Example \(\PageIndex{1}\): Collision Graph
revisited
Look again at the collision graph
from
Example 3.5.1 from the point of view of the kinetic energy of
the two carts.
- What is the initial kinetic energy of the system?
- How much of this is in the center of mass motion, and how much
of is convertible?
- Does the convertible kinetic energy go to zero at some point
during the collision? If so, when? Is it fully recovered after the
collision is over?
- What kind of collision is this? (Elastic, inelastic, etc.) What
is the coefficient of restitution?
Solution
(a) From the solution to
Example 3.5.1 we know that
\begin{aligned}
v_{1 i}=&-1 \: \frac{\mathrm{m}}{\mathrm{s}} & & v_{2
i}=0.5 \: \frac{\mathrm{m}}{\mathrm{s}} \\
v_{1 f}=&1 \: \frac{\mathrm{m}}{\mathrm{s}} & &
v_{2 f}=-0.5 \: \frac{\mathrm{m}}{\mathrm{s}}
\end{aligned}
and \(m_1\) = 1 kg and \(m_2\) = 2
kg. So the initial kinetic energy is
\[ K_{s y s, i}=\frac{1}{2}
m_{1} v_{1 i}^{2}+\frac{1}{2} m_{2} v_{2 i}^{2}=0.5 \:
\mathrm{J}+0.25 \: \mathrm{J}=0.75 \: \mathrm{J} \label{eq:4.17}
\]
(b) To calculate \(K_{cm} =
\frac{1}{2} (m_1 + m_2)v^2_{cm}\), we need \(v_{cm}\), which
in this case is equal to
\[ v_{c m}=\frac{m_{1} v_{1
i}+m_{2} v_{2 i}}{m_{1}+m_{2}}=\frac{-1+2 \times 0.5}{3}=0
\nonumber \]
so \(K_{cm}\) = 0, which means all
the kinetic energy is convertible. We can also calculate that
directly:
\[ K_{\text {conv},
i}=\frac{1}{2} \mu v_{12, i}^{2}=\frac{1}{2}\left(\frac{1 \times
2}{1+2} \: \mathrm{kg}\right) \times\left(0.5 \:
\frac{\mathrm{m}}{\mathrm{s}}-(-1) \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=\frac{1.5^{2}}{3} \:
\mathrm{J}=0.75 \: \mathrm{J} \label{eq:4.18} \]
(c) If we look at
figure 3.5.1, we can see that the carts do not pass through
each other, so their relative velocity must be zero at some point,
and with that, the convertible energy. In fact, the figure makes it
quite clear that both \(v_1\) and \(v_2\) are zero at
\(t\) = 5 s, so at that point also \(v_{12}\) = 0, and the
convertible energy \(K_{conv}\) = 0. (And so is the total
\(K_{sys}\) = 0 at that time, since \(K_{cm}\) = 0 throughout.)
On the other hand, it is also clear
that \(K_{conv}\) is fully recovered after the collision is over,
since the relative velocity just changes sign:
\begin{array}{l}
{v_{12, i}=v_{2 i}-v_{1 i}=0.5 \:
\frac{\mathrm{m}}{\mathrm{s}}-(-1) \:
\frac{\mathrm{m}}{\mathrm{s}}=1.5 \: \frac{\mathrm{m}}{\mathrm{s}}}
\\
{v_{12, f}=v_{2 f}-v_{1 f}=-0.5 \: \frac{\mathrm{m}}{\mathrm{s}}-1
\: \frac{\mathrm{m}}{\mathrm{s}}=-1.5 \:
\frac{\mathrm{m}}{\mathrm{s}}} \label{eq:4.19}
\end{array}
Therefore
\[ K_{\text {conv},
f}=\frac{1}{2} \mu v_{12, f}^{2}=\frac{1}{2} \mu v_{12,
i}^{2}=K_{\text {conv}, i} \nonumber \]
(d) Since the total kinetic energy
(which in this case is only convertible energy) is fully recovered
when the collision is over, the collision is elastic. Using
equation (\ref{eq:4.19}), we can see that the coefficient of
restitution is
\[ e=-\frac{v_{12, f}}{v_{12,
i}}=-\frac{-1.5}{1.5}=1 \nonumber \]
as it should be.
Example \(\PageIndex{2}\): Inelastic collision
and explosive separation
Analyze
Example 3.5.2 from the point of view of the system’s kinetic
energy. In particular, answer the following questions:
- What is the total kinetic energy of the system (i)
before the players collide, (ii) right after the
collision, when they are holding to one another, and (iii)
after they separate. How much of this energy is translational (that
is, center-of-mass kinetic energy), and how much is
convertible?
- Answer the same questions from the point of view of the player
who is skating at a constant 1.5 m/s to the right (player 3) (To
avoid needless repetition, you may use already established results,
such as conservation of momentum.)
Solution
(a) Before the players collide, we
have
\[ K_{s y s, i}=\frac{1}{2}
m_{1} v_{1 i}^{2}+\frac{1}{2} m_{2} v_{2 i}^{2}=\frac{1}{2}(80 \:
\mathrm{kg}) \times\left(3 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}+\frac{1}{2}(90 \:
\mathrm{kg}) \times\left(-2 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=540 \: \mathrm{J}
\label{eq:4.20} .\]
While they are still holding to each
other, we know from the solution to
Example 3.5.2 that their joint velocity is 0.353 , and that
this has to be also the velocity of their center of mass, which is
unchanged by the collision. So, we have
\[ K_{c
m}=\frac{1}{2}\left(m_{1}+m_{2}\right) v_{c m}^{2}=\frac{1}{2}(170
\: \mathrm{kg})\left(0.353 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=10.6 \: \mathrm{J}
\label{eq:4.21} .\]
This is \(K_{cm}\) throughout, as
well as \(K_{sys}\) right after the collision, since the collision
is totally inelastic and that means that \(K_{conv}\) drops to
zero. Also, subtracting this from (\ref{eq:4.20}) will give us the
initial value of the convertible energy, without the need for a
separate calculation, so
\[ K_{c o n v, i}=K_{s y s,
i}-K_{c m}=540 \: \mathrm{J}-10.6 \: \mathrm{J}=529.4 \: \mathrm{J}
\simeq 529 \: \mathrm{J} \label{eq:4.22} .\]
After the separation, the new total
kinetic energy (for which I will use the subscript f)
is
\[ K_{s y s, i}=\frac{1}{2}
m_{1} v_{1 f}^{2}+\frac{1}{2} m_{2} v_{2 f}^{2}=\frac{1}{2}(80 \:
\mathrm{kg}) \times\left(-0.176 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}+\frac{1}{2}(90 \:
\mathrm{kg}) \times\left(0.824 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=31.8 \: \mathrm{J}
\label{eq:4.23} \]
where I have gotten the values for
\(v_{1f}\) and \(v_{2f}\) from the solution to part (d) of
Example 3.5.2. Subtracting \(K_{cm}\) from this will give us
the final value of the convertible energy:
\[ K_{\text {conv}, f}=K_{\text
{sys,f}}-K_{c m}=31.8 \: \mathrm{J}-10.6 \: \mathrm{J}=21.2 \:
\mathrm{J} \label{eq:4.24} \]
To summarize, then, we have:
- Before the collision: \[ K_{s y s, i}=540 \: \mathrm{J},
\quad K_{c m}=10.6 \: \mathrm{J}, \quad K_{c o n v,
i}=529.4 \: \mathrm{J} \nonumber \]
- Right after the collision (players still holding to each
other): \[ K_{s y s}=K_{c m}=10.6 \: \mathrm{J}, \quad
K_{c o n v}=0 \nonumber \]
- After the (explosive) separation: \[ K_{s y s, f}=31.8 \:
\mathrm{J}, \quad K_{c m}=10.6 \: \mathrm{J}, \quad K_{c o n v,
i}=21.2 \: \mathrm{J} \nonumber \]
So, in the collision, approximately 529 J of kinetic energy
“disappeared” from the system (or, we could say, were “converted
into some form of internal energy”), whereas the players’ pushing
on each other managed to put about 21 J of kinetic energy back into
the system; we will explore these kinds of processes in more detail
in the following chapter!
(b) We need to repeat all the above calculations with all the
velocities shifted down by 1.5 m/s, to bring them to the reference
frame of player 3. Instead of putting a subscript “3” on all the
quantities, since we already have tons of subscripts to worry
about, I’m going to follow an alternative convention and use a
“prime” superscript (\(^{\prime}\)) to denote all the quantities in
this frame of reference. In brief, we have
\[ K_{s y s, i}^{\prime}=\frac{1}{2} m_{1}\left(v_{1
i}^{\prime}\right)^{2}+\frac{1}{2} m_{2}\left(v_{2
i}^{\prime}\right)^{2}=\frac{1}{2}(80 \: \mathrm{kg})
\times\left(1.5
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}+\frac{1}{2}(90 \:
\mathrm{kg}) \times\left(-3.5 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=641.3 \: \mathrm{J}
\label{eq:4.25} \]
\[ K_{c
m}^{\prime}=\frac{1}{2}\left(m_{1}+m_{2}\right)\left(v_{c
m}^{\prime}\right)^{2}=\frac{1}{2}(170 \: \mathrm{kg})\left(0.353
\: \frac{\mathrm{m}}{\mathrm{s}}-1.5 \:
\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}=111.8 \mathrm{J} \:
\label{eq:4.26} \]
\[ K_{\text
{conv,i}}^{\prime}=K_{s y s, i}^{\prime}-K_{c m}^{\prime}=641.3 \:
\mathrm{J}-111.8 \: \mathrm{J}=529.5 \: \mathrm{J} \simeq 529 \:
\mathrm{J} \label{eq:4.27} \]
This shows explicitly that the
convertible energy, as I pointed out earlier in this chapter, is
the same in every reference frame! (The equality is exact, if you
keep enough decimals in the calculation.)
Knowing this, we can simplify the
calculation of the final kinetic energy, after the explosive
separation: the convertible energy, \(K^{\prime}_{conv,f}\), will
be the same as in the earth reference frame, that is to say, 21.2
J, and the total kinetic energy will be \(K^{\prime}_{sys,f} =
K^{\prime}_cm + K^{\prime}_{conv,f}\) = 111.8 J +21.2 J = 133
J.
So, in this frame of reference, we
have (to three significant figures):
\begin{aligned}
K_{s y s, i}^{\prime}=641 \: \mathrm{J}, \quad K_{c m}^{\prime}=112
\: \mathrm{J}, \quad K_{\text {conv,i}}^{\prime}=529 \: \mathrm{J}
\quad &\text { (before the collision) }\\
K_{s y s}^{\prime}=K_{c m}^{\prime}=112 \: \mathrm{J}, \quad K_{c o
n v}^{\prime}=0 \quad &\text { (right after the collision)
}\\
K_{s y s, f}^{\prime}=133 \: \mathrm{J}, \quad K_{c m}^{\prime}=112
\: \mathrm{J}, \quad K_{c o n v, i}^{\prime}=21.2 \:
\mathrm{J} \quad &\text { (after the separation) }
\end{aligned}
So, even though the total kinetic
energy is different in the two reference frames, all the (inertial)
observers will agree as to the amount of kinetic energy “lost” in
the collision, as well as the amount of kinetic energy put back
into the system by the players’ pushing on each other.
