Can someone explain to me this process?

[Xuekai Du]

π- + p→Λ + K0
Why is this a strong interaction instead of weak? Since the quark content changed.

 

[mfb]

The strong interaction can produce (and eliminate) quark+antiquark pairs of the same type. Like gluon -> strange + antistrange.
The interaction here does not have to be strong but that is by far the most likely process.

 

[Vanadium 50]

Since the quark content changed.

But the net flavor did not.

 

[samalkhaiat]

π- + p→Λ + K0
Why is this a strong interaction instead of weak? Since the quark content changed.

1) It is pure hadronic interaction, and 2) strangeness is conserved: Strangeness-conserving hadronic interactions can not be weak.

 

[Envelope]

Strangeness-conserving hadronic interactions can not be weak.

This isn't a general rule. Hadronic-only strangeness-changing processes must be weak, but hadronic-only strangeness-conserving processes needn't be strong. The example in the original post can proceed via strong, weak, or electromagnetic diagrams (albeit that does mean it would be called "strong" since that diagram will dominate). The decay $B_s^0 \rightarrow D_s^{^-} \pi^+$ conserves strangeness but proceeds only weakly since bottomness and charmness are changed.

 

[samalkhaiat]

This isn't a general rule. Hadronic-only strangeness-changing processes must be weak, but hadronic-only strangeness-conserving processes needn't be strong. The example in the original post can proceed via strong, weak, or electromagnetic diagrams (albeit that does mean it would be called "strong" since that diagram will dominate). The decay $B_s^0 \rightarrow D_s^{^-} \pi^+$ conserves strangeness but proceeds only weakly since bottomness and charmness are changed.

My statement was about light hadrons, i.e., flavour $SU(3)$. Of course, if you include heavy quarks, then you can have weak decays in which the s-quark behaves as spectator giving $| \Delta S | = 0$.