The cart starts from rest at the top of a hill with height, H1. It rolls down the hill and at the bottom of the hill it has a speed of 8 m/s. Next it is at the top of same hill but pushed so that it starts at the top of the hill with a speed of 6 m/s. How fast is it going at the bottom of the hill?

Question

The cart starts from rest at the top of a hill with height, H1.  It rolls down the hill and at the bottom of the hill it has a speed of 8 m/s.  Next it is at the top of same hill but pushed so that it starts at the top of the hill with a speed of 6 m/s.  How fast is it going at the bottom of the hill?

drwls · Answer

The change in kinetic energy will be the same in both cases. (M/2)8^2 = M/2[V^2 - 6^2] cancel the M's V = sqrt[6^2 + 8^2] = 10 m/s