Question

A uniform solid ball first rolls purely along a floor,
then up a ramp inclined at 30°. It momentarily
stops when it has rolled 1.5 m along the ramp. Its
initial speed is approximately
(1) 2 m/s
(2) 3 m/s
(3) 4 m/s
(4) 5 m/s​

Answer 1

Answer:

Explanation:

According to the conservation of energy law :

     $E_{translation}$ + $E_{rotation}$ = $E_{potential}$

where $E_{translation}$ = $\frac{mv^{2}}{2}$ is initial translational kinetic energy of the ball,  = $E_{rotation}$ = $\frac{1w^{2}}{2}$ is initial rotational

kinetic energy of the ball, $E_{potential}$  = mgℎ = mgl sin 30 is final gravitational potential energy of the ball.

For a solid ball:

                                       I = $\frac{2}{5} mr^{2}$

where m  is a mass of ball, r  is a radius of ball. So,

           $E_{rotation}$   = $\frac{1w^{2}}{2}$

            = $\frac{1}{2}$ $\frac{2}{5} mr^{2}$ $ω^{2}$     = $\frac{mv^{2}}{5}$

Thus,   $\frac{mv^{2}}{2}$ + $\frac{mv^{2}}{5}$ = mgl sin 30

              $\frac{7}{10} v^{2}$ = gl . $\frac{1}{2}$

The initial speed is

               v = $\sqrt{\frac{5gl}{7} }$ = $\sqrt{\frac{5 * 9.8 * 1.5}{7} }$         = 3.24m/s

So we will go with 3m/s aa answer.

Answer 2

option 2

simply used conservation of total energy (translational+rotational)