Answer:
Velocity =3.72 [m/s]
Explanation:
We can solve this problem using the principle of energy conservation
If we take the reference point at the bottom of the ramp where the potential energy will be 0 and the top of the ramp where the mechanical energy is maximum.
In the attached image we have a detailed sketch with the conditions of the ball moving down without friction.
Then using the properties of a rectangle triangle we have:
[tex]sin(\alpha )= \frac{h}{1} \\h=sin (45)*1\\h = 0.707[m]\\where\\h= elevation of the ball with respect to the reference point[/tex]
the potential energy will be:
[tex]Ep=m*g*h\\where:\\m = mass of the ball = 0.013[kg]\\g=gravity = 9.81 [m/s^s]\\h = 0.707 [m][/tex]
[tex]Ep= 0.013 [kg]*9.81 [m/s^s]*0.707[m]\\Ep=0.090[J]\\[/tex]
This energy is transformed to kinetic energy
[tex]Ek=\frac{1}{2}*m*v^{2} \\where\\v=velocity of the ball [m/s]\\[/tex]
[tex]v=\sqrt{(\frac{2*Ek}{m} )} \\Ek=Ep\\v=\sqrt{(\frac{2*0.09}{0.013} )} \\\\v=3.72[m/s][/tex]
