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

Answer:

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

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,

$\frac{1}{2}I\omega_f^2+\frac{1}{2}mv_f^2= mgh_i$

MOI for Solid ball = $\frac{2}{5}mR^2$

$\frac{1}{2}\frac{2}{5}mR^2\frac{V}{R}^2+\frac{1}{2}mV^2= mgh_i$

=$\frac{7}{10}mV^2= mgh_i$

$V= \sqrt{\frac{10}{7}gh_i}$

Now substituting the values to get value of V

$V= \sqrt{\frac{10}{7}9.81\times4}$

V= 7.48 m/s

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

h= $\frac{1}{2}gt^2$

2= $\frac{1}{2}gt^2$

t=0.6385 sec

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