# Solutions to Fall 2016 Midterm 2

## Question 1 Solution (Average 23.5/30, 78.3%)

A 9.9 kg block is lying on a frictionless surface attached to a massless spring with spring constant 200N/m. A 100g bullet is fired in to the the block. Before the collision the bullet has a velocity of 120ms$^{-1}$.The collision between the block and the bullet takes 0.001 seconds. After the collision the block moves to the left with the bullet lodged in it.

A. (5 points) What is the momentum (magnitude and direction) of the bullet before the collision?

$p=mv=0.1\times120=12\,\mathrm{kg\,m\,s^{-1}}$ to the left

B. (5 points) What is the velocity (magnitude and direction) of the block, with the bullet lodged in it, immediately following the 0.001 seconds it takes for the collision to take place?

$v'=\frac{p}{m'}=\frac{12}{10}=1.2\,\mathrm{m\,s^{-1}}$

C. (5 points) What is the average force (magnitude and direction) exerted by the bullet on the block during the 0.001 second collision?

$F_{Ave}=\frac{\Delta P_{brick}}{t}=\frac{9.9\times1.2}{0.001}=11880\,\mathrm{N}$ left

D. (5 points) What is the average force (magnitude and direction) exerted by the block on the bullet during the 0.001 second collision?

$F_{Ave}=\frac{\Delta P_{bullet}}{t}=\frac{0.1\times118.8}{0.001}=11880\,\mathrm{N}$ right. This answer may also be found by applying Newton's Third Law

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

$KE_{lost}=-\Delta{KE}=-(\frac{1}{2}10\times1.2^{2}-\frac{1}{2}0.1\times120^{2})=712.8\,\mathrm{J}$

F. (5 points) What is the maximum compression of the spring from it's original length in the motion that takes place after the collision?

$KE_{final}=\frac{1}{2}10\times1.2^{2}=7.2=\frac{1}{2}200x^{2}$

$x=0.268\,\mathrm{m}$

## Question 2 Solution (Average:30.97/40=77.4%)

A spider hanging from a thread can be approximated as a solid sphere of radius 1.5 cm with 8 equally spaced legs, approximated as uniform rods of length 2 cm equally spaced around the equator of the sphere. The spider can either allow the legs to dangle vertically, or lift them so that they are horizontal. The spider's body has mass 10g and each leg has a mass of 1.5 g. Initially the legs of the spider are hanging vertically, as shown on the left, and the spider is spinning with an angular velocity of $\omega=5 \,\mathrm{rad/s}$. The spider would like to slow it's rate of spinning and so it raises its legs in to the horizontal configuration as shown on the right. In the following questions please give your answers to three significant figures and use scientific notation when appropriate.

A. (5 points) What is the moment of inertia of the spider for rotation around the thread when it's leg's hang vertically?

$I=I_{body}+I_{legs}=\frac{2}{5}m_{body}r^{2}+8m_{legs}r^{2}=\frac{2}{5}\times10\times10^{-3}\times0.015^{2}+8\times1.5\times10^{-3}\times0.015^{2}=3.6\times10^{-6}\,\mathrm{kg\,m^{2}}$

B. (5 points) What is the initial angular momentum of the spider?

$L=I\omega=3.6\times10^{-6}\times5=1.8\times10^{-5}\,\mathrm{kg\,m^{2}\,s^{-1}}$

C. (5 points) What is initial rotational kinetic energy of the spider?

$KE=\frac{1}{2}I\omega^{2}=\frac{1}{2}3.6\times10^{-6}\times5^{2}=4.5\times10^{-5}\,\mathrm{J}$

D. (5 points) What is the moment of inertia of the spider for rotation around the thread when it's leg's stick out horizontally?

$I'=I_{body}+I'_{legs}=\frac{2}{5}m_{body}r^{2}+8(\frac{1}{12}m_{legs}l^{2}+m_{legs}h^{2})=\frac{2}{5}\times10\times10^{-3}\times0.015^{2}+8(\frac{1}{12}\times1.5\times10^{-3}\times0.02^{2}+1.5\times10^{-3}\times0.025^{2})=8.8\times10^{-6}\,\mathrm{kg\,m^{2}}$

(e) (5 points) What is the final angular velocity of the spider, $\omega'$, when the spider sticks it's legs out horizontally? Give your answer in both $\mathrm{s^{-1}}$ and rpm.

$\omega'=\frac{I}{I'}\omega=2.05\,\mathrm{s^{-1}}$

To convert to rpm, multiply by 60 and divide by $2\pi$ giving $19.5 \mathrm{rpm}$

(f) (5 points) What is the final rotational kinetic energy of the spider?

$KE'=\frac{1}{2}I'\omega'^{2}=\frac{1}{2}\times8.8\times10^{-6}\times2.05^{2}=1.84\times10^{-5}\,\mathrm{J}$

(g) (5 points) How much work does the spider do when it extends it's legs?

$W=\Delta KE+ \Delta PE$

$\Delta KE=1.84\times10^{-5}-4.5\times10^{-5}=-2.66\times10^{-5}\,\mathrm{J}$

$\Delta PE=8m_{legs}g\Delta h=8\times1.5\times10^{-3}\times9.8\times0.01=1.18\times10^{-3}\,\mathrm{J}$

$W=1.15\times10^{-3}\,\mathrm{J}$

(h) (5 points) What is the name of the spider? (All suggestions accepted)

## Question 3 Solution (Average: 19.28/30=64.3%)

A 0.3kg metal ball of radius 2cm is suspended from a uniform metal bar of length 1.1m and mass 0.8kg by a string under tension $T_{1}$. The metal ball is completely submerged in a cylindrical glass containing water ($\rho_{water}=1000\,\mathrm{kg\,m^{-3}})$ that has a 10\,cm diameter. The glass contains 1L (0.001$\mathrm{m^{3}})$ of water. The string is tied to the bar 0.7m from the wall to which the bar is attached. A second string under tension $T_{2}$ is tied to the end of the bar and to the wall, making a 30$^{\circ}$ angle with the bar.

A. (5 points) What is the density of the metal used to make the ball?

$\rho=\frac{m}{V_{sphere}}=\frac{0.3}{4/3\pi0.02^{3}}=8953.5\,\mathrm{kg\,m^{-3}}$

$V_{sphere}=3.35\times10^{-5}\mathrm{m^{-3}}$

B. (5 points) What is the pressure at the bottom of the glass of water? (Atmospheric pressure is 101.3 kPa).

$\pi r_{cylinder}^{2}h=V_{water}+V_{sphere}=0.0010335$

$h=0.132\mathrm{m}$

$P=P_{0}+\rho_{water}gh=P_{0}+1293.6\mathrm{Pa}=102.6\mathrm{kPa}$

C. (5 points) What is the tension $T_{1}$?

$T_{1}=mg-F_{B}=0.3\times9.8-1000\times3.35\times10^{-5}=2.61\mathrm{N}$

D. (5 points) What is the tension $T_{2}$?

Sum of the torques around the point where the bar is attached to the wall

$\Sigma \tau=T_{2}\sin 30^{\circ}\times1.1-T_{1}\times0.7-0.8\times9.8\times0.55=0$

$T_{2}=\frac{2.61\times0.7+0.8\times9.8\times0.55}{\sin 30^{\circ}\times1.1}=11.17\,\mathrm{N}$

E. (5 points) What is the magnitude of the component of the force that the wall exerts on the bar in the vertical direction? Is it directed up or down?

Sum of the vertical forces is zero

$F_{wy}+T_{2}\sin 30^{\circ}-0.8g-T_{1}=0$

$F_{wy}=2.61+0.8\times9.8-11.17\sin 30^{\circ}=4.86\,\mathrm{N}$ up

(f) (5 points) What is the magnitude of the component of the force that the wall exerts on the bar in the horizontal direction? Is it directed towards the wall or away from the wall?

Sum of the horizontal forces is zero

$F_{wx}-T_{2}\cos 30^{\circ}=0$

$F_{wx}=9.67\,\mathrm{N}$ away from the wall