Question

A bowling ball encounters a 0.760-m vertical rise on

Answer 1

Answer:1.26 m/s

Explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com $I=\frac{2}{5}mr^2$

Initial Energy

$E_i=\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2(\omega =\frac{u}{r})$

Final Energy

$E_f=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh$

Equating energy as no energy loss take place

$E_i=E_f$

$\frac{1}{2}mu^2+\frac{1}{2}I\omega _i^2=\frac{1}{2}mv^2+\frac{1}{2}I\omega _f^2+mgh$

$\frac{1}{2}mu^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{u}{r}\right )^2=\frac{1}{2}mv^2+\frac{1}{2}\times \frac{2}{5}mr^2\times \left ( \frac{v}{r}\right )^2+mgh$

m term get cancel

$\left ( \frac{u^2}{2}\right )+\left ( \frac{2u^2}{10}\right )=\left ( \frac{v^2}{2}\right )+\left ( \frac{2v^2}{10}\right )+gh$

$\frac{7}{10}u^2=\frac{7}{10}v^2+gh$

$v^2=3.5^2-\frac{10}{7}\times 9.81\times 0.76$

$v=\sqrt{1.6}=1.26 m/s$