Question

A ball is rolling down a track as shown. How fast must it be going at A in order to just make it to the top of B? The ball is at Peak A which is 5.0 m tall and wants to go to Peak B, which is 10.0m tall

Answer 1

v = 98.1 m/s

Step-by-step explanation:

We will apply the law of conservation of energy here, between peak A and peak B.

Total Energy of Ball at A = Total Energy of Ball at B

`K.E_(A)+P.E_(A)=K.E_(B)+P.E_(B)\\where,\\K.E_(A) = Kinetic\ Energy\ of\ ball\ at\ Peak\ A = (1)/(2)mv^(2) \\K.E_(B) = Kinetic\ Energy\ of\ ball\ at\ Peak\ B = (1)/(2)m(0)^(2) = 0\ J (since, ball\ stops\ at\ last) \\`

`P.E_(A) = Potential\ Energy\ of\ ball\ at\ Peak\ A = mgh_(A)\\P.E_(B) = Potential\ Energy\ of\ ball\ at\ Peak\ B = mgh_(B)`

`(1)/(2)mv^(2) + mgh_(A) = 0\ J + mgh_(B)\\m((1)/(2)v^(2) + gh_(A)) = mgh_(B)\\v = \sqrt{2g(h_(B)-h_(A))}\\where,\\h_(A) = height\ pf\ peak\ A = 5\ m\\h_(B) = height\ pf\ peak\ B = 10\ m\\v = \sqrt{2(9.81\ m/s^(2))(10\ m-5\ m)}\\`

v = 98.1 m/s