What is the energy lost when a body falls onto a moving cart without friction?

[devanlevin]

a cart with a mass of M is moving at velocity V, a body with a mass of m falls onto the cart from a height of H and sticks to it, how much energy was lost (to heat), there is no friction.

i said that, since the mass freefalls, its velocity at the moment of impact is\sqrt{2gh}

the momentum doesn't change so
MV+0=(m+M)Ux
0+m\sqrt{2gh}=(m+M)Uy

is this correct? will the cart have velocity on y axis??
if there is no friction how is Ux<V

then find the total energy at the start, which is mgh+0.5MV^{2}, and subtract it from the energy at the end 0.5(m+M)[Uy^{2}+Ux^{2}]

is this correct

the answer in my book is

Q=mgh+V^2\frac{Mm}{2(m+M}

 

[alphysicist]

a cart with a mass of M is moving at velocity V, a body with a mass of m falls onto the cart from a height of H and sticks to it, how much energy was lost (to heat), there is no friction.

i said that, since the mass freefalls, its velocity at the moment of impact is\sqrt{2gh}

the momentum doesn't change so
MV+0=(m+M)Ux
0+m\sqrt{2gh}=(m+M)Uy

This last equation is not correct. These equations represent the momentum of the masses m and M, but there is an external force in the y direction (from the ground). So the y-momentum is not conserved because the ground prevents the cart from moving in the y-direction.

is this correct? will the cart have velocity on y axis??
if there is no friction how is Ux<V

then find the total energy at the start, which is mgh+0.5MV^{2}, and subtract it from the energy at the end 0.5(m+M)[Uy^{2}+Ux^{2}]

It's the other way around; energy lost is E_i-E_f