Simulation

Two-Body Explosion · SimulatorSpring release, inverse mass-speed law

Momentum & CollisionsConservation of momentum

Two stuck masses release a compressed spring and fly apart; momentum sums to zero in the COM frame

Published: August 24, 2026

Objective

Verify that when two initially stationary masses release a compressed spring, their total momentum remains exactly zero throughout the flight. The simulation lets you measure how the velocity ratio equals the inverse mass ratio (v₁/v₂ = m₂/m₁) and confirm that all stored spring energy converts to kinetic energy, with no energy lost because the explosion is perfectly internal to the system.

Setup

Leave the sliders at their defaults: Mass 1 = 1.0 kg, Mass 2 = 2.0 kg, Spring Energy = 10 J. Note the two bodies side-by-side at center with the compressed spring between them.
Read the momentum arrows before pressing Start. Both navy arrows point outward and are already equal in length, confirming that ΣP = 0 before Start.
Click Start. The spring releases instantaneously and the bodies fly outward. Watch the ΣP readout: it should read 0.00 kg·m/s throughout.
Record v₁, v₂, KE₁, KE₂ from the HUD after the bodies separate. Confirm that KE₁ + KE₂ = 10.00 J and that v₁ / v₂ ≈ 2.0 (the mass ratio m₂/m₁).
Click Reset, then drag Mass 1 to 2.0 kg (equal masses). Start again and observe that both bodies move at the same speed in opposite directions, and KE₁ = KE₂ = 5.00 J.
Finally, try the extreme ratio: set Mass 1 = 0.5 kg and Mass 2 = 4.0 kg. The lighter body should move 8 times faster, demonstrated by the trail length difference on screen.

Before Start at the default settings: the 1 kg and 2 kg bodies sit side by side at the centre with the compressed spring between them, and the momentum arrows already show equal lengths.

The end of a run at the default settings: the 1 kg body (sky, left) has travelled twice as far from the dashed centre-of-mass line as the 2 kg body (amber, right), while the two navy momentum arrows are equal in length.

With Mass 1 = 0.5 kg and Mass 2 = 4.0 kg, the light body has crossed almost the whole left half of the canvas while the heavy body has barely left the centre line: eight times the speed, eight times the trail.

Analytical Prediction

With m₁ = 1.0 kg, m₂ = 2.0 kg, and spring energy E = 10 J, the analytic closed form gives:

v₂=sqrt(2·E·m₁ / (m₂·(m₁+m₂)))

=sqrt(2·10·1.0 / (2.0·3.0))

=sqrt(20/6)

≈1.826 m/s

v₁=-sqrt(2·E·m₂ / (m₁·(m₁+m₂)))

=-sqrt(2·10·2.0 / (1.0·3.0))

=-sqrt(40/3)

≈-3.651 m/s

ΣP=m₁·v₁ + m₂·v₂

=1.0·(-3.651) + 2.0·1.826

≈0.000 kg·m/s

KE₁=½·1.0·3.651² ≈ 6.67 J

KE₂=½·2.0·1.826² ≈ 3.33 J

KE_total≈10.00 J = E

The speed ratio v₁/v₂ = 3.651/1.826 ≈ 2.0 = m₂/m₁, confirming the inverse mass-speed law. Total momentum is exactly zero because both conserved quantities are satisfied simultaneously.

Results Analysis

After clicking Start with the default settings, the |v₁| readout should show 3.65 m/s and |v₂| should show 1.83 m/s, giving a ratio of approximately 2.00. The ΣP readout should hold steady at 0.00 kg·m/s for the entire flight. The KE₁ and KE₂ readouts should show approximately 6.67 J and 3.33 J respectively, summing to 10.00 J. Any deviation from 0.00 in ΣP is purely display rounding (the velocities are computed analytically, not by numerical integration, so the true momentum is exactly zero). Compare the navy momentum arrow lengths: they should be visually identical in length, pointing in opposite directions, which is the graphical proof of ΣP = 0.

Source of Error

The simulation computes post-explosion velocities analytically from the closed-form energy-momentum equations, not by integrating forces through time. This means the model omits the finite spring-release duration (in a real explosion, the impulse is delivered over a short interval), any rotational or deformational modes of the bodies, and gravitational attraction between the masses. The bodies are treated as point masses riding on a frictionless, horizontal track in one dimension. Because the velocities are exact (not accumulated by a stepping integrator), the only source of residual in the ΣP readout is display rounding to two decimal places. The gap between the predicted KE partition and the readout values is therefore purely numerical.

Further Exploration

Set Mass 1 = Mass 2 = 2.0 kg and Spring Energy = 10 J, then Start. Are the trail lengths equal? Are v₁ and v₂ identical? What does symmetry in mass imply for symmetry in KE partition?
With Mass 1 = 0.5 kg and Mass 2 = 4.0 kg, predict the speed ratio before clicking Start, then verify against the HUD. Does the lighter body really travel 8 times farther before exiting?
Double the Spring Energy from 10 J to 20 J while keeping masses fixed. Do both velocities increase by the same factor? By what factor do they scale (check: sqrt(2) ≈ 1.414)?
Compare several ghost trails with different mass ratios layered on screen. Can you identify which trail belongs to which mass configuration from the relative lengths alone?
What happens to ΣP when you set the most extreme mass contrast allowed (0.5 kg vs 5.0 kg)? Does the readout ever leave 0.00? Why or why not?