The average on the exam was 72.6%. The histogram of scores is displayed below. Each bin label shows the bottom end of that bin, eg. to be in the top bin you needed a score of 95 or higher.

Two blocks of mass $m_{1}$ and $m_{2}$ are connected by a rope which runs over a frictionless pulley of negligible mass. Initially, $m_{1}$ rests on a surface with an incline of $\theta_{1}$ to the horizontal, while $m_{2}$ rests on a plane inclined at an angle of $\theta_{2}$ to the horizontal. Both surfaces have coefficients of static and kinetic friction $\mu_{s}$ and $\mu_{k}$.

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 $\theta$.

$m_{1}a=T+m_{1}g\sin\theta_1-\mu_{k}m_{1}g\cos\theta_1$

$m_{2}a=-T+m_{2}g\sin\theta_2-\mu_{k}m_{2}g\cos\theta_2$

$(m_{1}+m_{2})a=m_{2}g\sin\theta+m_{1}g\sin\theta_1-\mu_{k}m_{2}g\cos\theta-\mu_{k}m_{1}g\cos\theta_1$

$a=\frac{m_{2}g\sin\theta_2+m_{1}g\sin\theta_1-\mu_{k}m_{2}g\cos\theta_2-\mu_{k}m_{1}g\cos\theta_1}{m_{1}+m_{2}}$

For parts C-G consider a case where $m_{1}=$2 kg and $m_{2}$=5 kg, $\mu_{s}$=0.25 and $\mu_{k}$=0.2.

C. (5 points) If $\theta_{1}=15^{o}$ and $\theta_{2}=30^{o}$, what is the speed of $m_{1}$, 0.5\,s after the system has been released from rest? How far has $m_{1}$ travelled 0.5\,s after the system has been released from rest?

$a=2.47 \mathrm{m\,s^{-2}}$

$v=at=2.47\times0.5=1.24 \mathrm{m\,s^{-1}}$

$x=\frac{1}{2}at^{2}=0.5\times2.47\times0.5^2=0.31 \mathrm{m}$

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

Using either

$m_{1}a=T+m_{1}g\sin\theta_1-\mu_{k}m_{1}g\cos\theta_1$

or

$m_{2}a=-T+m_{2}g\sin\theta_2-\mu_{k}m_{2}g\cos\theta_2$

with $a=2.47 \mathrm{m\,s^{-2}}$

gives

$T=3.66\mathrm{N}$

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

$W_{g}=m_{2}gd\sin\theta_2+m_{1}gd\sin\theta_1=9.14\mathrm{J}$

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

$W_{Fr}=-\mu_{k}m_{2}gd\cos\theta_2-\mu_{k}m_{1}gd\cos\theta_1=-3.79\mathrm{J}$

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

$0\mathrm{J}$

A cannon fires a tennis ball with an initial velocity $v_{0}$ at an angle of $\theta$ above the horizontal at time $t=0$. A robot bucket is supposed to catch the ball. Once it starts moving the bucket moves to the left with a constant speed of $v_{B}$. The bucket is initially $x_{B0}$ m to the right of the cannon and can only move to the left.

A. (5 points) Write down equations for the $x$ and $y$ coordinates of the tennis ball as a function of time in terms of $v_{0}$, $\theta$ and $g$.

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

B. (5 points) From the equations you wrote down in part A, derive an expression for the horizontal distance $R$ from the cannon at which the tennis ball will hit the ground in terms of $v_{0}$, $\theta$ and $g$.

When the ball returns to the ground $y=0$ so

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

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

Substituting for $t$ in the $x$ equation gives

$x=R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$

C. (5 points) If the bucket starts moving at a time $t=t_{B}$ write an expression for the $x$ position of the bucket for times $t>t_{B}$ in terms of $x_{B0}$, $v_{B}$ and $t_{B}$.

$x=x_{B0}-v_{B}(t-t_{B})$

D. (5 points) If $\theta=30^{\circ}$ and $x_{B0}=40$m what is the maximum tennis ball launch velocity $v_{0}$ for which the robot bucket can successfully catch the ball?

If the ball overshoots the bucket, the bucket cannot catch it so

$\frac{2v_{0}^2\sin\theta\cos\theta}{g}<40$

$v_{0}^2<\frac{20g}{\sin\theta\cos\theta}$

$v_{0}<21.29\mathrm{\,m\,s^{-1}}$

E. (10 points) Find the time $t_{B}$ that the bucket should start moving so that it catches the ball if $v_{0}=20$m/s, $\theta=30^{\circ}$, $v_{B}=4$m/s and $x_{B0}=40$m.

The bucket needs to reach the point at which the ball hits the ground at the same time as it hits the ground. ie.

$x=R=\frac{2v_{0}^2\sin\theta\cos\theta}{g}$

and

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

Putting this in to the equation of motion for the bucket gives

$\frac{2v_{0}^2\sin\theta\cos\theta}{g}=x_{B0}-v_{B}(\frac{2v_{0}\sin\theta}{g}-t_{B})$

which can be solved for

$t_{B}=0.87\mathrm{s}$

The diagram shows the motion of a block of mass 100 grams. The block is given an initial velocity $v_0$ by a spring which was previously compressed by a distance $x_0$ from it’s equilibrium length. The block slides along a flat surface, goes through a loop and then continues up a ramp that has an angle of 45$^{\circ}$ to the horizontal, eventually coming to rest at height $h$ before turning around and sliding back down the slope. All of the surfaces that the block slides along are frictionless. The diameter of the loop is 40cm.

A. (5 points) Add arrows to the pictures of the block to indicate the directions of the forces acting on the block at each of the 5 positions shown in the diagram. Label each arrow to indicate which force it corresponds to.

B. (10 points) What is the minimum value of $v_{0}$ required such that the block does not fall off the track at the top of the loop?

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

$\frac{1}{2}mv^{2}=\frac{1}{2}mv_{0}^{2}-2mgr$

$\frac{1}{2}mgr=\frac{1}{2}mv_{0}^{2}-2mgr$

$\frac{1}{2}mv_{0}^{2}=\frac{5}{2}mgr$

$v_{0}^{2}=5gr=5\times9.8\times0.2=9.8$

$v_{0}=3.13\mathrm{\,m\,s^{-1}}$

C. (5 points) If the block has this value of $v_{0}$ what is the maximum height $h$ reached by the block?

$\frac{1}{2}mv_{0}^{2}=mgh$

$h=\frac{v_{0}^{2}}{2g}=\frac{3.13^2}{2\times9.8}=0.50\mathrm{m}$

D. (5 points) If the spring constant $k$ of the spring is 30N/m by what distance $x_{0}$ was the spring compressed initially to provide the value of $v_{0}$ you found earlier?

$\frac{1}{2}mv_{0}^{2}=\frac{1}{2}kx_{0}^{2}$

$x_{0}^{2}=\frac{mv_{0}^{2}}{k}=\frac{0.1\times9.8}{30}=0.0326$

$x_{0}=0.18\mathrm{m}$

E. (5 points) How much work did gravity do on the block between the time the spring was released and the time when it reached height $h$?

$-\frac{1}{2}mv_{0}^{2}=-mgh=W_{g}=-0.49\mathrm{J}$

F. (5 points) The ramp exerts a normal force on the block at all times. How much work does the normal force do on the block between the time the spring was released and the time when it reached height $h$?

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