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

 

[thomate1]

Your approach to solving this problem is mostly correct. However, there are a few things that need to be clarified and corrected.

First, at the moment of impact, the body's velocity will not be equal to \sqrt{2gh}. The body is falling from a height of H, but it is also moving horizontally with the cart's velocity, V. This means that the body's velocity at the moment of impact will be the vector sum of the two velocities, given by the equation \sqrt{V^2+(2gh)^2}.

Second, the momentum does change at the moment of impact. Before the impact, the cart has momentum MV and the body has momentum 0. After the impact, the combined system of the cart and body has momentum (m+M)U, where U is the final velocity of the system. This is due to the conservation of momentum, which states that the total momentum of a system remains constant in the absence of external forces.

Third, the final velocity of the system, U, will have both x and y components. The x component will be equal to V, as there is no friction to slow down the cart's motion. The y component can be calculated using the equation you provided: 0 + m\sqrt{2gh} = (m+M)Uy.

Finally, to calculate the energy lost to heat, you need to subtract the final energy of the system from the initial energy. The final energy can be calculated using the equation 0.5(m+M)(Ux^2+Uy^2). The initial energy is equal to mgh + 0.5MV^2. Subtracting these two values will give you the energy lost to heat.

In summary, your approach is on the right track, but there are a few corrections and clarifications that need to be made. The correct answer can be found by using the equation Q = mgh + 0.5MV^2 - 0.5(m+M)(Ux^2+Uy^2).