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 %

The speed decreases 75%.

Explanation:

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:

       

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

       

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:

       

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

       

From (4) and (5) we get:v₂ = v₀/3  (6)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:

     

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

       

From (7) and (8) we get:v₃ = v₀/4This 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.