A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that the maximum height reached by the top of the ball is 45 cm, as shown.
80 60 40 20
If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
.45-.04=0.41
.80-.04=.76
(.41/.76)×0.75
=0.39
Unless I misunderstood the numbers, the value 45 cm does not match the sequence 80,60,40,20.
Please check or clarify.
the 45cm is supposed to be between 60cm to 40cm. The question is correct. Anyone who can help me in this too?
It says top of the ball, so you need to minus both of the heights by 8cm.
h1 = 0.8 -0.08 = 0.72
It is given that kinetic energy before striking the surface is 0.75 J. So you just nid to use potential energy formula(mgh) to calculate the mass of the ball .
h2 = 0.45 - 0.08 = 0.37
Use the mass that you had calculate and formula mgh again to count the kinetic energy after it leaves the surface.
To find the kinetic energy of the ball just after it leaves the surface, we can make use of the principle of conservation of energy. According to this principle, the total mechanical energy of a system remains constant if no external forces are acting on it.
In this case, we can consider the ball as the system. At the topmost point of its motion, the ball has a certain potential energy due to its height above the ground (PE = mgh). At the bottommost point, just before it strikes the surface, the ball has a combination of potential energy and kinetic energy (KE = 0.75 J).
When the ball bounces back up, it converts its potential energy to kinetic energy. At the highest point of its bounce, the ball will have maximum potential energy and zero kinetic energy. But since the ball still has a certain amount of kinetic energy just before it strikes the surface, it must have the same kinetic energy just after it leaves the surface.
Therefore, the kinetic energy of the ball just after it leaves the surface is 0.75 J.