Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of v = 3.10 m/s, as shown in the figure below.

To reach the rack, the ball rolls up a ramp that rises through a vertical distance of h = 0.529 m. What is the linear speed of the ball when it reaches the top of the ramp?

Please provide a brief explanation

Answer 1

here,

initial speed , u = 3.1 m/s

h = 0.529 m

let the linear speed of the ball when it reaches the top of ramp be v

using conservation of energy

initial kinetic energy = final kinetic energy + final potential energy

0.5 * I * w0^2 + 0.5 * m * u^2 = 0.5 * I * w^2 + 0.5 * m * v^2 + m * g * h

0.5 * (0.4 * m * r^2) * (u/r)^2 + 0.5 * m * u^2 = 0.5 * 0.4 * m * r^2 * (v/r)^2 + 0.5 * m * v^2 + m * g * h

0.7 * u^2 = 0.7 * v^2 + g * h

0.7 * 3.1^2 = 0.7 * v^2 + 9.81 * 0.529

solving for v

v = 1.48 m/s

the linear speed of the ball when it reaches the top of ramp is 1.48 m/s