MHB Momentum of Falling Ball X & Ball Y: A Physics Puzzle

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Ball X, with a mass of 0.03 kg, falls from a height of 30 m, while Ball Y, with a mass of 0.01 kg, is projected upwards at 20 m/s. To find the downward momentum of each ball just before they meet, the equations of motion for both balls are used, leading to the determination of their velocities at the point of collision. After they coalesce, the combined object’s velocity is calculated using conservation of momentum. The time taken for the combined object to reach the ground is shown to be 2.68 seconds after Ball X started to fall. This problem illustrates the principles of momentum and kinematics in a dynamic system.
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Ball X has mass 0.03kg. It falls vertically from rest from a window that is 30 m above the ground. Ball Y has mass 0.01kg. At the same time that Ball X starts to fall, Ball Y is projected vertically upwards from ground level directly towards Ball X. The initial speed of Ball Y is 20 m/s vertically upwards.

a) Find the downward momentum of each Ball just before they meet.

The Ball coalesce and the combined object falls to the ground.
b) show that the combined object reaches the ground 2.68 s after Ball X started to fall.
Pls help as I don't know how to solve.
 
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let $d$ be the distance above ground the two masses collide ...

$d = 20t - 5t^2$
$-(30-d) = -5t^2$

solve for $t$, the time of collision, and $d$

falling mass, $v_{f1} = 0 - 10t$
rising mass, $v_{f2} = 20 - 10t$

conservation of momentum ...

$Mv_{f1} + mv_{f2} = (M+m)V$, where $V$ is the velocity of the combined masses after the collision

you should have the position of collision and $V$ the initial velocity after the collision ... calculate the time necessary to hit the ground and add to the time of collision
 
skeeter said:
let $d$ be the distance above ground the two masses collide ...

$d = 20t - 5t^2$
$-(30-d) = -5t^2$

solve for $t$, the time of collision, and $d$

falling mass, $v_{f1} = 0 - 10t$
rising mass, $v_{f2} = 20 - 10t$

conservation of momentum ...

$Mv_{f1} + mv_{f2} = (M+m)V$, where $V$ is the velocity of the combined masses after the collision

you should have the position of collision and $V$ the initial velocity after the collision ... calculate the time necessary to hit the ground and add to the time of collision
Thank you very much!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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