Question
A $\Sigma^{+}$particle decays from rest into a neutron and another particle $X$ according to the reaction
$
\Sigma^{+} \rightarrow \mathrm{n}+\mathrm{X}
$
The following data are available.
$
\begin{array}{ll}
\text { Rest mass of } \Sigma^{+} & =1190 \mathrm{MeV} \mathrm{c}^{-2} \\
\text { Momentum of neutron } & =185 \mathrm{MeV} \mathrm{c}^{-1}
\end{array}
$
Calculate, for the neutron,
a(i)the total energy.[1]
a(iithe speed.[2]
b. Determine the rest mass of $X$.[3]
▶️Answer/Explanation
Ans:
a(i)neutron energy $=\sqrt{185^2+940^2}=958 « \mathrm{MeV} »$
NOTE: Allow $1.5 \times 10^{-10} « \mathrm{~J} »$
a(ii)ALTERNATIVE 1
«use of $E=\gamma E_0$ »
$
\begin{aligned}
& « 958=940 \gamma \text { so» } \gamma=1.019 \\
& v=0.193 c
\end{aligned}
$
ALTERNATIVE 2
«use of $p=\gamma m v »$
$
\begin{aligned}
& 185=940 \frac{\frac{v}{c}}{\sqrt{1-\left(\frac{v}{c}\right)^2}} \checkmark \\
& v=0.193 c
\end{aligned}
$
ALTERNATIVE 3
«use of $p=\gamma \quad m v$ »
$
\begin{aligned}
& v=\frac{p c}{E} \boldsymbol{V} \\
& v=\frac{185}{958}=0.193 c
\end{aligned}
$
NOTE: Allow $v=5.8 \times 10^7 \ll \mathrm{ms}^{-1} \gg$
b. momentum of $\mathrm{X}=185 \ll \mathrm{MeV} \mathrm{c}^{-1} » \mathscr{V}$ energy of $X=1190-958=232 \ll \mathrm{MeV} \gg \boldsymbol{V}$ $m_0=« \sqrt{232^2-185^2}=» 140 \ll \mathrm{MeV} \mathrm{c}^{-2} »$ NOTE: Allow mass in kg – gives $2.5 \times 10^{-28} \ll \mathrm{kg} »$
Question
The particle omega minus $\left(\Omega^{-}\right)$decays at rest into a neutral pion $\left(\pi^0\right)$ and the xi baryon $\left(\Xi^{-}\right)$according to
$
\Omega^{-} \rightarrow \pi^0+\Xi^{-}
$
The pion momentum is $289.7 \mathrm{MeV} \mathrm{c}^{-1}$.
The rest masses of the particles are:
$
\begin{aligned}
& \Omega^{-}: 1672 \mathrm{MeV} \mathrm{c}^{-2} \\
& \pi^0: 135.0 \mathrm{MeV} \mathrm{c}^{-2} \\
& \Xi^{-}: 1321 \mathrm{MeV} \mathrm{c}^{-2}
\end{aligned}
$
a. Show that energy is conserved in this decay.[3]
b. Calculate the speed of the pion.[2]
▶️Answer/Explanation
Ans:
a. momentum of $x i$ baryon is also $289.7 « \mathrm{MeVc}^{-1} »$
total energy of xi baryon and pion is $\sqrt{289.7^2+1321^2}+\sqrt{289.7^2+135.0^2}=1672$ «eV»
which equals the rest energy of the omega
Allow a backwards argument, assuming the energy is equal.
b. $\gamma \ll=\frac{\sqrt{289.7^2+135.0^2}}{135.0} »=2.367$
$
v \ll=\sqrt{1-\frac{1}{2.367^2}} c »=0.903 c
$
Award [2] for bald correct answer.
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