If the objects are circular, then the probability of them colliding is the probability that their distance is below a threshold.  The difference between two independent normally distributed vectors is itself a normally distributed vector.  The length of such a vector has a non-central chi-square distribution.  So the problem boils down to computing the cumulative distribution of a non-central chi-squared variate.  See the page on [Probability of collision (two bivariate normal distributions)](http://stats.stackexchange.com/questions/41900/probability-of-collision-two-bivariate-normal-distributions) for more details.