The first midterm will take place in class on Monday October 3rd in class starting at 8:30 AM. We will begin at 8:30 AM and finish promptly at 9:25 AM so please be on time.

The exam will consist of 3 problems similar in style to the homework problems from the first 3 homework sets. Anything covered in the first 12 lectures may be on the exam, though material that is also in the textbook and has been covered by the homework problems is emphasized (ie. you do not need to memorize Wikipedia).

You may bring one handwritten sheet of notes which can have anything you want on them, and you can write on both sides of the sheet of paper.

You should bring a calculator. Any kind of calculator is acceptable. It should be able to perform scientific functions (especially trig), but does not need to have graphing capabilities. Your calculation device should not allow you to access the internet, send and receive messages or play Angry Birds.

There are two practice exams on Mastering Physics. These have 4 problems instead of 3 to ensure complete coverage of the material which may be on the exam, but the actual exam will only have 3 problems.

You can also take a look at last year's midterm. That also has 4 problems (OK, so I'll admit it, last year I made the first midterm too long, but I won't do that this semester.) Solutions to this exam and the average score the class achieved on each problem in the exam are below.

There will be a review session from 1:30PM-3:30PM on Wednesday 28 September in B131 on the B level of the Physics building. The review session will focus on solving the practice exam problems so to get the maximum benefit from the review session it's a good idea to attempt (or at least look at) the practice exams before the review session.

The average on the exam was 72%. Here is a histogram of the scores. You can find your score on Blackboard under “My Grades”.

To give you an idea of how you are doing overall here are **very rough** grade estimates based on work done to date. This graph is based on the weighted total you can find on Blackboard, which takes into account the weighting advertised in the course syllabus. Of course, we still have most of the assessment ahead of us, so your score can change greatly. Also, I will never under any circumstances guarantee any particular grade for any particular overall score. But this should give you an idea of how you and the class overall is doing. Be warned, the next few topics are harder than what we have been doing so far, so even those of you who are doing well should not sit back and relax…

**Question 1.** (30 points) A smooth block of mass 100g is sliding along the edge of a smooth cone with constant speed. The height of the cone is 20cm, and half of it's apex angle is 30$^{o}$.

**A.** (5 points) Draw a free body diagram which represents all the forces acting on the block.

**B.** (5 points) What is the magnitude of the gravitational force acting on the block?

$mg=0.1\mathrm{kg}\times9.81\mathrm{ms^{-2}}=0.981\mathrm{N}$

**C.** (5 points) What is the magnitude of the component of the gravitational force on the block which points down the slope of the cone?

$mg\sin(60^{o})=0.85N$

**D.** (5 points) What is the magnitude of the normal force acting on the block?

$F_{N}\sin(30^{o})=mg$

$F_{N}=\frac{0.981\mathrm{ms^{-2}}}{0.5}=1.962\mathrm{N}$

**E.** (10 points) What is the speed of the block?

$\frac{mv^{2}}{r}=F_{N}\cos(30^{o})=\frac{mg}{\tan(30^{o})}$

$r=0.2\tan{30^{o}}$

$v^{2}=0.2g$

$v=1.4\mathrm{ms^{-1}}$

**Question 2.** (35 points) A plane is flying horizontally with a constant speed of 100m/s at a height $h$ above the ground, and drops a 50kg bomb with the intention of hitting a car that has just begun driving up a 10$^{o}$ incline which starts a distance $l$ in front of the plane. The speed of the car is a constant 30m/s. For the following questions use the coordinate axes defined in the figure, where the origin is taken to be the initial position of the car. (Note: The car has been stolen by a Martian trying to get hands on experience with our GPS system and our planet's survival depends on us stopping the Martian).

**A.** (5 points) What is the initial velocity of the bomb relative to the car? Write your answer in unit vector notation.

$v_{x}=100-30\cos(10^{o})=70.46\mathrm{ms^{-1}}$

$v_{y}=-30\sin(10^{o})=-5.21\mathrm{ms^{-1}}$

$\vec{v}=70.46\mathrm{ms^{-1}}\hat{i}-5.21\mathrm{ms^{-1}}\hat{j}$

**B.** (5 points) Write equations for both components ($x$ and $y$) of the car's displacement as a function of time, taking t=0s to be the time the bomb is released.

$x=30\cos(10^{o})t=29.54t\,\mathrm{m}$

$y=30\sin(10^{o})t=5.21t\,\mathrm{m}$

**C.** (5 points) Write equations for both components ($x$ and $y$) of the bomb's displacement as a function of time, taking t=0s to be the time the bomb is released.

$x=100t-l\,\mathrm{m}$

$y=h-\frac{1}{2}gt^{2}\,\mathrm{m}$

**D.** (5 points) If the bomb hits the car at time t=10s what was the height of the plane above the ground $h$ when it dropped the bomb?

$y=52.1\mathrm{m}$

$52.1=h-\frac{1}{2}g10^{2}$

$h=52.1+50\times9.81=542.61\mathrm{m}$

**E.** (5 points) What is the horizontal displacement of the plane relative to the car when the bomb hits the car at t=10s.

$0\mathrm{m}$

**F.** (5 points) How much work did gravity do on the bomb while it was falling?

$mg\frac{1}{2}g10^{2}=50\times50\times9.81=240590\mathrm{J}$

**G.** (5 points) How much kinetic energy does the bomb have when it hits the car?

$\frac{1}{2}mv_{0}^{2}+24525=\frac{1}{2}\times50\times100^{2}+240590=250000+240590=490590\mathrm{J}$

**Question 3.** (35 points) The weight of a mass $m_{1}$ suspended over a pulley is used to drag another mass $m_{2}$ along a surface with coefficient of friction $\mu$. The angle $\theta$ is the angle between the rope and the vertical direction as shown in the diagram.

**A.** (5 points) Draw free body diagrams which represent the forces on each object.

**B.** (5 points) Write equations which represent Newton's second law for each object. For the object on the surface write two equations, one for the horizontal direction and one for the vertical direction.

$m_{1}a=m_{1}g-T$

$m_{2}a=T\sin\theta-\mu F_{N}$

$m_{2}g-F_{N}-T\cos\theta=0$

$m_{2}a=T\sin\theta+\mu T\cos\theta-\mu m_{2}g$

**C.** (10 points) Find an expression for the magnitude of the acceleration of $m_{2}$ along the surface in terms of $\theta$, $g$, $\mu$, $m_{1}$ and $m_{2}$.

$m_{2}a=T\sin\theta+\mu T \cos\theta-\mu m_{2}g$

$m_{2}a=(m_{1}g-m_{1}a)(\sin\theta+\mu \cos\theta)-\mu m_{2}g$

$a(m_{2}+m_{1}(\sin\theta+\mu \cos\theta))=m_{1}g(\sin\theta+\mu \cos\theta)-\mu m_{2}g$

$a=\frac{m_{1}g(\sin\theta+\mu \cos\theta)-\mu m_{2}g}{m_{2}+m_{1}(\sin\theta+\mu \cos\theta)}$

**D.** (5 points) If $m_{1}=m_{2}$ find the acceleration in terms of $\mu$ and $g$ at $\theta=0$, $\theta=45^{o}$ and $\theta=90^{o}$.

$a=\frac{g(\sin\theta+\mu \cos\theta)-\mu g}{1+(\sin\theta+\mu \cos\theta)}$

$\theta=0$

$a=0\,\mathrm{ms^{-2}}$

$\theta=45$

$a=\frac{\frac{1}{\sqrt{2}}-\frac{\sqrt{2}-1}{\sqrt{2}}\mu}{1+\frac{1}{\sqrt{2}}+\frac{\mu}{\sqrt{2}}}g\,\mathrm{ms^{-2}}$

$\theta=90$

$a=\frac{g-\mu g}{2}\mathrm{ms^{-2}}$

**E.** (10 points) For the case where $\mu=0$ show that, independent of the values of $m_{1}$ and $m_{2}$, the acceleration is maximum when $\theta=90^{o}$. (Hint: recall that if $f(x)=\frac{u(x)}{v(x)}$ then $\frac{df}{dx}=\frac{\frac{du}{dx}v-\frac{dv}{dx}u}{v^{2}}$ )

$a=\frac{m_{1}g(\sin\theta)}{m_{2}+m_{1}(\sin\theta)}$

$\frac{da}{d\theta}=g\frac{m_{1}m_{2}\cos\theta+m_{1}^{2}\sin\theta\cos\theta-m_{1}^{2}\sin\theta\cos\theta}{(m_{2}+m_{1}(\sin\theta))^{2}}$

$\frac{da}{d\theta}=g\frac{m_{1}m_{2}\cos\theta}{(m_{2}+m_{1}(\sin\theta))^{2}}$

$\frac{da}{d\theta}=0$ when $\cos\theta=0$ which is when $\theta=90^{o}$

The denominator of $\frac{da}{d\theta}$ is always positive, and we can see that for a small change $\delta$, $\cos(90+\delta)$ is negative and $\cos(90-\delta)$ is positive, which implies this is a maximum.

Vertical motion of stone $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$ | Horizontal motion of stone $x=v_{0}\cos\theta\, t$ |

At time of impact $t_{i}$

$0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$

$t_{i}=\frac{2v_{0}\sin\theta}{g}$

$100-10t_{i}=v_{0}\cos\theta\, t_{i}$

$(v_{0}\cos\theta+10)\frac{2v_{0}\sin\theta}{g}=100$

$\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}+\frac{20v_{0}\sin\theta}{g}=100$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}=100g$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}+50=100g+50$

$(v_{0}+\frac{10}{\sqrt{2}})^2=100g+50$

$v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$

$v_{0}=25.0 \mathrm{ms^{-1}}$ or $v_{0}=-39.2\mathrm{ms^{-1}}$ but we discard the negative solution.

A. $\tan\theta=\frac{400}{400}$ → $\theta=45^{o}$

B. $v_{BWx}^2+v_{BWy}^2=25$

$v_{BWx}+1=v_{BWy}$

This can either be solved by inspection (ie, if you see that its just a 3,4,5 triangle) or

$2v_{BWx}^2+2v_{BWx}+1=25$

$v_{BWx}^2+v_{BWx}-12=0$

$(v_{BWx}-3)(V_{BWx}+4)=0$

$v_{Bwx}=3$ (negative solution doesn't work) and it follows that $v_{Bwy}=4$

so $\vec{v}_{BW}=(3\hat{i}+4\hat{j})\mathrm{ms^{-1}}$

C. $\tan\phi=\frac{4}{3}=53.13^{o}$

D. Total work done is proportional to change in velocity, which is zero, so no work is done.

A. Write Newton's 2nd Law for each body

For the skier $m_{s}a=m_{s}g\sin\theta-T$

For the tree $m_{t}a=T+m_{t}g\sin\theta-\mu_{t}m_{t}g\cos\theta$

Add the two equations together

$(m_{s}+m_{t})a=(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta$

$a=\frac{(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta}{m_{s}+m_{t}}$

B. If $v$ is constant $a=0$

$(m_{s}+m_{t})g\sin\theta-\mu_{t}m_{t}g\cos\theta=0$

$(m_{s}+m_{t})g\sin\theta=\mu_{t}m_{t}g\cos\theta$

$\tan\theta=\frac{\mu_{t}m_{t}}{(m_{s}+m_{t})}=\frac{0.2\times400}{400+80}=\frac{1}{6}$

$\theta=9.46^{o}$

C. From our first equation, when $a=0$, $m_{s}g\sin\theta-T=0$

$T=80\mathrm{kg}\times9.8\mathrm{ms^{-2}}\times\sin(9.46^{0})=129N$

For an orbit of a satellite of mass of $m$ at distance $r$ around a body of mass $M$

$ma=\frac{GMm}{r^2}$

where $a=\frac{mv^{2}}{r}$

$\frac{v^2}{r}=\frac{GM}{r^2}$ → $v^2=\frac{GM}{r}$

We need to find out what $v$ should be.

The period of the motion is $\frac{24\times60\times60}{2}\mathrm{s}=43,200\mathrm{s}$

$v=\frac{2\pi r}{T}$ (Watch out, here r is the distance from the center of the planet NOT the radius of the planet!)

$\frac{4\pi^2 r^2}{T^2}=\frac{GM}{r}$

$r^{3}=\frac{GMT^2}{4\pi^2}=\frac{G\times0.1M_{E}\times(43,200)^2}{4\pi^2}$

$r=1.23\times10^{7}$

height above surface = $r-R_{M}=r-\frac{R_{E}}{2}=9.11\times10^6\mathrm{m}$