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tags: homework
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# Physics homework 1

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## 2


$\text{let }p(t)=\text{the position of a projectile after }t\text{ second}$

$p_x(t)=v_{0x}t=(v\cos\theta)t$

$p_y(t)=v_{0y}t-\frac12a_yt^2=(v\sin\theta)t-\frac12gt^2$

$p_y'(t)=v\sin\theta-gt$

$\text{let }t_m=\text{the time that x-axis of the projectile equals to }2h$

$p_x(t_m)=\sqrt{4gh}\cos\theta t_m=2h$

$t_m=\frac{2h}{\sqrt{4gh}\cos\theta}$

$p_y(t_m)=\sqrt{4gh}\sin\theta t_m-\frac12g{t_m}^2=h$

$\Rightarrow \sqrt{4gh}\sin\theta\frac{2h}{\sqrt{4gh}\cos\theta}-\frac12g(\frac{2h}{\sqrt{4gh}\cos\theta})^2=h$
$\Rightarrow\frac{\sqrt{4gh}\sin\theta2h(\sqrt{4gh}\cos\theta)-2gh^2}{(\sqrt{4gh}\cos\theta)^2}=h$
$\Rightarrow 4gh^2\sin(2\theta)-4gh^2\cos^2\theta=2gh^2$
$\Rightarrow gh^2(4\sin(2\theta)-4(\frac{1+\cos2\theta}2))=2gh^2$
$\Rightarrow4\sin(2\theta)-2\cos(2\theta)=2+2=4$
$\Rightarrow 2\sin(2\theta)-\cos(2\theta)=2$
$\Rightarrow \sqrt{5}(\frac{2}{\sqrt{5}}\sin(2\theta)-\frac{1}{\sqrt{5}}\cos(2\theta))=2$

$\text{Let }\sin\phi=\frac1{\sqrt5},\ \cos\phi=\frac2{\sqrt5}$

$\Rightarrow\sin(2\theta-\phi)=\frac{2}{\sqrt{5}}=
\pm\cos\phi$

$2\theta-\phi=\phi-\frac{\pi}{2}$
$\Rightarrow\theta=\phi-\frac{\pi}4\text{ which < 0, conflict}$

$2\theta-\phi=\frac{\pi}{2}-\phi$
$\Rightarrow \theta=\frac{\pi}{4}$

$\text{therefore, }\theta=\frac{\pi}4 \text{ is the only answer to this question}$

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## 3


### a.


$\text{Let }T=\text{the time of flight}$

$\vec v_{sum}=\vec v+\vec u$

$T=\frac{L}{|\vec v_{sum}|}=\frac{L}{|\vec v+\vec u|}=\frac{L}{v+u}$

### b.


$|\vec v_{sum}|=\sqrt{v^2-u^2}$

$T=\frac{L}{|\vec v_{sum}|}=\frac{L}{\sqrt{v^2-u^2}}$

### c.


$h=u\sin\theta$
$|\vec v_{sum}|=u\cos\theta+\sqrt{v^2-h^2}=u\cos\theta+\sqrt{v^2-u^2\sin^2\theta}$

$T=\frac{L}{|\vec v_{sum}|}=\frac{L}{u\cos\theta+\sqrt{v^2-u^2\sin^2\theta}}$

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## 4


### a.

$a_r=\frac{v^2}{r}=25m/s^2$

### b.

$a_t=g$

### c.


### d.

magnitude:

$a=\sqrt{{a_r}^2+{a_t}^2}=\sqrt{25+g^2}$

direction $\phi$ : $\arctan(\frac{a_r}{a_t})$

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## 5


### a.



### b.

$\text{consider the 5kg block}$
$N=5g$
$T=f_kN=0.2(5g)=g$

$\text{consider the 10kg block}$

$N=5g+10g=15g$
$a=\frac{F_{sum}}{m}=\frac{F-f_kN-f_k(5g)}{10}=\frac{45N-4g}{10}$

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## 6


### a.

$v_{m1}=2v_{m2}=8m/s$

$T_{s1}=m_1a=m_1(\frac{{v_{m1}}^2}{2l})=4(\frac{64}1)=256(N)$

$\text{the force that }m_2\text{ need to keep rotating}=m_2\frac{{v_{m2}}^2}{l}=3(\frac{4}{0.5})=24(N)$

$T_{s2}=256+24=280(N)$

### b.

String 2, because the tension in string 2 is bigger than string 1.

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## 7


### a.

$-bv=ma$
$\Rightarrow -bv=m\frac{dv}{dt}$
$\Rightarrow m\frac{dv}{dt}+bv=0\rightarrow \text{first order differential equation}$

$\text{Let }v_h(t)=Ae^{\alpha t}$

$\Rightarrow mA\alpha e^{\alpha t}+bAe^{\alpha t}=0$
$\Rightarrow m\alpha+b=0$
$\alpha=-\frac{b}m$

$v(t)=Ae^{-\frac{b}{m}t}$

$v(0)=v_i=Ae^0=A$
$\Rightarrow A=v_i$

$v(t)=v_i\cdot e^{-\frac{b}{m}t}$

### a. (another solution)

$-bv=m(dv/dt)$
$\Rightarrow -\frac{b}{m}dt=\frac{dv}{v}$
$\Rightarrow \frac{-b}{m}\int_0^t dt=\int_{v_i}^v\frac{1}{v}dv$
$\Rightarrow-\frac{b}{m}t=\ln v|^v_{v_i}=\ln v-\ln v_i=\ln\frac{v}{v_i}$
$\Rightarrow \frac{v}{v_i}=e^{-\frac{bt}{m}}$
$\Rightarrow v=v_i\cdot e^{-\frac{bt}{m}}$

### b.


### c.

$e^{-\frac{bt}{m}}$ will not be $0$ unless $t=\infty$, therefore the answer is no.

### d.

$\begin{align} &\int^\infty_0v_ie^{-\frac{bt}{m}}dt\\=&v_i\int^\infty_0e^{-\frac{bt}{m}}dt\\=&v_i(-\frac mbe^{\frac{-bt}m})|^\infty_0\\=&0-(-\frac{mv_i}{b})=\frac{mv_i}{b}\text{ which is constant}\end{align}$

Therefore, the answer is yes.

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## 8


### a

$F-bv=ma$
$\Rightarrow F-bv=m\frac{dv}{dt}$
$\Rightarrow m\frac{dv}{dt}+bv=F\rightarrow \text{first order differential equation}$

$v(t)=v_h(t)+v_p(t)$

$\text{while }t\rightarrow\infty,\ \frac{dv(t)}{dt}\rightarrow 0$
$\Rightarrow \frac{dv(t)}{dt}bv(t)=F\rightarrow bv_p=F$

$\text{therefore, }v_p=\frac{F}b$

$\text{while }m\frac{dv(t)}{dt}+bv(t)=0$

$m\frac{dv(t)}{dt}+bv(t)=0$

$\text{Let }v_h(t)=Ae^{\alpha t}$

$\Rightarrow mA\alpha e^{\alpha t}+bAe^{\alpha t}=0$
$\Rightarrow m\alpha+b=0$
$\alpha=-\frac{b}m$

$v(t)=v_h(t)+v_p(t)=Ae^{-\frac{b}{m}t}+\frac{F}b$

$v(0)=v_0=Ae^0+\frac{F}{b}=A+\frac{F}{b}$
$\Rightarrow A=v_0-\frac{F}b$

$\begin{align}
&v(t)=v_h(t)+v_p(t)\\
=&(v_0-\frac Fb)e^{-\frac{b}{m}t}+\frac{F}b\\
\end{align}$

$\because F=mg$
$v(t)=(v_0-\frac{mg}{b})e^{-\frac{b}{m}t}+\frac{mg}b$

### a. (another solution)

$\begin{align}
&F_{all}=mg-bv=ma\\
\Rightarrow&a=\frac{mg-bv}{m}
\end{align}$

$\begin{align}
&\frac{dv}{dt}=\frac{mg-bv}m\\
\Rightarrow & \int^v_{v_0}\frac{dv}{mg-bv}=\frac1m\int^t_0dt\\
\Rightarrow&(-\frac{1}{b}\ln(mg-bv))|^v_{v_0}=\frac{t}{m}\\
\Rightarrow&-\frac{1}{b}\ln(mg-bv)-(-\frac1b\ln(mg-bv_0))=\frac{t}m\\
\Rightarrow&\ln(mg-bv)=\frac{-bt+m\ln(mg-bv_0)}{m}\\
\Rightarrow&v=-\frac{1}{b}(-mg+e^{\frac{-bt}{m}+\ln(mg-bv_0)})\\
\Rightarrow&v=\frac{mg}{b}+(v_0-\frac{mg}{b})e^{\frac{-bt}{m}}
\end{align}$

### b.

$v_{ternimal}=\lim_{t\rightarrow \infty}\frac{mg}{b}$

$\tau=\frac{m}{b}$

### c.

$\text{same as a., however }F=-mg$

$\text{Therefore, }v(t)=(v_0+\frac{mg}{b})e^{-\frac{b}{m}t}-\frac{mg}b$