Considering the wheels to be a small part of the vehicle mass, and neglecting external force, here air resistance, Law of conservation of energy is applicable. Mgh = ½ Mv^2 + ½ Iω^2 assuming ½Iω^2 to be a very small quantity considering mass of the vehicle, we can neglect the same. Thus, velocity square is directly proportional to twice the product of g*h, as mass cancels out in the above equation. Thus, velocity is seen to be independent of the mass of the vehicle, and thus both of them will reach at the same velocity.

from an energy standpt PE = KE + Work PE= KE(tran) + KE(rot) + work m(overall)gh= 1/2m(overall)v^2+#wheels*(1/2(m(wheel)*radius^2)*angularvelocity^2)+ m(overall)*g*h*cos(angle of slope)*dist so if we take out all mass overall... and some algebra v^2 = 2[ gh - #wheel*(1/2*(mwheel/moverall)*radius^2*angularvelocity^2)- ghcos(theta)*d] terminal velocity is dependent on mwheel/moverall, if car is heavy on body, theoretically it should be faster than a lighter car.

The vehicles are no longer identical.

how long is the hill? if short - light car will reach first. if long enough, then heavy car (which has lower acceleration, but can attain higher terminal velocity) -- heavy car must overcome greater inertial delay at top of hill, so gets a slower start.

Light weigh car reach first, because the heavier body vehicle will stop the rotation by its mass itself

They reach at the same time. Assuming the mass is included in the body and minimal friction and no air resistance, mass will not be a factor. PE will be completely transformed to KE by the bottom of hill meaning mgh=.5mv^2, mass cancels out meaning mass doesn't matter. This is similar to dropping a feather and a bowling ball on the moon. No air resistance so they hit the ground at same time. Don't over think it. That's the worst thing you can do in an interview.

Same time... Don't listen to the rest of the shmos what we are really worried about here are the cars acceleration From FDB: F = ma = mgcos(theta) - FrictionForce = mgcos(theta) - mu*NormalForce = mgcos(theta) - mu*mgsin(theta) Divide both sides by m, and there is no mass left in equation... just a = gcos(theta) - mu*gsin(theta) Therefore, the equation is independent of mass and they arrive at the same time.

They should reach to the bottom at the same time. The solution is as follows: Based on "Law of conservation of energy", assuming the mass of both vehicles are M1 = m, M2 = 2m, where, the velocity for both vehicles are v_1 and v_2, and angular velocity for both vehicles are omega_1 and omega_2, the inertia for both vehicles are J1 and J2. The height of hill is h The equations are listed as: M1*g*h = 1/2 * M1 * (v_1^2) + 1/2 * J1 * (omega_1^2) (1) M2*g*h = 1/2 * M2 * (v_2^2) + 1/2 * J2 * (omega_2^2) (2) Note that rolling is in the plane of X-Z about Y-axis, based on "Parallel Axis Theorem", Jyy = 1/12 * M * R^2, where R is defined as the half width of the vehicle. Thus, J1 = 1/12 * M1 * R^2; J2 = 1/12 * M2 * R^2, due to both vehicle are identical in shape. Also, v_1 = R*omega_1; v_2 = R*omega_2; substituting M1, M2, J1, J2, v_1, v_2, into (1) and (2), such that: omega_1 = omega_2 = sqrt(2*g*h/(R+1/12)) (unit: rad/sec). Therefore, v_1 = v_2 = R*sqrt(2*g*h/(R+1/12)) (unit: m/sec); Assuming the length of slope from the top to the bottom is "L", the time ("T") for both vehicle rolling to the bottom is: T = L/v_1 = L/v_2 = L/(R*sqrt(2*g*h/(R+1/12))) (unit: sec). In conclusion, these two vehicles will spend identical time on rolling to the bottom.