Question

13. 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​

Answer 1

Answer:

Explanation:

Given  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

We know that according to law of conservation of energy we get

E pot = E translation + E rotation

We know that E translation = mv^2 / 2 (initial translational kinetic energy of ball, E rotation = lω^2 / 2 is initial rotational kinetic energy of the ball.

Now E pot = mgh = mgl sin 30 is final gravitational potential energy of ball.

For solid ball l = 2/5 mr^2 where m is mass and r is radius.

So E rotation = lω ^2 / 2 = l / 2 2/5 mr^2 ω^2

                                        = mv^2 / 5

So mv^2 /2 + mv^2 / 5 = mgl sin 30

7/10 v^2 = g l ½

Now initial speed

        V = √5 g l / 7

             = √5 x 9.8 x 1.5 / 7

             = 3.24 m / s

Answer 2

Explanation:

this is the best i can do☄☄☄☄