Question

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 8.57 m/s at the bottom of the rise. Find the translational speed at the top.

Answer 1

Answer:

The velocity at the top is 7.65 m/s

Solution:

As per the question:

Height of the ball, h = 0.760 m

Linear speed of the ball at bottom, v = 8.57 m/s

Let the speed at the top be v' m/s

Now,

Using the principle of the work energy theorem:

The total mechanical energy of the system remain conserved.

$PE + KE = 0$

where

PE = mgh = potential energy

KE = $\frac{1}{2}m(v'^{2} - v^{2})$ =  change in kinetic energy

Thus

$mgh + \frac{1}{2}m(v'^{2} - v^{2})$ = 0[/tex]

$v'^{2} = v^{2} - 2gh$

$v' = \sqrt{8.57^{2} - 2\times 9.8\times 0.760} = 7.65\ m/s$