2nd part of a physics problem?

krissy

I need help with the 2nd part of this question. Here is the first part.

For a steel ball rolling down a 45 degree slope from a height of 1 meter above the table, how far will the ball travel in the x direction if it launches horizontally from the table which is 1 meter high?

For the same ball rolling down a 45 degree slope from one meter high, what maximum diameter of loop the loop will the ball go around without falling? Show work

fireman

With an assumption that there is no friction:-

A) By the law of energy conservation:-

=>PE(initial) = KE(final)

=>mgh = 1/2mv^2

=>v = √2gh

=>v = √[2 x 9.8 x 1]

=>v = 4.43 m/s

Let the ball take t sec to fall 1 meter

=>By s = ut + 1/2gt^2

=>1 = 0 + 1/2 x 9.8 x t^2

=>t = √0.20

=>t = 0.45 sec

Thus R = [Ux] x t

=>R = 4.43 x 0.45

=>R = 2.0 m

By the law of energy conservation:-

=>maximum height gain = 1 m

=>By maximum diameter of loop the loop will be 1 m