Adventurous Mary tries a new game: she puts on her rollerblades and throws the ball towards the wall, lets it bounce off, catches it again, throws it right back, and so on. She notices a strange thing: after the second throw, she is unable to catch the ball anymore since the ball never reaches her. What is the maximum ratio of Mary's mass to the ball's mass in order for this to happen?
To determine the maximum ratio of Mary's mass to the ball's mass, we need to analyze the physics involved in this scenario.
Let's break down the events step by step:
1. Mary throws the ball towards the wall.
2. The ball bounces off the wall.
3. Mary tries to catch the ball.
For Mary to successfully catch the ball, the ball needs to travel back to her after bouncing off the wall. This implies that the ball's momentum needs to be sufficient to overcome the momentum of Mary on rollerblades.
The momentum of an object is given by the product of its mass and velocity. Since the ball bounces off the wall, it changes direction but retains its initial velocity magnitude.
Let's denote Mary's mass as M and the ball's mass as m. Initially, Mary's velocity is zero, and the ball's velocity is v (which we consider positive in the direction towards the wall).
After the ball bounces off the wall, its velocity becomes -v (reversed direction). Thus, the change in momentum for the ball is given by 2mv.
For Mary to be able to catch the ball, her change in momentum should be less than or equal to the change in momentum of the ball. This implies:
|2mv| ≤ MΔv,
where Δv is the final change in velocity for Mary.
Now, let's consider the situation after the second throw when Mary cannot catch the ball. At this point, Mary would have thrown the ball twice, resulting in the following changes:
Ball's momentum: 2mv
Mary's momentum: MΔv
Since Mary cannot catch the ball, it means these two momenta are equal in magnitude but have opposite directions:
|2mv| = MΔv.
Substituting the equation above, we get:
2mv = MΔv.
We can simplify this equation by canceling out the v and rearranging:
2m = M.
Thus, the maximum ratio of Mary's mass to the ball's mass for her to be unable to catch the ball is 2:1.
In other words, Mary's mass should not exceed twice the mass of the ball for her to be unable to catch it after the second throw.