Answer:
A) True B) false C) True D) False E) False
Explanation:
A) We know that the potential energy is transformed into kinetic energy since the ball is released.
[tex]EP1=m*g*h = 1 * 9.81 * 6 = 58.86J\\[/tex]
This energy will be the same at the moment that the ball is about to hit the floor. Therefore there was a transformation from potential to kinetic energy
EK1 = EP1
[tex]EK=\frac{1}{2}*m*v^{2} \\v1=\sqrt{(58.86*2)/1} \\v1=10,84 m/s[/tex]
With the 2kg ball happens the same
[tex]EP2=m*g*h2=2*9.81*3=58.86J\\[/tex]
Velocity at the moment when is about to hit the floor (due to the energy transform) will be:
[tex]EK2=\frac{1}{2}*m*v^{2} \\v1=\sqrt{(58.86*2)/2} \\v2=7.67 m/s[/tex]
A) Since both balls have the same potential energy, they will have the same kinetic energy
B) The velocities are differents as they were calculated before.
C) Therefore the 1 kg ball moves faster as it reaches the ground.
D) The kinetic energy is the same for both balls as it was calculated before.
E) The 2 kg ball will reach the ground first, since it is closer to the ground
With the velocities calculated in the previous steps we have now:
[tex]1 kg ball\\v=v0+g*t\\t1=10.85/9.81 = 1.1s\\\\2 kg ball\\t2=7.672/9.81 = 0.782s[/tex]
