Answer
a). (1) the smaller car will have longer travel distance.
b). The ratio of stopping distance of smaller car to larger car is $4$.
Work Step by Step
By conservation of energy,
Work done = Change in KE
Final KE =0.
Thus, Work done = Initial KE
$Fd=\frac{1}{2}mv^{2}$
Let coefficient of friction = k
For smaller car, $kmgd_{s}=\frac{1}{2}m(2v)^{2}$
For larger car, $k2mgd_{l}=\frac{1}{2}2m(v)^{2}$
Therefore, from above 2 equations $d_{s}=4d_{l}$
So, the smaller car travels longer distance to come to rest.
a). (1) the smaller car will have longer travel distance.
b). The ratio of stopping distance of smaller car to larger car is 4.
