When the wire is bent in the form of a square and connected between $M$ and $N$ as shown in fig.$(2)$, the effective resistance between $M$ and $N$ decreases to one fourth of the value in fig.$(1)$. The current increases four times the initial value according to the relation $V=I R$. Since $H=I^2 R t$, the decrease in the value of resistance is more than compensated by the increases in the value of current. Hence heat produced increases. Percentage loss in energy during the collision $\simeq 56 \%$