To solve this problem, we can apply the law of conservation of momentum and the law of conservation of kinetic energy.

The law of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. Mathematically, this can be expressed as:

m1v1i + m2v2i + m3v3i = m1v1f + m2v2f + m3v3f

where:
m1, m2, m3 are the masses of the objects (2 kg, 3 kg, and 5 kg, respectively),
v1i, v2i, v3i are the initial velocities of the objects (before the collision),
v1f, v2f, v3f are the final velocities of the objects (after the collision).

The law of conservation of kinetic energy states that the total kinetic energy of an isolated system before and after a collision remains constant for an elastic collision. Mathematically, this can be expressed as:

(1/2)m1(v1i)^2 + (1/2)m2(v2i)^2 + (1/2)m3(v3i)^2 = (1/2)m1(v1f)^2 + (1/2)m2(v2f)^2 + (1/2)m3(v3f)^2

We can solve these equations simultaneously to find the final velocities of the objects after the collision.

Given:
m1 = 2 kg
m2 = 3 kg
m3 = 5 kg

Let's assume the initial velocities as:
v1i = 1 m/s
v2i = 0 m/s (the second object is at rest)
v3i = -2 m/s (moving in the opposite direction)

Substituting the values into the momentum equation:
2(1) + 3(0) + 5(-2) = 2(v1f) + 3(v2f) + 5(v3f)
2 - 10 = 2(v1f) + 3(v2f) + 5(v3f)
-8 = 2v1f + 3v2f + 5v3f ----(1)

Now, substituting the values into the kinetic energy equation (the equation simplifies since v2i = 0):
(1/2)(2)(1)^2 + (1/2)(3)(0^2) + (1/2)(5)(-2)^2 = (1/2)(2)(v1f)^2 + (1/2)(3)(v2f)^2 + (1/2)(5)(v3f)^2
2 + 0 + 10 = (1/2)(2)(v1f)^2 + (1/2)(3)(v2f)^2 + (1/2)(5)(v3f)^2
12 = (1/2)(2)(v1f)^2 + (1/2)(5)(v3f)^2 ----(2)

We now have a system of equations (1) and (2) with three unknowns (v1f, v2f, v3f). Solving these equations will give us the final velocities.

Using equation (1):
-8 = 2v1f + 3v2f + 5v3f ----(3)

Using equation (2):
12 = v1f^2 + 5v3f^2 ----(4)

Now, we can solve equations (3) and (4) simultaneously. Since equation (3) has v2f = 0 (object 2 is at rest), we can substitute v2f = 0 in equation (3):

-8 = 2v1f + 5v3f ----(5)

Simplifying equation (4):
12 = v1f^2 + 5v3f^2 ----(6)

Solving equations (5) and (6):
Multiply equation (5) by 5 and add it to equation (6) to eliminate v3f:
-40 = 10v1f + 25v3f ----(7) (after multiplying equation (5) by 5)
12 = v1f^2 + 5v3f^2 ----(8)

Multiply equation (8) by 10 and subtract it from equation (7):
-40 - 120 = 10v1f + 25v3f - (10v1f^2 + 50v3f^2)
-160 = 25v3f - 50v3f^2 ----(9)

Rearranging equation (9):
50v3f^2 - 25v3f - 160 = 0

Solving this quadratic equation will give us the possible values of v3f (final velocity of object 3).

Using the quadratic formula:
v3f = (-(-25) ± √((-25)^2 - 4(50)(-160))) / (2(50))
v3f = (25 ± √(625 + 32000)) / 100
v3f = (25 ± √32325) / 100

Calculating √32325 and substituting the values back:
v3f = (25 ± 179.86) / 100

v3f = (25 + 179.86) / 100 = 2.04 m/s (approx.)
v3f = (25 - 179.86) / 100 = -1.55 m/s (approx.)

Therefore, the final velocity of object 3 is approximately 2.04 m/s or -1.55 m/s.

Now, substituting the value of v3f in equation (7) and solving for v1f:
-40 = 10v1f + 25(2.04) (taking v3f = 2.04 m/s)
-40 = 10v1f + 51
10v1f = -91
v1f = -9.1 m/s

-40 = 10v1f + 25(-1.55) (taking v3f = -1.55 m/s)
-40 = 10v1f - 38.75
10v1f = -1.25
v1f = -0.125 m/s

Therefore, the final velocity of object 1 is approximately -9.1 m/s or -0.125 m/s.

Since object 2 was initially at rest, it will remain at rest after the collision, so v2f = 0 m/s.

In conclusion:

The final velocities of the objects after the collision are as follows:
Object 1: approximately -9.1 m/s or -0.125 m/s
Object 2: 0 m/s
Object 3: approximately 2.04 m/s or -1.55 m/s

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