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

Answer:

v = 98.1 m/s

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

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

$P.E_{A} = Potential\ Energy\ of\ ball\ at\ Peak\ A = mgh_{A}\\P.E_{B} = Potential\ Energy\ of\ ball\ at\ Peak\ B = mgh_{B}$

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

v = 98.1 m/s