Given:
- Mass of the proton (mp)
- Proton bounces back with 4/9 of its initial kinetic energy
- Collision is perfectly elastic

To Find:
- Mass of the heavy particle

Explanation:
In an elastic collision, both momentum and kinetic energy are conserved.

Step 1: Conservation of Momentum:
Momentum is defined as the product of an object's mass and velocity. In a collision, the total momentum before the collision is equal to the total momentum after the collision.

Let the mass of the heavy particle be M and its initial velocity be v0.
The momentum of the proton before the collision is mp * (-v0) = -mpv0.
The momentum of the proton after the collision is mp * v1, where v1 is the velocity of the proton after the collision (bouncing back).
The momentum of the heavy particle after the collision is M * v2, where v2 is the velocity of the heavy particle after the collision.

Using the principle of conservation of momentum:
Initial momentum = Final momentum

(-mpv0) = mp * v1 + M * v2

Step 2: Conservation of Kinetic Energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The initial kinetic energy of the proton is given by (1/2) * mp * v0^2.
The final kinetic energy of the proton is given by (1/2) * mp * v1^2.
The initial kinetic energy of the heavy particle is (1/2) * M * v0^2.
The final kinetic energy of the heavy particle is (1/2) * M * v2^2.

Using the principle of conservation of kinetic energy:
Initial kinetic energy = Final kinetic energy

(1/2) * mp * v0^2 = (1/2) * mp * v1^2 + (1/2) * M * v2^2

Step 3: Solving the Equations:
We have two equations from the conservation of momentum and conservation of kinetic energy:

Equation 1: -mpv0 = mp * v1 + M * v2
Equation 2: (1/2) * mp * v0^2 = (1/2) * mp * v1^2 + (1/2) * M * v2^2

We need to solve these equations to find the mass of the heavy particle (M).

Step 4: Substituting the Given Information:
From the given information, we know that the proton bounces back with 4/9 of its initial kinetic energy. This means v1 = -(4/9)v0.

Substituting this value in Equation 1:
-mpv0 = mp * (-(4/9)v0) + M * v2
Simplifying, we get:
(5/9)mpv0 = M * v2

Substituting v1 = -(4/9)v0 in Equation 2:
(1/2) * mp * v0^2 = (1/2) * mp * (-(4/9)v0)^2 + (1/2) * M * v2^

Is it 5mp ?