# Midterm 1 Information

The first midterm will take place in class on Thursday October 3 in the recitation class starting at 2:30 PM. We will begin at 2:30 PM and finish promptly at 3:50 PM so please be on time.

The exam will consist of 3 problems similar in style to the homework problems from the first 4 homework sets. Anything covered in the first 13 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 will be a review session on Tuesday at 9AM in B131 in the physics building. It will be most useful if you have had a look at the past exams before coming to the session. The session should last about 2 hours and will be be recorded and posted online (here).

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.

Fall 2010 midterm. Note this has 4 problems instead of 3. That was a mistake that won't be repeated!

Fall 2011 midterm. The final problem on this exam has been modified from the one that was actually given as that had errors that will cause needless confusion….

## Results

The class performed very well on the exam, the average score was 86%. Here is the histogram of the scores.

## Question 1 Solution (Average: 31.2/35)

Two blocks of mass $m_{1}$ and $m_{2}$ are connected by a rope which runs over a frictionless pulley of negligible mass. One block hangs from the rope, while the other rests on a plane with coefficients of static and kinetic friction $\mu_{s}$ and $\mu_{k}$ inclined at an angle of $\theta$ to the horizontal.

A. (5 points) Add arrows indicating the direction of all of the forces acting on both $m_{1}$ and $m_{2}$ to the diagram.

B. (5 points) Assuming that the incline is sufficiently steep that the blocks move when they are released from rest, find an expression for the downwards acceleration, $a$ in terms of $m_{1}$, $m_{2}$, $g$, $\mu_{s}$ or $\mu_{k}$ and $\theta$.

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

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

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

$a=\frac{m_{1}g+m_{2}g\sin\theta-\mu_{k}g\cos\theta}{m_{1}+m_{2}}$

For parts C-G consider a case where $m_{1}=$4 kg and $m_{2}$=1 kg, $\mu_{s}$=0.2 and $\mu_{k}$=0.15.

C. (5 points) If $\theta=30^{o}$ what is the velocity and displacement of $m_{1}$, 0.5\,s after the system has been released from rest?

$a=\frac{4+1\sin 30^{o}-0.15\cos30^{o}}{4+1}\times9.8=8.57\mathrm{ ms^{-2}}$

$v=at=8.57\times0.5=4.28\mathrm{ms^{-1}}$ down

$y=\frac{1}{2}at^{2}=0.5\times8.57\times0.5^{2}=1.07m$ down

D. (5 points) What is the magnitude of the tension in the rope during the motion in part (C)?

$T=m_{1}g-m_{1}a=4\times(9.8-8.57)=4.92\mathrm{N}$

E. (5 points) How much work is done by gravity on the two block system during the motion in part (C)?

$W_{g}=m_{1}gd+m_{2}gd\sin 30^{o}=4\times9.8\times1.07+1\times9.8\times1.07\times\sin 30^{o}=47.187\mathrm{J}$

F. (5 points) How much work is done by friction during the motion in part (C)?

$W_{Fr}=\mu_{k}m_{2}g\cos\theta d=0.15\times 1 \times 9.8\times\cos30^{o}\times 1.07=1.36\mathrm{J}$

G. (5 points) How much work is done by the normal force on $m_{2}$ during the motion in part (C) ?

$W_{N}=0\mathrm{J}$

## Question 2 Solution (Average 29.4/35)

A 200 kg rocket is fired straight up from the ground with an initial velocity of 100 $\mathrm{ms^{-1}}$ after which it is subject to a constant gravitational acceleration of 9.8 $\mathrm{ms^{-2}}$ down. A 15kg projectile is fired from a cannon 1 km away at the same time as the rocket is launched. 10 seconds later the projectile hits the rocket.

A. (5 points) At what height above the ground does the collision take place?

Solve using rocket equation of motion

$y=v_{0}-\frac{1}{2}gt^{2}=100\times10-0.5\times9.8\times10^{2}=510\mathrm{m}$

B. (5 points) At what angle $\theta$ above the horizontal must the cannon be fired in order to hit the rocket?

Equations of motion of projectile

$x=v_{0}\cos\theta t$

$y=v_{0}\sin\theta t -\frac{1}{2}gt^{2}$

These give

$1000=v_{0}\cos\theta \times 10$

$1000=v_{0}\sin\theta \times 10$

So

$\theta=45^{o}$

C. (5 points) What is the initial velocity of the projectile launched from the cannon?

$v_{0x}=100\mathrm{ms^{-1}}$

$v_{0}=\frac{v_{0x}}{\cos45^{o}}=141\mathrm{ms^{-1}}$

D. (5 points) What is the kinetic energy of the rocket when it is hit by the projectile?

Can use either conservation of mechanical energy

$KE=\frac{1}{2}mv_{0}^{2}-mgh=0.5\times200\times100^{2}-200\times9.8\times510=400\mathrm{J}$

of find velocity of rocket when it is hit

$v=100-9.8\times10=2\mathrm{ms^{-1}}$

$KE=\frac{1}{2}mv^{2}=400\mathrm{J}$

E. (5 points) What is the kinetic energy of the projectile when it hits the rocket?

$KE=\frac{1}{2}mv_{0}^{2}-mgh=0.5\times15\times100^{2}\times2-15\times9.8\times510=75030\mathrm{J}$

F. (10 points) If instead of starting with the velocity found in part B the projectile was fired with a velocity of 500 ms$^{-1}$ find the required angle $\theta$ above the horizontal for the projectile to the hit the rocket and the height at which the collision occurs in this case.

Equations of motion for projectile are now:

$x=500\cos\theta t$

$y=500\sin\theta t-\frac{1}{2}gt^{2}$

The angle can be found by considering when the y coordinates of the two objects are the same

$500\sin\theta t-\frac{1}{2}gt^{2}=100t-\frac{1}{2}gt^{2}$

$\sin\theta=\frac{1}{5}$

$\theta=11.54^{o}$

The time at which the collision takes place can now be found using the equation for the x motion of the projectile.

$t=\frac{1000}{500\cos\theta}=2.04\mathrm{s}$

The height can now be found by substituting back in the y equation of motion

$y=500\times\frac{1}{5}\times2.04-0.5\times9.8\times2.04^{2}=183.6\mathrm{m}$

## Question 3 Solution (Average: 25.4/30)

A 500 kg car is traveling with a constant speed of 50 $\mathrm{ms^{-1}}$ in a circular path around the inside edge of a cylinder. The radius of the curve is 40 m. The coefficient of static friction between the road and the tires is $\mu=0.2$.

A. (5 points) How long does it take for the car to go around the cylinder once?

$T=\frac{2\pi r}{v}=\frac{2\pi \times40}{50}=5.03\mathrm{s}$

B. (5 points) What is the magnitude of the normal force exerted by the wall of the cylinder on the car.

$F_{N}=\frac{mv^{2}}{r}=\frac{500\times50^{2}}{40}=31250\mathrm{N}$

C. (5 points) What is the magnitude and direction of the frictional force on the car. Give the direction as one of the following, up, down, towards the center of the cylinder, radially outwards, in the direction the car is moving, opposite to the direction the car is moving.

$F_{Fr}=mg=500\times9.8=4900\mathrm{N}$ up

D. (5 points) How much work is done by the normal force in one complete circuit of the track?

$O \mathrm{J}$

E. (5 points) How much work is done by the frictional force in one complete circuit of the track?

$O \mathrm{J}$

F. (5 points) The car starts to slow down. At what speed does it start to slip down the wall of the cylinder?

The car starts to slip when the maximum frictional force is less than the weight of the car. The minimum velocity for this not to happen can be found from

$\mu\frac{mv^{2}}{r}=mg$

$v^{2}=\frac{rg}{\mu}=\frac{40\times9.8}{0.2}$

$v=44.3\mathrm{ms^{-1}}$

## Recitation problem 4 solution

Question 1. 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 normal force acting on the block?

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

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

C. What is the magnitude of the horizontal component of the normal force acting on the block?

$F_{Nx}=F_{N}\cos(30^{o})=1.7N$

D. (5 points) What is the speed of the block?

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

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

$v^{2}=0.2g$

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