Movement of electron in ac current
You cannot say that there are more collisions in AC current. The power dissipated in a resistor with voltage \$v(t)\$ and current \$i(t)\$ is given by:
$$P(t) = v(t)i(t)$$
For an AC source, \$v(t) = Vsin(\omega t)\$ and \$i(t) = v(t)/R\$, thus,
$$P(t) = \frac{V^2sin^2(\omega t)}{R}$$
Average power is thus:
$$P_a = \frac{V^2}{2R}$$
For a DC voltage \$V_{DC}\$, the average power dissipated is:
$$P_a^\prime = \frac{V^2_{DC}}{R}$$
Since the power dissipated is converted into heat. The AC current with magnitude \$V\$ dissipates the same power as DC current with magnitude $$V_{DC} = \frac{V}{\sqrt{2}}$$
So actually if both AC and DC currents have same magnitude the DC current will draw more power as the AC current.
This can be expected since the current is flowing all the time for a DC voltage. But in case of AC voltage, current is zero at certain instants contributing to no power dissipation. Thus the average power dissipation is less for the AC source.