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\begin{center}
{\Large {Solution Derivations for Capa \#9\\[20pt]} }
\end{center}
1) A $48\ kg$ shell is fired from a gun with a muzzle velocity $115\ m/s$ at
$55^{\circ}$ above the horizontal. At the top of the trajectory, the shell
explodes into two fragments of equal mass. One fragment, whose speed
immediately after the explosion is zero, falls vertically. What is the
horizontal speed of the other fragment?
\begin{quote}
$m$ = Given\\
$v$ = Given\\
$\theta $ = Given\\
$v_{1x}$ = ?\\
Right before the explosion, the velocity of the projectile is only the
horizontal component of the initial velocity, or $v_{x}=v\cos \theta $.
Momentum is always conserved, so the initial momentum must equal the final
momentum. To simplify the problem (since momentum is a vector quantity),
we can consider only the horizontal momentum. Since the particle splits
into two equally massed particles, the final momentum is the sum of momenta
of the two particles. So,
\[
mv_{0x}=\frac{1}{2}mv_{1x}+\frac{1}{2}mv_{2x}.
\]
Since one particle completely stops, its horizontal velocity is zero.
The equation becomes
\[
mv_{0x}=\frac{1}{2}mv_{1x}.
\]
The $m$'s cancel, and we're left with
\[
v_{0x}=\frac{1}{2}v_{1x}.
\]
So,
\begin{eqnarray*}
v_{1x} & = & 2v_{0x}\\
& = & 2v_{0}\cos \theta
\end{eqnarray*}
\end{quote}
2) Two identical steel balls, each of mass $2.6\ kg$, are suspended from
strings of length $34\ cm$ so that they touch when in their equilibrium
position. We pull one of the balls back until its string makes an angle
$\theta = 54^{\circ}$ with the vertical and let it go. It collides
elastically with the other ball. How high will the other ball rise above its
starting point?
\begin{quote}
$m$ = Given, equal for both balls.\\
$L$ = Given\\
$\theta $ = Given\\
Since the collision is elastic, the second ball will rise to the same height
as the first ball. The height from the ball to the mass can be found from
a triangle and the height of the ball can be found by subtracting the
previous height from the string height.
\[
h=L-L\cos \theta =L(1-\cos \theta )
\]
This can be shown to be true by conservation of energy:
\[
KE_{i}+PE_{i} = KE_{f}+PE_{f}
\]
Since the masses are equal,
\begin{eqnarray*}
0+m_{1}gh_{1} & = & 0+m_{2}gh_{2}\\
mgh_{1} & = & mgh_{2}\\
h_{1} & = & h_{2}
\end{eqnarray*}
\end{quote}
3) Suppose that instead of steel balls we use putty balls . They will collide
inelastically and remain stuck together after the collision. How high will
the balls rise after the collision?
\begin{quote}
$m$ = Given\\
$L$ = Given\\
$\theta $ = Given\\
Since this collision is inelastic, only conservation of momentum applies.
However, we can still use the energy equations but only for the particles
before and after (not during) the collision. Momentum is always conserved,
so we can always use that equation. In an inelastic collision,
\[
m_{1}v_{1i}+m_{2}v_{2i}=\left( m_{1}+m_{2}\right) v_{f}
\]
Here, the masses are the same and the second particle has no initial
velocity.
\[
mv_{1i}=2mv_{f}
\]
The $m$'s cancel, leaving
\begin{equation}
\frac{1}{2}v_{1i}=v_{f} \label{eqn:1}
\end{equation}
In order to get $v_{1i}$, we can use the energy equation
\begin{eqnarray*}
\frac{1}{2}mv_{1i}^{2} & = & mgh\\
v_{1i} & = & \sqrt{2gh}
\end{eqnarray*}
We found an equation for $h$ in the previous problem, so
\[
v_{1i}=\sqrt{2gL(1-\cos \theta )}
\]
Plugging this into (\ref{eqn:1}), we can solve for $v_{f}$
\begin{eqnarray*}
\frac{1}{2}v_{1i} & = & v_{f}\\
\frac{\sqrt{2gL(1-\cos \theta )}}{2} & = & v_{f}
\end{eqnarray*}
Now that we know the final velocity of the two masses when they stick
together, we can find the height by using the energy equation once again,
this time solving for $h$.
\begin{eqnarray*}
\frac{1}{2}mv_{f}^{2} & = & mgh\\
h & = & \frac{v_{f}^{2}}{2g}=\frac{\frac{2gL(1-\cos \theta )}{4}}{2g}
\end{eqnarray*}
Simplifying,
\[
h=\frac{2gL(1-\cos \theta )}{8g}=\frac{L(1-\cos \theta )}{4}
\]
Remember to convert the units of $L$ to meters or else enter your answer
into CAPA\ as centimeters.
\end{quote}
4) A car mass $m_{1}$ is moving to the right on a frictionless air track. It
collides with a second car, mass $m_{2}$, which is initially at rest. Which
of the following statements are true? (If A and E are true, and the others
are not, enter TFFFT).
\begin{quote}
QUESTION:\\
A) If car 1 is much lighter than $m_{2}$, and the collision is perfectly
elastic, car 1 will continue to the right with nearly its original speed
after the collision.\\
B) If car 1 sticks to car 2, the car1-car2 system must be at rest after the
collision.\\
C) If car 1 sticks to car 2, the final kinetic energy of the car1-car2
system is less than the initial total kinetic energy of the two cars.\\
D) If the collision is elastic, car 1 must always come to a stop after the
collision.\\
E) Suppose (for this statement only) that car 2 was NOT initially at rest,
but was instead heading towards car 1 with equal (but opposite) momentum
before the collision. After the collision, the cars must both be at rest.\\
ANSWER:\\
a) False. Car 1 will rebound in the opposite direction.\\
b) False. The system must move to preserve momentum.\\
c) True. KE not conserved in inelastic collisions.\\
d) False. It may, but not necessarily.\\
e) False. Momentum must be conserved, so if the cars had different masses,
they would not come to rest.\\
\end{quote}
5) A ball of mass $9.8\ g$ with a speed of $27.5\ m/s$ strikes a wall at an
angle $17.0^{\circ}$ and then rebounds with the same speed and angle. It is
in contact with the wall for $41.0\ ms$. What is the magnitude of the impulse
associated with the collision force?
\begin{quote}
$m$ = Given\\
$v$ = Given\\
$\theta $ = Given\\
$t$ = Given\\
Impulse = $I$ = ?\\
\begin{eqnarray*}
I & = & \int_{t_{1}}^{t_{2}}F\cdot dt=\Delta p\\
\Delta \mathbf{p} & = & mv_{f}-mv_{i}
\end{eqnarray*}
Add the components of the velocities to get the resultant momentum. Assume
right is positive $x$ and up is positive $y$.
\begin{eqnarray*}
\Delta \mathbf{p} & = & mv_{fx}+mv_{fy}-(mv_{ix}+mv_{iy})\\
& = & m(v_{fx}+v_{fy}-v_{ix}-v_{iy})
\end{eqnarray*}
Since the initial speed was equal to the final speed, $v=v_{f}=-v_{i}$. Since
the velocity was broken into components, the sign will determine the
direction.
\begin{eqnarray*}
\Delta \mathbf{p} & = & m(-v\sin \theta +v\cos \theta -v\sin \theta -v\cos \theta)\\
& = & -2mv\sin \theta
\end{eqnarray*}
Note that your mass may be given in grams which means you'll
need to convert it to kilograms for this problem. Momentum has units
$kg\cdot m/s$. CAPA is also looking for the magnitude of the momentum.
\end{quote}
6) What is the average force exerted by the ball on the wall?
\begin{quote}
The average force is (Equation 11-2, pg. 256)
\[\bar{F}=\frac{\Delta p}{\Delta t}
\]
Beware! The time may be given in milliseconds (ms), thus you may have to
convert it to seconds. Just remember that $1000\ ms = 1\ s$. So,
\[
\bar{F}=\frac{-2mv\sin \theta }{\Delta t}.
\]
Force is in Newtons (N). Enter the magnitude on this problem, also.
\end{quote}
7) A steel ball of mass $0.400\ kg$ is fastened to a cord $61.0\ cm$ long and
fixed at the far end. It is released when the cord is horizontal. At the
bottom of its path, the ball strikes a $3.80\ kg$ steel block initially at rest
on a frictionless surface. The collision is elastic. Find the speed of the
ball just after the collision.
\begin{quote}
$m_{1}$ = Given\\
$L$ = Given\\
$m_{2}$ = Given\\
$v_{1f}$ = ?\\
In an elastic collision, momentum is conserved but so is kinetic energy.
The two equations are
\[
m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{1f}+m_{2}v_{2f}
\]
and
\[
\frac{1}{2}m_{1}v_{1i}^{2}+\frac{1}{2}m_{2}v_{2i}^{2}=\frac{1}{2}%
m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{2f}^{2}
\]
From equations 11-9a and 11-9b,
\begin{eqnarray*}
v_{1f} & = & \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}+\frac{2m_{2}}{m_{1}+m_{2}}v_{2i}\\
v_{2f} & = & \frac{2m_{1}}{m_{1}+m_{2}}v_{1i}+\frac{m_{2}-m_{1}}{m_{1}+m_{2}}v_{2i}
\end{eqnarray*}
Since the second particle is initially at rest, and we are solving for the
final velocity of the ball, the equations become
\[
v_{1f}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}
\]
As before, we can solve for the initial velocity by conservation of energy.
Initially, the first mass has only potential energy, but at the bottom of
its swing (before it hits the second mass) it has only kinetic energy.
\begin{eqnarray*}
KE_{i}+PE_{i} & = & KE_{f}+PE_{f}\\
0+mgh & = & \frac{1}{2}mv_{1i}^{2}+0\\
2mgh & = & mv_{1i}^{2}\\
\sqrt{2gh} & = & v_{1i}
\end{eqnarray*}
The ball is released when the string is vertical, so the height is the
length of the string (convert to meters, if necessary). So,
\[
v_{1f}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\sqrt{2gh}
\]
CAPA is looking for the speed, so enter the magnitude of your answer.
\end{quote}
8) In a pool game, the cue ball, which has an initial speed of $1.9\ m/s$,
makes an elastic collision with the 8-ball, which is initially at rest. After
the collision, the 8-ball moves at an angle $\theta = 17.2^{\circ}$ with
respect to the original direction of the cue ball, as shown in the figure. At
what angle $\phi$ does the cue ball travel after the collision? Assume the
pool table is frictionless and the masses of the cue ball and the 8-ball are
equal. Use ``deg'' as units.
\begin{quote}
$v_{1i}$ = Given (Cue ball, $m_{1}$)\\
$v_{2i}$ = $0$ = Given (8 ball, $m_{2}$)\\
$\theta $ = Given\\
$\phi $ = ?\\
The collision is elastic. Since momentum is a vector, we can split it into
horizontal and vertical components. The vertical components must sum to
zero and the horizontal components must equal the initial horizontal
component. The final kinetic energy must be equal to the initial kinetic
energy also. Since the masses are equal, we can write them as a single $m$.
The equations are
\begin{eqnarray*}
mv_{1ix}+mv_{2ix} & = & mv_{1fx}+mv_{2fx}\\
mv_{1iy}+mv_{2iy} & = & mv_{1fy}+mv_{2fy}\\
\frac{1}{2}mv_{1i}^{2}+\frac{1}{2}mv_{2i}^{2} & = & \frac{1}{2}mv_{1f}^{2}+\frac{1
}{2}mv_{2f}^{2}
\end{eqnarray*}
(Since kinetic energy is scalar, there are no components). Notice that in all
equations the masses cancel and in the last equation the $\frac{1}{2}$ cancels.
Since the second ball (the 8 ball) is initially not moving and the first ball
(cue ball) is moving only horizontally, the equations become
\begin{eqnarray*}
v_{1ix} & = & v_{1fx}+v_{2fx}\\
0 & = & v_{1fy}+v_{2fy}\\
v_{1i}^{2} & = & v_{1f}^{2}+v_{2f}^{2}
\end{eqnarray*}
We now have three equations and three unknowns. In order to solve this linear
system, we must first relate the final component velocities in terms of the
vector sum final velocity. That is, assume the axis (angle $\beta =0$) runs
down the middle of the table. If the angle $\beta $ is confusing, just
remember that the cue ball is only moving in the $x$ (horizontal) direction.
\begin{eqnarray*}
v_{1ix} & = & v_{1i}\cos \beta =v_{1i}\\
v_{1fx} & = & v_{1f}\cos \phi\\
v_{1fy} & = & v_{1f}\sin \phi \\
v_{2fx} & = & v_{2f}\cos \theta\\
v_{2fy} & = & v_{2f}\sin \theta
\end{eqnarray*}
Plugging these new numbers back into the system,
\begin{eqnarray}
v_{1i} & = & v_{1f}\cos \phi +v_{2f}\cos \theta \label{eqn:system1}\\
0 & = & v_{1f}\sin \phi +v_{2f}\sin \theta \label{eqn:system2}\\
v_{1i}^{2} & = & v_{1f}^{2}+v_{2f}^{2} \label{eqn:system3}
\end{eqnarray}
The first equation is already solved for $v_{1i}$, and the second can easily
be solved for $v_{1f}$ leaving only $v_{2f}$ in the 3rd equation.
\begin{eqnarray*}
v_{1f} & = & \frac{v_{2f}\sin \theta }{\sin \phi }\\
\left( \frac{v_{2f}\sin \theta }{\sin \phi }\cos \phi +v_{2f}\cos \theta
\right) ^{2} & = & \left( \frac{v_{2f}\sin \theta }{\sin \phi }\right)
^{2}+v_{2f}^{2}
\end{eqnarray*}
Expanding this equation,
\[
v_{2f}^{2}\frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi
+2v_{2f}^{2}\frac{\sin \theta }{\sin \phi }\cos \phi \cos \theta
+v_{2f}^{2}\allowbreak \cos ^{2}\theta =v_{2f}^{2}\frac{\sin ^{2}\theta }{
\sin ^{2}\phi }+v_{2f}^{2}
\]
Solving for $v_{2f}^{2}$,
\begin{eqnarray*}
v_{2f}^{2}\frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi +2v_{2f}^{2}
\frac{\sin \theta }{\sin \phi }\cos \phi \cos \theta +v_{2f}^{2}\allowbreak
\cos ^{2}\theta & = & v_{2f}^{2}\frac{\sin ^{2}\theta }{\sin ^{2}\phi }
+v_{2f}^{2} \\
v_{2f}^{2}\left( \frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi +2
\frac{\sin \theta }{\sin \phi }\cos \phi \cos \theta +\allowbreak \cos
^{2}\theta \right) & = & v_{2f}^{2}\left( \frac{\sin ^{2}\theta }{\sin ^{2}\phi }
+1\right)\\
v_{2f}^{2}\left( \frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi +2
\frac{\sin \theta }{\sin \phi }\cos \phi \cos \theta +\allowbreak \cos
^{2}\theta -\frac{\sin ^{2}\theta }{\sin ^{2}\phi }-1\right) & = & 0
\end{eqnarray*}
Since $v_{2f}^{2}$ does not equal zero (the 8-ball certainly moves after the
collision), the equation becomes
\begin{eqnarray*}
\frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi +2\frac{\sin \theta }{
\sin \phi }\cos \phi \cos \theta +\allowbreak \cos ^{2}\theta -\frac{\sin
^{2}\theta }{\sin ^{2}\phi }-1 & = & 0\\
\frac{\sin ^{2}\theta }{\sin ^{2}\phi }\cos ^{2}\phi +2\frac{\sin \theta }{
\sin \phi }\cos \phi \cos \theta -\frac{\sin ^{2}\theta }{\sin ^{2}\phi }
& = & 1-\allowbreak \cos ^{2}\theta\\
\sin ^{2}\theta \cot ^{2}\phi +2\sin \theta \cos \theta \cot \phi -\frac{%
\sin ^{2}\theta }{\sin ^{2}\phi } & = & \allowbreak \sin ^{2}\theta
\end{eqnarray*}
Trig identities $\frac{\cos x}{\sin x}=\cot x$ and $1-\cos ^{2}\theta
=\sin ^{2}\theta $. Divide by $\sin ^{2}\theta $ since $\theta \neq 0$.
\begin{eqnarray*}
\cot ^{2}\phi +2\frac{\cos \theta }{\sin \theta }\cot \phi -\csc ^{2}\phi & = & 1\\
\cot ^{2}\phi +2\cot \theta \cot \phi -\csc ^{2}\phi & = & 1
\end{eqnarray*}
Trig identity $\frac{\cos x}{\sin x}=\cot x$
\[
\csc ^{2}\phi -1+2\cot \theta \cot \phi -\csc ^{2}\phi =1
\]
Trig identity $\cot ^{2}x=\csc ^{2}x-1$
\begin{eqnarray*}
2\cot \theta \cot \phi & = & 2\\
\cot \theta \cot \phi & = & 1\\
\phi & = & \cot^{-1}\left( \tan \theta \right)
\end{eqnarray*}
Remember that $\cot^{-1}x$ can be entered in the calculator as
$\arctan \left( \frac{1}{x}\right) $. So,
\[
\phi =\arctan \left( \frac{1}{\tan \theta }\right)
\]
Or, just remember from lecture (and the vector proof) that $\phi =90-\theta $.
\end{quote}
9) What is the speed of the cue ball after the collision?
\begin{quote}
In this problem, we want $v_{1f}$ from the previous problem. We can use
the first and last equations of the system above (equations
(\ref{eqn:system1}) and (\ref{eqn:system3}) marked above)
\begin{eqnarray*}
v_{1i} & = & v_{1f}\cos \phi +v_{2f}\cos \theta\\
v_{1i}^{2} & = & v_{1f}^{2}+v_{2f}^{2}
\end{eqnarray*}
Solving the first one for $v_{2f}$,
\begin{eqnarray*}
v_{1i}-v_{1f}\cos \phi & = & v_{2f}\cos \theta\\
\frac{v_{1i}-v_{1f}\cos \phi }{\cos \theta } & = & v_{2f}
\end{eqnarray*}
Plugging $v_{2f}$ into the second equation and solving for $v_{1f}$,
\begin{eqnarray*}
v_{1i}^{2} & = & v_{1f}^{2}+\left( \frac{v_{1i}-v_{1f}\cos \phi }{\cos \theta }\right) ^{2}\\
v_{1i}^{2} & = & v_{1f}^{2}+\left( \frac{v_{1i}}{\cos \theta }-\frac{
v_{1f}\cos \phi }{\cos \theta }\right) ^{2}\\
v_{1i}^{2} & = & v_{1f}^{2}+\frac{v_{1i}^{2}}{\cos ^{2}\theta }-2\frac{
v_{1i}v_{1f}\cos \phi }{\cos ^{2}\theta }+\frac{v_{1f}^{2}\cos ^{2}\phi }{
\cos ^{2}\theta }\\
0 & = & v_{1f}^{2}\left( 1+\frac{\cos ^{2}\phi }{\cos ^{2}\theta }\right)
-v_{1f}\left( 2\frac{v_{1i}\cos \phi }{\cos ^{2}\theta }\right) \vspace{1pt}+
\frac{v_{1i}^{2}}{\cos ^{2}\theta }\vspace{1pt}\vspace{1pt}-v_{1i}^{2}\\
v_{1f} & = & \frac{\left( 2\frac{v_{1i}\cos \phi }{\cos ^{2}\theta }\right)
\pm \sqrt{\left( 2\frac{v_{1i}\cos \phi }{\cos ^{2}\theta }
\right) ^{2}-4\left( 1+\frac{\cos ^{2}\phi }{\cos ^{2}\theta }\right) \left(
\frac{v_{1i}^{2}}{\cos ^{2}\theta }\vspace{1pt}\vspace{1pt}
-v_{1i}^{2}\right) }}{2\left( 1+\frac{\cos ^{2}\phi }{\cos ^{2}\theta }
\right) }\\
v_{1f} & = & v_{i}\frac{\cos \phi +\left( \cos \theta \right) \sqrt{
\left( -1+\cos ^{2}\theta +\cos ^{2}\phi \right) }}{\cos ^{2}\theta +\cos
^{2}\phi }
\end{eqnarray*}
(Trust me, it does)
\end{quote}
10) Suppose $m$ is a certain mass and $v$ is a certain speed $v > 0$. A mass
$m_1 = 5m$ moving to the right with a velocity $v_1 = +9v$ collides head-on
with a second mass $m_2 = 5m$ moving to the left with velocity $v_2 = -2v$.
The two masses stick together to form a single mass moving with velocity
$v_f = x\;v$ where $x$ is a number (no units). What is $x$?
\begin{quote}
$m_{1}$ = Given\\
$m_{2}$ = Given\\
$v_{1i}$ = Given\\
$v_{2i}$ = Given\\
Since the masses stick, there is not conservation of kinetic energy. However,
momentum is conserved.
\[
m_{1}v_{1}+m_{2}v_{2}=\left( m_{1}+m_{2}\right) v_{f}
\]
So,
\[
\frac{m_{1}v_{1}+m_{2}v_{2}}{\left( m_{1}+m_{2}\right) }=v_{f}
\]
So just plug in the expressions you have for each variable. The $m$'s will
end up canceling (leaving only the proportionality constant) and you will
end up with a number times $v$. Your answer is just the number (the
coefficient of $v$).
\end{quote}
\end{document}