A highly evolved Australian in the back of a stationary pickup truck (a.k.a. a ute) fires his rifle in to the air at an angle $\theta$ to the horizontal (note that the angle shown in the diagram is not to scale). Assume that the bullet is not subject to drag forces. At the same time as the gun is fired the driver of the pickup truck begins to accelerate the truck to the right with a constant acceleration $a$.

A. (5 points) If after 15s the pickup truck is 320m down the road from where it started, what is the value of the acceleration $a$?

$x=x_{0}+v_{0}t+\frac{1}{2}at^{2}$

$320=0.5\times a\times15^2$

$a=2.84\,\mathrm{ms^{-2}}$

B. (5 points) How fast is the pickup truck going at this point?

$v=v_{0}+at$

$v=2.84\times15=42.7\,\mathrm{ms^{-1}}$

C. (5 points) At this time the highly evolved Australian is hit by the bullet he fired into the air. What is the initial velocity of the bullet? Express your answer using unit vector notation, using $\hat{i}$ for the direction the truck moves in and $\hat{j}$ for vertically upward.

$y=y_{0}+v_{y0}t-\frac{1}{2}gt^{2}$

$0=15v_{y0}-\frac{1}{2}\times9.8\times15^{2}$

$v_{y0}=73.5\,\mathrm{ms^{-1}}$

$v_{x0}=\frac{320}{15}=21.3\,\mathrm{ms^{-1}}$

$\vec{v}_{0}=21.3\hat{i}+73.5\hat{j}\,\mathrm{ms^{-1}}$

D. (5 points) What is the angle $\theta$ at which the rifle is fired?

$\tan\theta=\frac{v_{y0}}{v_{x0}}=\frac{73.5}{21.3}$

$\theta=73.83^{o}$

E. (5 points) What is the maximum height that the bullet reaches before it turns around?

$v_{y}^{2}=v_{y0}^2-2g(y-y_{0})$

$0=73.5^{2}-2\times9.8\times y$

$y=\frac{73.5^{2}}{2\times9.8}=275.6\,\mathrm{m}$

A smooth block of mass 200g is sliding along the edge of a smooth cone with constant speed. The height of the cone is 23cm, and half of it's apex angle is 40$^{o}$.

**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?

$mg=0.2\mathrm{kg}\times9.81\mathrm{ms^{-2}}=1.962\mathrm{N}$

**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?

$mg\sin(50^{o})=1.5N$

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

$F_{N}\sin(40^{o})=mg$

$F_{N}=\frac{1.962\mathrm{ms^{-2}}}{\sin(40^{o})}=3.05\mathrm{N}$

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

$\frac{mv^{2}}{r}=F_{N}\cos(40^{o})=\frac{mg}{\tan(40^{o})}$

$r=0.3\tan{40^{o}}$

$v^{2}=0.3g$

$v=1.71\mathrm{ms^{-1}}$

A. (5 points) Considering a case where the system is accelerating to the right, add arrows to the diagram indicating the forces acting on $m_{1}$ and $m_{2}$

B. (5 points) If $m_{2}=m_{1}$ and $\theta=30^{o}$ find the acceleration of the system in terms of $g$ for $\mu=\frac{1}{2}$.

If $m_{2}=m_{1}$, $a=0\,\mathrm{ms^{-2}}$

C. (5 points) If $m_{2}=2m_{1}$ and $\theta=30^{o}$ find the acceleration of the system in terms of $g$ for $\mu=\frac{1}{3\sqrt{3}}$.

D. (5 points) If $m_{2}=2m_{1}$ and $\theta=30^{o}$ find the acceleration of the system in terms of $g$ for $\mu=\frac{1}{9}$.

E. (5 points) If $m_{2}=2m_{1}$ and $\theta=30^{o}$ find the acceleration of the system in terms of $g$ for $\mu=\frac{1}{\sqrt{3}}$.

C,D,E. If the system accelerates it will do it to the right. We can consider the case where the frictional force has it's maximum value.

Newton's second law for $m_{2}$

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

Newton's second law for $m_{1}$

$m_{1}a=-m_{1}g\sin\theta-\mu m_{1}g\cos\theta+T$

Combining these equations gives

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

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

Substitute in mass and angles

$a=\frac{g\sin\theta-3\mu g\cos\theta}{3}=\frac{0.5-\frac{3\sqrt{3}}{2}\mu}{3}$

Now we can test for various $\mu$. If we get a negative value this means that the frictional force is sufficient that motion does not occur and acceleration will be zero.

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

D. $a=0.07\,\mathrm{ms^{-2}}$

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

A solid cube of mass 4kg is partially submerged in water. The density of the cube is 400kg/m$^{3}$ and the density of water is 1000kg/m$^{3}$

A. (5 points) What is the volume of the cube and the length of each side?

$\rho=\frac{m}{V}$

$V=\frac{4}{400}=.01\,\mathrm{m^{3}}$

$l=V^{\frac{1}{3}}=0.01^{\frac{1}{3}}=.215\,\mathrm{m}$

B. (5 points) What fraction of the volume of the cube is submerged?

$F_{B}=F_{G}$

$f$ denotes the submerged fraction

$\rho_{water}gfV=\rho_{cube}gV$

$f=\frac{\rho_{cube}}{\rho_{water}}=0.4$

C. (5 points) What is the pressure on the bottom face of the cube?

Depth at bottom of cube is $0.4\times0.215=.086\,\mathrm{m}$

$P=P_{0}+\rho gh$

$P=1.013\times10^{5}+1000\times9.81\times0.086=1.018\times10^{5}\,\mathrm{Pa}$

D. (10 points) If the cube is pushed down and then released it executes simple harmonic motion. What is the frequency of this simple harmonic motion?

If the block is pushed down a distance $x$ from the equilibrium position the restoring force is given by the excess bouyant force

$F=rho_{water}gAx$

where $A$ is the area of one of the cube faces.

This means there is an effective spring constant $k=\rho_{water}gA=1000\times9.8\times0.215^2=453\,\mathrm{N/m}$

The frequency of the simple harmonic motion is $f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{453}{4}}=1.69\,\mathrm{s}$

A 9.9 kg block is lying on a frictionless surface attached to a massless spring. A 100g bullet is fired in to the the block. Before the collision the bullet has a velocity of 120ms$^{-1}$. Following the collision the block (with the bullet embedded inside it) executes simple harmonic motion with an amplitude of 15cm.

A. (5 points) What is the maximum velocity of the simple harmonic motion?

The maximum velocity of the SHM will be the velocity of the block+bullet immediately after the collision. This is found using conservation of momentum

$(9.9+0.1)v=0.1\times120$

$v_{max}=1.2\mathrm{ms^{-1}}$

B. (5 points) How much kinetic energy is lost in the collision?

$\Delta KE = \frac{1}{2}(m_{block}+m_{bullet})v_{f}^{2}-\frac{1}{2}m_{bulllet}v_{i}^{2}=\frac{1}{2}10\times1.2^2-\frac{1}{2}0.1\times120^2=-712.8\mathrm{J}$

C. (5 points) What is the spring constant of the spring attached to the block?

From conservation of energy

$\frac{1}{2}kA^2=\frac{1}{2}(m_{block}+m_{bullet})v_{max}^2$

$k=\frac{10\times1.2^2}{0.15^2}=640\,\mathrm{N/m}$

D. (5 points) What is the frequency of the simple harmonic motion?

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{640}{10}}=1.273\,\mathrm{s}$

E. (5 points) What is the maximum acceleration of the simple harmonic motion?

This can be found either from the maximum force divided by the mass

$a_{max}=\frac{kA}{(m_{block}+m_{bullet})}=\frac{640\times0.15}{10}=9.6\,\mathrm{ms^{-2}}$

or from the relation

$a_{max}=\omega^{2}A=(2\pi)^2\times1.273^2\times0.15=9.6\,\mathrm{ms^{-2}}$

3 moles of a ideal monatomic gas are heated from atmospheric pressure 20$^{\circ}$C to 120$^{\circ}$C at constant volume. The gas is then further heated from 120$^{\circ}$C to 220$^{\circ}$C at constant pressure.

A. (5 points) How much does the internal energy of the gas change in each process?

For both processes

$\Delta E_{int}=\frac{3}{2}nR\Delta T=\frac{3}{2}\times3\times8.314\times100=3741.3\mathrm{J}$

B. (5 points) How much heat is added to the gas in each process?

For an ideal monatomic gas expanded at constant volume

$Q=\frac{3}{2}nR\Delta T=\frac{3}{2}\times3\times8.314\times100=3741.3\mathrm{J}$

At constant pressure

$Q=\frac{5}{2}nR\Delta T=\frac{5}{2}\times3\times8.314\times100=6235.5\mathrm{J}$

C. (5 points) How much work is done by the gas in each process?

At constant volume $W=0J$

At constant pressure $W=nR\Delta T=2494.2\mathrm{J}$

D. (5 points) What is the entropy change of the gas in each process?

For the constant volume process

$\Delta S=\int \frac{dQ}{T}=\frac{3}{2}nR\int_{T_{1}}^{T_{2}}\frac{dT}{T}=\frac{3}{2}nR\ln\frac{T_{2}}{T_{1}}=\frac{3}{2}\times3\times8.314\ln\frac{393}{293}=10.985\mathrm{J/K}$

For the constant pressure process

$\Delta S=\int \frac{dQ}{T}=\frac{5}{2}nR\int_{T_{1}}^{T_{2}}\frac{dT}{T}=\frac{5}{2}nR\ln\frac{T_{2}}{T_{1}}=\frac{5}{2}\times3\times8.314\ln\frac{493}{393}=14.136\mathrm{J/K}$

E. (5 points) What is the final pressure and volume of the gas?

$PV=nRT$

$P_{initial}=1.013\times10^{5}\mathrm{Pa}$

$V_{initial}=\frac{3\times8.314\times293}{1.013\times10^{5}}=0.072\mathrm{m^{3}}$

$P_{final}=\frac{3\times8.314\times393}{V_{initIal}}=1.386\times10^{5}\mathrm{Pa}$

$V_{final}=\frac{3\times8.314\times493}{P_{final}}=0.090\mathrm{m^{3}}$