Since cars A and B are at same height above the starting point. ∴h=
1
2
aAsinθAt2=
1
2
aBsinθBt2
aA
aB
=
sinθB
sinθA
Given, sinθA>sinθB ∴
aA
aB
<1 ∴aA<aB.....(1) Along the plane,vA=aA×t vB=aB×t.....[Using (1)] ∴
vA
vB
=
aA
aB
<1........[Using (i)] ∴vA<vB T⋅EA=mgh+
1
2
mvA2 Tâ‹…EB=mgh+
1
2
mvB2 At a given instant, height is same so potential energywill be same. But due to large speed of B, kinetic energy of B will be more. ∴T⋅EB>T⋅EA