Answer
$5.3\;\rm MeV$
$32.4\;\rm MeV$
Work Step by Step
The author here told us to use relativistic formulas.
a) We need to find the energy released from the given decay and we know that the energy released is given by
$$Q=Q_{\rm \Lambda^0}-(Q_{\rm p}+Q_{\pi^{-}})$$
Thus;
$$Q= m_{\rm \Lambda^0}c^2-m_{\rm p}c^2-m_{\pi^{-}}c^2 $$
Plugging the known; see table 32-2;
$$Q=1115.6-938.3-139.6 =\color{red}{\bf 37.7}\;\rm MeV$$
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b) We are told that $\Lambda ^0$ decays from rest.
We assume that the system is isolated and hence the energy and the momentum are conserved.
$$E_{\Lambda^0}=E_{\rm p}+E_{\rm \pi^-}$$
$$m_{\Lambda^{0}}\;c^2=E_{\rm p}+E_{\rm \pi^-}\tag 1$$
Hence the net momentum of the products is zero since $\Lambda$ was at rest.
This means that the momentum of the two products is equal in magnitude.
$$p_{\rm p}=p_{\rm \pi^-}$$
Recalling that $E^2=p^2c^2+m^2c^4$, and hence, $p^2c^2=E^2-m^2c^4$.
Applying that by squaring both sides;
$$p_{\rm p}^2c^2 =p^2_{\rm \pi^-}c^2 $$
Now plugging;
$$E_{\rm p}^2-m_{\rm p}^2c^4=E_{\rm \pi^-}^2-m_{\rm \pi^-}^2\;c^4$$
Plugging $E_{\rm p}$ from (1); where $m_{\Lambda^{0}}\;c^2-E_{\rm \pi^-}=E_{\rm p}$
$$(m_{\Lambda^{0}}\;c^2-E_{\rm \pi^-})^2-m_{\rm p}^2c^4=E_{\rm \pi^-}^2-m_{\rm \pi^-}^2\;c^4$$
Solving for $E_{\rm \pi^-}$;
$$ m^2_{\Lambda^{0}}\;c^4-2E _{\rm \pi^-}m_{\Lambda^{0}}\;c^2+\color{red}{\bf\not}E^2_{\rm \pi^-} -m_{\rm p}^2c^4=\color{red}{\bf\not}E_{\rm \pi^-}^2-m_{\rm \pi^-}^2\;c^4$$
$$ m^2_{\Lambda^{0}}\;c^4 -m_{\rm p}^2c^4+m_{\rm \pi^-}^2\;c^4=2E _{\rm \pi^-}m_{\Lambda^{0}}\;c^2$$
$$ (m^2_{\Lambda^{0}} -m_{\rm p}^2 +m_{\rm \pi^-}^2)\;\color{red}{\bf\not}c^4=2E _{\rm \pi^-}m_{\Lambda^{0}}\;\color{red}{\bf\not}c^2$$
$$ \dfrac{(m^2_{\Lambda^{0}} -m_{\rm p}^2 +m_{\rm \pi^-}^2) c^2}{2m_{\Lambda^{0}}}= E _{\rm \pi^-} $$
Plugging the known;
$$E _{\rm \pi^-}= \dfrac{1115.6^2 \;c^2 - 938.3^2 + 139.6^2 }{2\times 1115.6} $$
$$E _{\rm \pi^-}= \bf 172\;\rm MeV\tag 2$$
Thus,
$$KE_{\rm \pi^-}=E_{\rm \pi^-}-m_{\rm \pi^-}c^2=172- 139.6 $$
$$KE_{\rm \pi^-}=\color{red}{\bf 32.4}\;\rm MeV$$
Plugging (2) into (1) and solving for $E_{\rm p}$
$$E_{\rm p}=m_{\Lambda^{0}}\;c^2-E_{\rm \pi^-}=1115.6-172=\bf 943.6\;\rm MeV $$
Thus,
$$KE_{\rm p}=E_{\rm p}-m_{\rm p}c^2=943.6- 938.3$$
$$KE_{\rm p}=\color{red}{\bf 5.3}\;\rm MeV$$
