Answer
All answers will be same
Work Step by Step
Total energy of car (kinetic + potential) is:
$E_{tot}=E_{kin}+E_{pot}=mgh+\dfrac {mv^{2}_{0}}{2}(1)$
We know that the total energy of the car is conserved. Lets write total energy equations for point A,B,C:
For A)
$E_{tot}=E_{kin}+E_{pot}=mgh+\dfrac {mv^{2}_{A}}{2} (2)$
For B)
$E_{tot}=E_{kin}+E_{pot}=mg\frac{h}{2}+\dfrac {mv^{2}_{B}}{2}(3)$
For C)
$E_{tot}=E_{kin}+E_{pot}=mg\times 0+\dfrac {mv^{2}_{C}}{2} (4)$
From (1) and (2) we get:
$v_{A}=v_{0}$
From (1)and (3) we get:
$v_{B}=\sqrt {v^{2}_{0}+gh}$
From (1) and (4) we get:
$v_{C}=\sqrt {v^{2}_{0}+2gh}$
As seen above, the speed of car at different points doesn't depend on the mass of car (as the equations do not contain $m$). So, if we change the mass, the speed of car at point A,B,C will not change. For D, at the last heel, we can write this equation: $E_{tot}=E_{kin}+E_{pot}=mgD+\dfrac {mv^{2}}{2};v=0\Rightarrow E_{tot}=mgD=mgh+\dfrac {mv^{2}_{0}}{2}\Rightarrow D=h+\dfrac {v^{2}_{0}}{2g} (5)$
Since D doesn't depend on the mass of the car, it will also not change.
