Where Do Two Bouncing Balls Meet?

[pmd28]

Bob has two balls. He releases one ball from a platform of height H. Just as that ball strikes the ground he releases the next ball from the same height H. Assuming that the first ball bounces perfectly, (i.e. reversing only the direction of its velocity when it strikes the ground), at what height from the ground do the balls strike each other?
a) H/4
b) h/2
c) h/3
d) 3H/4
e) none of the above

the answer is d

I thought that when the first ball strikes the ground it has a momentary velocity of zero. And since it bounces "perfectly" it should return with the same velocity it was released with therefore the ball that was released would have the same kinematic properties. ergo they should meet in the middle. I'm so confused.

 

[azizlwl]

Use vertlcal motion for both downward and upward direction balls.
2 equations 2 unknown, y and t.

 

[pmd28]

Use vertlcal motion for both downward and upward direction balls.
2 equations 2 unknown, y and t.

So for ball 1 (the ball that is returning up) my variables are:
v= (dunno, I think they would be the same for both.)
v(initial)= 0
t=?
a=g
y=?

and variables for ball 2 are the same. And when I solve the equation i still get H/2

 

[azizlwl]

It is a 1D motion
One from top with y₀=H, v₀=0
On from bottom, with kinetic energy of equal mgh. You can convert this to v₀.

 

[Nickie]

Hi there,

I can provide a response to this question. The correct answer is d) 3H/4.

To understand why this is the correct answer, we need to first understand the concept of conservation of energy. When the first ball is released from the platform at height H, it has a certain amount of potential energy due to its position above the ground. As it falls, this potential energy is converted into kinetic energy, and when it strikes the ground, all of its potential energy is converted into kinetic energy. However, due to the perfect bounce, the kinetic energy is conserved and the ball bounces back up to its original height H.

Now, when the second ball is released from the same height H, it also has the same amount of potential energy. However, it is released at the exact same time that the first ball is returning to its original height. This means that the second ball will have the same amount of potential energy as the first ball when it reaches the ground. Since we know that the first ball strikes the ground at a height of 3H/4, we can conclude that the second ball will also strike the ground at the same height of 3H/4. Therefore, the balls will strike each other at a height of 3H/4 from the ground.

I hope this explanation helps to clarify any confusion. Keep exploring and asking questions!