Question

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 %

Answer 1

Answer:

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:

       $p_{i} = p_{f} = m*v_{o} (1)$

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

       $p_{f1} = 2*m*v_{1} (2)$

• 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:

       $p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)$

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

        $p_{2} = 3*m*v_{2} (5)$

• 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:

      $p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)$

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

       $p_{3} = 4*m*v_{3} (8)$

• 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.