Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of v = 2.85 m/s. To reach the rack, the ball rolls up a ramp that gives the ball a h = 0.47 m vertical rise. What is the speed of the ball when it reaches the top of the ramp?

Answer 1

Apply conservation of energy

PEi + KEi = PEf + KEf

0 + (1/2)*m*vi^2 + (1/2)*I*wi^2 = m*g*h + (1/2)*m*vf^2 + (1/2)*I*wf^2

(1/2)*m*vi^2 + (1/2)*(2/5)*m*r^2*wi^2 = m*g*h + (1/2)*m*vf^2 + (1/2)*(2/5)*m*r^2*wf^2

(1/2)*m*vi^2 + (1/5)*m*vi^2 = m*g*h + (1/2)*m*vi^2 + (1/5)*m*vf^2

(7/10)*m*vi^2 = m*g*h + (7/10)*m*vf^2

v^2 = 10*g*h/7 + vf^2

==> vf = sqrt(vi^2 - 10*g*h/7)

= sqrt(2.85^2 - 10*9.8*0.47/7)

= 1.24 m/s <<<<<<<<<<<<----------------------Answer