Answer
The gravitational force between the cars can not cause the collision, as this force is very feeble compared to the gravitational force exerted by the Earth on these cars.
Work Step by Step
We know the gravitation force depends on the mass of the objects and the distance between them. Let us assume that the two cars be GMC Hummer EV, one of the heaviest available cars on the market, weighing approximately 4100 kg. Assume both of them are going side by side in parallel lanes at a distance of 2m apart as shown in the figure.
The gravitational force between the cars is given by
$F_{C}=\frac{GM_{c}M_{c}}{r^{2}}$
where $G$ is the gravitational constant, $M_{c}$ is the mass of the car and $r$ is the distance between them.
We know,
$M_{c}=4100 kg$
$r=2 m$
$G=6.674\times10^{-11} N.m^{2}.kg^{-2}$
Therefore,
$F_{C}=\frac{6.674\times10^{-11} N.m^{2}.kg^{-2}\times4100 kg\times4100 kg}{(2 m)^{2}}$
$F_{C}=0.00112 N$
Now, The gravitational force with which the Earth attracts the car is given by
$F_{E}=\frac{GM_{E}M_{c}}{r^{2}}$
where $M_{E}$ is the mass of the Earth, $M_{c}$ is the mass of the car and $r$ is the distance between them. Since the car is near Earth's surface, the distance between the car and the Earth's center is nearly equal to the Earth's mean radius, $R_{E}=6.37\times10^{6}$.
We know,
$M_{E}=5.97\times10^{24} kg$
Therefore,
$F_{E}=\frac{6.674\times10^{-11} N.m^{2}.kg^{-2}\times5.97\times10^{24} kg\times4100 kg}{(6.37\times10^{6} m)^{2} }$
$F_{E}=40 N$
$\frac{F_{E}}{F_{C}}=\frac{40}{0.00112}=36363$
Thus, the gravitational force of attraction between the cars is very small compared to the gravitational force of attraction between Earth and the car. In fact, the force of attraction by the Earth on the car is 36363 times stronger than that between the cars.
