Question

A solid ball rolls without slipping from rest (starting at height H = 6.00 m) until it leaves the horizontal section at the end of the track, at height h = 2.00 m above the floor. How far horizontally from point A does the ball hit the floor? Hint: the ball will be rotating as well as moving horizontally, and the horizontal speed will be what matters.

Answer 1

How far horizontally from point A does the ball hit the floor d= 5 m

Step-by-step explanation:

By law of conservation of energy we can say that

final Kinetic energy kf+ final potential energy Uf= initial Kinetic energy ki+ initial potential energy Ui

We know that the ball has Uf, the ball at the bottom and no Ki ,the ball at initial state

therefore,

Kf= Ui

therefore,

`(1)/(2)I\omega_f^2+(1)/(2)mv_f^2= mgh_i`

MOI for Solid ball =
`(2)/(5)mR^2`

`(1)/(2)(2)/(5)mR^2(V)/(R)^2+(1)/(2)mV^2= mgh_i`

=
`(7)/(10)mV^2= mgh_i`

`V= \sqrt{(10)/(7)gh_i}`

Now substituting the values to get value of V

`V= \sqrt{(10)/(7)9.81*4}`

V= 7.48 m/s

now to calculate How far horizontally from point A does the ball hit the floor

h=
`(1)/(2)gt^2`

2=
`(1)/(2)gt^2`

t=0.6385 sec

now horizontal distance d= vt = 0.6385×7.48= 5.00 m