Midterm 2 Results

The exam was clearly too long, probably by about 20%. Therefore I graded the exam as if it was out of 80 and converted that to a percentage. When that was done the average turned out to be almost the same as the first midterm. (M1 average: 69%, M2 average 66%).

Adjusted midterm score

Originally it was advertised that midterm 1 would be worth 10% and midterm 2 would be 15% of the grade. Several students clearly had a “bad day” on one or the other of the midterms. To try to take that in to account in a way fair to everyone I have given everyone a “transferable 5%” allowing you to transfer 5% based on your best midterm to replace 5% of your worst, which means that for every student the 25% of the overall score due to midterms is determined by either 5%/20% M1/M2 or 15%/10%, whichever gives a higher score.

Current standings and grade projections

Here are the current weighted averages and the grades I would give if I was going to give grades today. NOTE: About half the assessment for the course is still to come! I will give as many A's as I think appropriate (ie. all the people on A- could get an A if they improve enough, but they can also slip to the B's..) Technically people on B+ could also get an A if they do very well for the rest of the course, and an A- is definitely achievable. Everyone should pay special attention to upcoming homework, lab reports..and of course the final exam.

Q1 Solutions (Ave: 12.5/20)

a) $v=at$

$1=a\times0.5$ $a=2\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}Mr^2\alpha$ → $F-T=\frac{1}{2}Ma$

$T-mg=ma$ → $T=mg+ma$


$F=0.5\times 0.2\times 2+1\times(9.8+2)=12\mathrm{N}$


c) At constant speed $a=0$ so $F=mg=1\times9.8=9.8\mathrm{N}$

d) $\omega=\frac{v}{r}=\frac{1}{0.1}=10\mathrm{rad\,s^{-1}}=10\times\frac{60}{2\pi}=95.5\mathrm{rpm}$

Q2 Solutions (Ave 17.8/25)

a) Start from projectile motion

$y=\frac{1}{2}gt^{2}$ → $t^{2}=\frac{2}{g}$ ($y=1$)

Use conservation of energy to get the horizontal velocity

$\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}$ → $\frac{1}{2}mv^{2}+\frac{1}{2}mfv^{2}$

where we have expressed the moment of inertia as $I=fmr^{2}$

$\frac{1}{2}m(1+f)v_{x}^{2}=mgh$ → $v_{x}^2=\frac{2gh}{1+f}$

$x^2=v_{x}^2\frac{2}{g}$ → $x^{2}=\frac{4h}{1+f}$ → $f=\frac{4h}{x^2}-1$

$f=\frac{4\times0.09375}{0.5^2}-1=\frac{1}{2}$ so we should choose a disk

b) g cancels out in the above equation so your choice should not change.

c) Total kinetic energy is the same as the potential energy lost $mgh=0.1\times9.8\times1.09375=1.07\mathrm{J}$

d) Rotational kinetic energy is only gained while the object is rolling. You can calculate it a few ways, for example, $\frac{1}{4}mv_{x}^{2}$ or $\frac{1}{3}0.1\times9.8\times0.09375=0.031\mathrm{J}$

Q3 Solutions (Ave: 7.6/30..Ooops)

a) Conservation of Momentum

$mv=mv'+\Delta p_{rod}=mv'+I\frac{\omega'}{r}$

This comes from the fact that $\vec{L}=\vec{r}\times\vec{p}$ and as $\vec{r}$ and $\vec{p}$ are at right angles to each other $L=rp$ so $\Delta p =\frac{\Delta L}{r}$ aand $\Delta L= I\omega'$

Taking in to account the moment of inertia of the rod is $I=\frac{1}{12}Ml^{2}$


This is not enough to solve! You also need to use conservation of energy


Rearrange the two equations


$\frac{1}{2}mv^{2}-\frac{1}{2}mv'^{2}=\frac{1}{2}\frac{1}{12}Ml^{2}\omega'^{2}$ → $m(v-v')(v+v')=\frac{1}{12}Ml^{2}\omega'^{2}$

Divide the second equation by the first

$v+v'=\omega'r$ → $\omega=\frac{v+v'}{r}$

b) Now to get v' we can substitute our formula for $\omega'$ in to our first equation


Group terms with $v$ on one side, terms with $v$' on the other and the rearrange to find


Q3 solutions part 2


c) Inspection shows that when $m$ is much less than $\frac{1}{12}M\frac{l^{2}}{r^{2}}$ the ball will have negative $v'$, when it is greater it will have positive $v'$. We are looking for when it is zero. The smallest value $\frac{1}{12}M\frac{l^{2}}{r^{2}}$ can take is when $r=\frac{l}{2}$ so we want to find the mass ratio when


which is $\frac{m}{M}=\frac{1}{3}$

d) When the ball hits the tip of the rod at the mass ratio above $v'=0$ which means from our first result $\omega=\frac{v+v'}{r}$, $\omega'=\frac{v}{r}$ and as the tangential velocity of the tip of the rod is $\omega'r$ we find that the tangential velocity of the tip of the rod is $v$, or $10\mathrm{ms^{-1}}$

Q4 Solutions (Ave 15.3/25)

a) Balance of vertical forces $T\sin\theta-mg+F_{Fr}=0$

Balance of horizontal forces $T\cos\theta-F_{N}=0$

Balance of torques around end of rod on wall $mg\frac{l}{2}=T\sin\theta l$ → $mg=2T\sin\theta$

Combining the vertical force equation and the torque equation gives

$T\sin\theta=F_{Fr}=\mu F_{N}$

Now use the horizontal force equation

$T\sin\theta=\mu T\cos\theta$



b) $T=\frac{mg}{2\sin\theta}=\frac{5\times9.8}{2\sin(11.3^{o})}=126\mathrm{N}$

c) $F_{N}=T\cos\theta=126\cos(11.3^{0})=123.6\mathrm{N}$

Midterm 2 information

The 2nd midterm was on Monday November 1st in class at 8:30 AM. Four questions.

Emphasis on

  • Energy
  • Momentum
  • Rotational Motion
  • Equilibrium

Exam is cumulative. No fluids. No integration (ie. you won't be asked to derive any moments of inertia or centers of mass).

2 Practice exams are posted, review questions will be posted soon.

Thursday's recitation - problem solving with emphasis on the type of questions I may ask on the exam.

Friday morning lecture will review the topics covered so far - a look at how to put it all together.

Review Session 2:30PM-4:30PM Friday, Physics B-131.

Video of Review Session

Part I

The battery in the camera ran out at the end of the above video, the last half an hour was recorded on my webcam and is below (not as good video..sorry)

Part II

Homework Set 8 will be combined with Set 9 and due on Monday November 8, so you don't have to worry about fluids problems till after the exam. (Don't wait till the last minute though, it will be a BIG homework set.)

phy141/examprep/m2.txt · Last modified: 2010/11/04 19:21 by mdawber
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