Answer:
When balloon moves in the downward direction two forces acts on it.
i) Force exerted by air in the upward direction
ii) Weight
According to newton’s second law of motion:
Sum of forces = Ma
W – F = Ma
Mg – F = Ma …….. (i)
when some of the mass m is dropped and balloon is moving in upward direction with acceleration a/2 then,
F – W = (M-m)a/2
F – (M-m)g = (M-m)a/2
F – Mg + mg = Ma/2 – ma/2 ….. (ii)
Adding equation (i) and (ii)
mg = M(3a/2) – ma/2
m(g + a/2) = M(3a/2)
m = M(3a/2)/(g + a/2)
Compared with an Earth year, a galactic year represents time on a grand scale but its not a consistent measurement across the galaxy
To calculate the average acceleration of the ball we use the formula,

Here,
is the final velocity of the ball and
is the initial velocity of the ball t is the time in contact with the wall.
Given
towards the wall and 
away from the wall and
.
Substituting these values in above formula , we get

Here
is negative because ball is moving away from the wall.

Therefore, average acceleration of the ball is
(away from the wall).
Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Amplitude does not mean a note has a high pitch