
$$
\begin{aligned}
W & =\frac{G m M_m}{D^3}\left[\int_r^0(-y) d y+\int_0^r 2 x d x\right] \\
& =\frac{G m M_m}{D^3}\left(\frac{r^2}{2}+r^2\right)=\frac{3 G m M_m r^2}{2 D^3}
\end{aligned}
$$
Because this work is equal to $m g h$, we have
$$
\begin{aligned}
m g h & =\frac{3 G m M_m r^2}{2 D^3} \\
h & =\frac{3 G M_m r^2}{2 g D^3}
\end{aligned}
$$
Note that the mass $m$ cancels, and the value of $h$ does not depend on $m$. Nor does it depend on the substance, so to the extent Earth is plastic, similar tidal effects should be (and are) observed for the surface land. If we insert the known values of the constants into the Equation, we find
$$
h=\frac{3\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(7.350 \times 10^{22} \mathrm{~kg}\right)\left(6.37 \times 10^6 \mathrm{~m}\right)^2}{2\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)\left(3.84 \times 10^8 \mathrm{~m}\right)^3}=0.54 \mathrm{~m}
$$
| Calculate the maximum height change in the ocean tides caused by the Moon. | 0.54 | 0.54 | 5.5 | $\mathrm{m}$ | class |
|
Because we are considering the possibility of the particle leaving the hemisphere, we choose the generalized coordinates to be $r$ and $\theta$. The constraint equation is
$$
f(r, \theta)=r-a=0
$$
The Lagrangian is determined from the kinetic and potential energies:
$$
\begin{aligned}
T & =\frac{m}{2}\left(\dot{r}^2+r^2 \dot{\theta}^2\right) \\
U & =m g r \cos \theta \\
L & =T-U \\
L & =\frac{m}{2}\left(\dot{r}^2+r^2 \dot{\theta}^2\right)-m g r \cos \theta
\end{aligned}
$$
where the potential energy is zero at the bottom of the hemisphere. The Lagrange equations are
$$
\begin{aligned}
& \frac{\partial L}{\partial r}-\frac{d}{d t} \frac{\partial L}{\partial \dot{r}}+\lambda \frac{\partial f}{\partial r}=0 \\
& \frac{\partial L}{\partial \theta}-\frac{d}{d t} \frac{\partial L}{\partial \dot{\theta}}+\lambda \frac{\partial f}{\partial \theta}=0
\end{aligned}
$$
Performing the differentiations gives
$$
\frac{\partial f}{\partial r}=1, \quad \frac{\partial f}{\partial \theta}=0
$$Then equations become
$$
\begin{gathered}
m r \dot{\theta}^2-m g \cos \theta-m \ddot{r}+\lambda=0 \\
m g r \sin \theta-m r^2 \ddot{\theta}-2 m r \dot{r} \dot{\theta}=0
\end{gathered}
$$
Next, we apply the constraint $r=a$ to these equations of motion:
$$
r=a, \quad \dot{r}=0=\ddot{r}
$$Then $$
\begin{array}{r}
m a \dot{\theta}^2-m g \cos \theta+\lambda=0 \\
m g a \sin \theta-m a^2 \ddot{\theta}=0
\end{array}
$$We have
$$
\ddot{\theta}=\frac{g}{a} \sin \theta
$$
We can integrate Equation to determine $\dot{\theta}^2$.
$$
\ddot{\theta}=\frac{d}{d t} \frac{d \theta}{d t}=\frac{d \dot{\theta}}{d t}=\frac{d \dot{\theta}}{d \theta} \frac{d \theta}{d t}=\dot{\theta} \frac{d \dot{\theta}}{d \theta}
$$
We integrate Equation $$
\int \dot{\theta} d \dot{\theta}=\frac{g}{a} \int \sin \theta d \theta
$$ which results in
$$
\frac{\dot{\theta}^2}{2}=\frac{-g}{a} \cos \theta+\frac{g}{a}
$$
where the integration constant is $g / a$, because $\dot{\theta}=0$ at $t=0$ when $\theta=0$. Substituting $\dot{\theta}^2$ from Equation, after solving for $\lambda$,
$$
\lambda=m g(3 \cos \theta-2)
$$
which is the force of constraint. The particle falls off the hemisphere at angle $\theta_0$ when $\lambda=0$.
$$
\begin{aligned}
\lambda & =0=m g\left(3 \cos \theta_0-2\right) \\
\theta_0 & =\cos ^{-1}\left(\frac{2}{3}\right)
\end{aligned}
$$
| A particle of mass $m$ starts at rest on top of a smooth fixed hemisphere of radius $a$. Determine the angle at which the particle leaves the hemisphere. | $\cos ^{-1}\left(\frac{2}{3}\right)$ | 48.189685 | 7.10 | $^\circ$ | class |
|
From the thrust, we can determine the fuel burn rate:$$\frac{d m}{d t}=\frac{\text { thrust }}{-u}=\frac{37 \times 10^6 \mathrm{~N}}{-2600 \mathrm{~m} / \mathrm{s}}=-1.42 \times 10^4 \mathrm{~kg} / \mathrm{s}
$$The final rocket mass is $\left(2.8 \times 10^6 \mathrm{~kg}-2.1 \times 10^6 \mathrm{~kg}\right)$ or $0.7 \times 10^6 \mathrm{~kg}$. We can determine the rocket speed at burnout $\left(v_b\right)$.
$$
\begin{aligned}
v_b & =-\frac{9.8 \mathrm{~m} / \mathrm{s}^2\left(2.1 \times 10^6 \mathrm{~kg}\right)}{1.42 \times 10^4 \mathrm{~kg} / \mathrm{s}}+(2600 \mathrm{~m} / \mathrm{s}) \ln \left[\frac{2.8 \times 10^6 \mathrm{~kg}}{0.7 \times 10^6 \mathrm{~kg}}\right] \\
v_b & =2.16 \times 10^3 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
| Consider the first stage of a Saturn $V$ rocket used for the Apollo moon program. The initial mass is $2.8 \times 10^6 \mathrm{~kg}$, and the mass of the first-stage fuel is $2.1 \times 10^6$ kg. Assume a mean thrust of $37 \times 10^6 \mathrm{~N}$. The exhaust velocity is $2600 \mathrm{~m} / \mathrm{s}$. Calculate the final speed of the first stage at burnout. | 2.16 | 2.16 | 9.12 | $10^3 \mathrm{~m} / \mathrm{s}$ | class |
|
Find the effective annual yield of a bank account that pays interest at a rate of 7%, compounded daily; that is, divide the difference between the final and initial balances by the initial balance. | 7.25 | 7.25 | page 130-7 | % | diff |
||
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains $200 \mathrm{~L}$ of a dye solution with a concentration of $1 \mathrm{~g} / \mathrm{L}$. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of $2 \mathrm{~L} / \mathrm{min}$, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches $1 \%$ of its original value. | 460.5 | 460.5 | Page 59-1 | min | diff |
||
A certain vibrating system satisfies the equation $u^{\prime \prime}+\gamma u^{\prime}+u=0$. Find the value of the damping coefficient $\gamma$ for which the quasi period of the damped motion is $50 \%$ greater than the period of the corresponding undamped motion. | 1.4907 | 1.4907 | page203-13 | diff |
|||
Find the value of $y_0$ for which the solution of the initial value problem
$$
y^{\prime}-y=1+3 \sin t, \quad y(0)=y_0
$$
remains finite as $t \rightarrow \infty$ | -2.5 | -2.5 | Page 40-30 | diff |
|||
A certain spring-mass system satisfies the initial value problem
$$
u^{\prime \prime}+\frac{1}{4} u^{\prime}+u=k g(t), \quad u(0)=0, \quad u^{\prime}(0)=0
$$
where $g(t)=u_{3 / 2}(t)-u_{5 / 2}(t)$ and $k>0$ is a parameter.
Suppose $k=2$. Find the time $\tau$ after which $|u(t)|<0.1$ for all $t>\tau$. | 25.6773 | 25.6773 | page336-16 | diff |
|||
Suppose that a sum $S_0$ is invested at an annual rate of return $r$ compounded continuously.
Determine $T$ if $r=7 \%$. | 9.90 | 9.90 | page 60-7 | year | diff |
||
A mass weighing $2 \mathrm{lb}$ stretches a spring 6 in. If the mass is pulled down an additional 3 in. and then released, and if there is no damping, by determining the position $u$ of the mass at any time $t$, find the frequency of the motion | $\pi/4$ | 0.7854 | page202-5 | s | diff |
||
If $\mathbf{x}=\left(\begin{array}{c}2 \\ 3 i \\ 1-i\end{array}\right)$ and $\mathbf{y}=\left(\begin{array}{c}-1+i \\ 2 \\ 3-i\end{array}\right)$, find $(\mathbf{y}, \mathbf{y})$. | 16 | 16 | page372-8 | diff |
|||
Consider the initial value problem
$$
4 y^{\prime \prime}+12 y^{\prime}+9 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-4 .
$$
Determine where the solution has the value zero. | 0.4 | 0.4 | page172-15 | diff |
|||
A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. What monthly payment rate is required to pay off the loan in 3 years? | 258.14 | 258.14 | page131-9 | $ | diff |
||
Consider the initial value problem
$$
y^{\prime \prime}+\gamma y^{\prime}+y=k \delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0
$$
where $k$ is the magnitude of an impulse at $t=1$ and $\gamma$ is the damping coefficient (or resistance).
Let $\gamma=\frac{1}{2}$. Find the value of $k$ for which the response has a peak value of 2 ; call this value $k_1$. | 2.8108 | 2.8108 | page344-15 | diff |
|||
If a series circuit has a capacitor of $C=0.8 \times 10^{-6} \mathrm{~F}$ and an inductor of $L=0.2 \mathrm{H}$, find the resistance $R$ so that the circuit is critically damped. | 1000 | 1000 | page203-18 | $\Omega$ | diff |
||
If $y_1$ and $y_2$ are a fundamental set of solutions of $t y^{\prime \prime}+2 y^{\prime}+t e^t y=0$ and if $W\left(y_1, y_2\right)(1)=2$, find the value of $W\left(y_1, y_2\right)(5)$. | 2/25 | 0.08 | page156-34 | diff |
|||
Consider the initial value problem
$$
5 u^{\prime \prime}+2 u^{\prime}+7 u=0, \quad u(0)=2, \quad u^{\prime}(0)=1
$$
Find the smallest $T$ such that $|u(t)| \leq 0.1$ for all $t>T$. | 14.5115 | 14.5115 | page163-24 | diff |
|||
Consider the initial value problem
$$
y^{\prime}=t y(4-y) / 3, \quad y(0)=y_0
$$
Suppose that $y_0=0.5$. Find the time $T$ at which the solution first reaches the value 3.98. | 3.29527 | 3.29527 | Page 49 27 | diff |
|||
Consider the initial value problem
$$
2 y^{\prime \prime}+3 y^{\prime}-2 y=0, \quad y(0)=1, \quad y^{\prime}(0)=-\beta,
$$
where $\beta>0$.
Find the smallest value of $\beta$ for which the solution has no minimum point. | 2 | 2 | page144-25 | diff |
|||
Consider the initial value problem
$$
y^{\prime}=t y(4-y) /(1+t), \quad y(0)=y_0>0 .
$$
If $y_0=2$, find the time $T$ at which the solution first reaches the value 3.99. | 2.84367 | 2.84367 | Page 49 28 | diff |