We may consider the entire mass of the stick to be concentrated as a point mass at the center of mass of the stick. As the center of mass moves as a projectile, it will move along a parabolic path.

Let A₁ = area of complete circle = πa²

(x₁, y₁) = coordinates of center of mass of the large circle = (0, 0)

(x₂, y₂) = coordinates of center of mass of the small circle = 

Using X_CM = (A₁x₁, - A₂x₂)/(A₁ - A₂), 

y_CM = 0 (as y₁ and y₂ both are zero)
Therefore, coordinates of center of mass of the lamina are (- a/6, 0).

Mass of the cut-out disc is

The center of the disc are at the origin of the coordinates.

Then the center of mass of the disc system will be

Let us consider incline surfaces of wedge as x-axis and y-axis respectively. We can write accelerations of blocks A and B.

The acceleration of center of mass of two blocks can be written as

Let m₁, m₂ and m₃ be the masses of the three pieces.
m₁ = 1.0 kg, m₂ = 2.0 kg
Let v₁ = 12 m/s, v₂ = 8 m/s, v₃ = 40 m/s.
Let v₁ and v₂ be directed along x- and y-axes, respectively, and v₃ be directed as shown.

By the principle of conservation of momentum, initial momentum is zero. Hence,
along x-axis: 0 = m₁v₁ - m₃v₃ cos θ
along y-axis: 0 = m₂v₂ - m₃v₃ sin θ
⇒ m₁v₁ = m₃v₃ cos θ 
and, m₂v₂ = m₃v₃ sin θ
By squaring and adding, 

The initial position of center of mass is

Initial velocity of center of mass

Here H is the maximum height reach by center of mass of two balls from initial level.
∴ 0² = 40² - 2 × 10 × H

Hence, maximum height reach by center of mass from ground level will be
h_max = (h_cm)_initial + H = 20 + 80 = 100 m

The velocity of the shell at the highest point of trajectory is
V_M = u cos θ = 100 cos 60° = 50 m/s
Let v₁ be the speed of the fragment which moves along the negative x-direction and the other fragment has speed v₂, which must be along the positive x-direction. 
Now from conservation of momentum, 

or, v₂ = 2v + v₁ = (2 × 50) + 50 
= 150 m/s

There are no external forces is acting on the system in horizontal direction, hence momentum remains conserved in this direction. After the jump, car attains a velocity v₂ in the same direction, which is less than v₁, due to backward push of the child for jumping. After the jump, child attains a velocity u + v₂ in the direction of motion of car with respect to ground.

According to conservation of momentum,
(M + m)v₁ = Mv₂ + m(u + v₂)

Let the required speed be V. Further, let J₁, be the impulse between the particle and the pan and J₂ be the impulse imparted to the block and the pan by the string.
Using impulse as the change in momentum, 
For particle, J₁ = mv - mV ..................(i)
For pan, J₁ - J₂ = mV.......................... (ii)
For block, J₂ = mV............................. (iii)
Solving Eqs. ................(i)-(iii), 
V = v/3

The string will become taut when the particle will fall through a distance 2 m in downward direction.

So the required impulse: