A bowling ball encounters a 0.760-m vertical rise on

Physics

1 answer:

matrenka [14]3 years ago

7 0

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

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

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

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

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

You might be interested in

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)