That's an average power of
Power = 5A5 A \$\times\$ 4V SAY10.5 V \$\times\$ 100 \$\mu\$Ss / 10s = 20 W \$\times\$ 1/100,00010 ms = 0.2 milliWatt ! :-)525 W.
Average power is easy for almost any battery. You just need a store to accomodate the pulse.
A capacitor that will "droop" say 0.5V in 100 \$\mu\$Ss needs to be
C = I \$\times\$ V \$\times\$ t / V = 5 \$\times\$ 0.5A \$\times\$ 100/1000000 = 250 \$\mu\$Fs / 0.
Use a say5 V= 1000 \$\mu\$F and it should do well.
A supercap would do well here if voltage rating is OK.
E&OE