Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three. As the masses collide they stick together. Mass 1 sticks to 2, then 1 2 sticks to 3, then 1 2 3 sticks to 4. When the combined 1 2 3 mass strikes mass 4, by what percentage does the speed decrease in %

Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:

[tex]p_{i} = p_{f} = m*v_{o} (1)[/tex]

Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:

[tex]p_{f1} = 2*m*v_{1} (2)[/tex]

From (1) and (2) we get:
v₁ = v₀/2 (3)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:

[tex]p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)[/tex]

Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:

[tex]p_{2} = 3*m*v_{2} (5)[/tex]

Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:

[tex]p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)[/tex]

Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:

[tex]p_{3} = 4*m*v_{3} (8)[/tex]

From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.