Two blocks on frictionless surface forced apart by spring. 4 kg moves left at 3 m/s; 6 kg moves right. Find 6 kg speed.
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A
2 m/s
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B
3 m/s
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C
4 m/s
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D
6 m/s
The 6 kg block moves with a speed of 2 m/s.
Because the surface is frictionless and the blocks are pushed apart by a spring, no external horizontal forces act on the system. This means total momentum is conserved. Since the system starts from rest, the initial total momentum is zero, and the final total momentum must also be zero.
A) 2 m/s
Momentum before separation is zero. After separation, the momenta of the two blocks must be equal in magnitude and opposite in direction. Using conservation of momentum:
total momentum equals zero
zero equals (mass of first block multiplied by its velocity) plus (mass of second block multiplied by its velocity)
zero equals (4 kg multiplied by −3 m per second) plus (6 kg multiplied by v)
zero equals −12 plus 6v
6v equals 12
v equals 2 m per second
The negative sign simply indicates opposite direction. The 6 kg block therefore moves to the right at a speed of 2 m/s.
B) 3 m/s
This would imply equal speeds despite different masses, which would violate conservation of momentum. The heavier block must move more slowly than the lighter one.
C) 4 m/s
A speed of 4 m/s would give the 6 kg block too much momentum, exceeding the momentum of the 4 kg block and breaking momentum balance.
D) 6 m/s
This would produce an even larger momentum mismatch and is inconsistent with a closed system starting from rest.
Conclusion
With no external forces, momentum is conserved. The lighter 4 kg block moves faster, and the heavier 6 kg block moves more slowly. Solving the momentum equation shows the 6 kg block’s speed is 2 m/s.

Topic Flashcards
Click to FlipWhat fundamental physics principle allows you to solve this problem, and what are the conditions required for it to apply?
The Law of Conservation of Momentum. It applies when the net external force on a system is zero. Here, the frictionless surface and the internal spring force mean no external horizontal forces act on the two-block system.
Write the momentum conservation equation for this specific scenario, defining your sign convention.
Let right be positive. Initial momentum = 0 (both blocks at rest). Final momentum: (m₁ * v₁) + (m₂ * v₂) = 0. So, (4 kg * (-3 m/s)) + (6 kg * v₂) = 0.
Solve the equation from the previous flashcard for the velocity (v₂) of the 6 kg block.
(4 * -3) + (6 * v₂) = 0 → -12 + 6v₂ = 0 → 6v₂ = 12 → v₂ = 2 m/s (positive, so moving right).
In this spring-block system, is kinetic energy conserved? Why or why not?
No. The spring's stored elastic potential energy is converted into the kinetic energy of the blocks. The total energy (elastic potential + kinetic) is conserved, but kinetic energy alone is not. You can calculate the spring's initial energy from the final KE of both blocks.
Calculate the total kinetic energy of the system after the blocks are pushed apart
KE_total = KE₄kg + KE₆kg = (½ * 4 kg * (3 m/s)²) + (½ * 6 kg * (2 m/s)²) = (0.5*4*9) + (0.5*6*4) = 18 J + 12 J = 30 J. This was the elastic potential energy stored in the spring.