Question

Question #59aa4

Answer 1

`sqrt2v^`

Explanation:

At height `h` total energy of the cart`=KE+PE=1/2mv^2+mgh`
where `m` is mass of cart and `g` is acceleration due to gravity.

After it looses its (`50%`) half mass it climbs to height `2h`
Let velocity at that height `=v_n`

At height `2h` total energy of the cart`=1/2(m/2)v_n^2+(m/2)g(2h)`
`=(m/4)v_n^2+mgh`

Assuming no loss of energy due to friction, and applying law of conservation of energy we get

`1/2mv^2+mgh=(m/4)v_n^2+mgh`
`=>v_n^2/4=v^2/2`
`=>v_n=sqrt2v^`