Simulation

1D Elastic Collision · SimulatorMomentum and KE Both Conserved

Momentum & CollisionsElastic collisions

Two carts collide head-on with adjustable masses and velocities; KE and momentum both conserved

Published: August 30, 2026

Objective

Verify that a 1D elastic collision conserves both linear momentum and kinetic energy simultaneously. Explore how the mass ratio between two carts governs the post-collision velocity split: equal masses swap velocities exactly, while unequal masses produce asymmetric outcomes. The model assumes a frictionless track and perfectly elastic impact with no energy lost to deformation.

Setup

Leave the sliders at their default values: Mass 1 = 1.0 kg, Mass 2 = 3.0 kg, Velocity 1 = +4.0 m/s, Velocity 2 = −2.0 m/s. Note the momentum readout (p = −2.00 kg·m/s) and kinetic energy readout (KE = 14.00 J) before starting.
Press Start. Watch the blue cart (left) and amber cart (right) approach each other. At the moment of impact a crimson ring briefly marks the collision point.
After the collision, record the v₁ and v₂ readouts. The momentum and KE readouts should remain unchanged at the pre-collision values, confirming conservation.
Press Reset to return both carts to their starting positions with the same slider settings. Set Mass 1 = 2.0 kg and Mass 2 = 2.0 kg, then press Start to observe the equal-mass velocity-swap case.

Two carts on the track at their default starting positions: the lighter 1 kg cart (blue, left) approaching at 4 m/s and the heavier 3 kg cart (amber, right) at 2 m/s.

After the elastic collision, the lighter cart bounces back at 5 m/s while the heavier cart moves forward at 1 m/s, with momentum and kinetic energy both conserved.

With equal masses, the two carts swap velocities exactly: cart 1 exits at the initial speed of cart 2 and vice versa, the clearest demonstration of the momentum-exchange rule.

Analytical Prediction

For m₁ = 1.0 kg, m₂ = 3.0 kg, v₁ = +4.0 m/s, v₂ = −2.0 m/s, the closed-form elastic collision formulas give:

v₁'=((m₁ − m₂)·v₁ + 2·m₂·v₂) / (m₁ + m₂)

=((1 − 3)·4 + 2·3·(−2)) / (1 + 3)

=(−8 − 12) / 4

=−5.00 m/s

v₂'=((m₂ − m₁)·v₂ + 2·m₁·v₁) / (m₁ + m₂)

=((3 − 1)·(−2) + 2·1·4) / (1 + 3)

=(−4 + 8) / 4

=+1.00 m/s

The lighter cart reverses direction and exits at 5.00 m/s; the heavier cart slows slightly and continues at 1.00 m/s. Total momentum before: 1·4 + 3·(−2) = −2.00 kg·m/s. Total momentum after: 1·(−5) + 3·1 = −2.00 kg·m/s (conserved). KE before: 0.5·1·16 + 0.5·3·4 = 14.00 J. KE after: 0.5·1·25 + 0.5·3·1 = 14.00 J (conserved).

Results Analysis

After pressing Start with the default sliders, watch the v₁ readout flip from +4.00 to −5.00 m/s at the collision frame, and v₂ jump from −2.00 to +1.00 m/s. The Momentum p and Kinetic Energy readouts remain at −2.00 kg·m/s and 14.00 J throughout, confirming both conservation laws. The momentum bar chart on the right side of the canvas shows p₁ (blue bar) and p₂ (amber bar) swap in magnitude while the navy total bar stays level. The dashed reference line on the total momentum bar confirms the pre-collision value is unchanged post-impact. If the equal-mass case (both 2.0 kg) is run instead, v₁ and v₂ swap exactly, which the readouts confirm to two decimal places.

Source of Error

The model assumes a perfectly frictionless track: no rolling resistance, no air drag, and no energy lost to cart deformation at impact. Real carts always lose a fraction of kinetic energy to sound, vibration, and heat even in near-elastic collisions, so measured post-collision speeds will fall slightly short of the predicted values. The closed-form formulas applied here are exact solutions to the two conservation equations for a point-mass system, so the residual gap between the predicted and displayed velocities is purely numerical, not physical.

Further Exploration

Set Mass 1 = Mass 2 (try 2.0 kg each) and give both carts non-zero velocities. Do the carts always swap velocities exactly? What does the v₁ readout show immediately after impact compared to the initial v₂?
Set Velocity 2 = 0 (stationary target) and vary the mass ratio from 0.5:5.0 to 5.0:0.5. How does the fraction of kinetic energy transferred to the target depend on the mass ratio?
Can you find a slider combination where cart 1 passes through cart 2 without triggering a collision? (Hint: what happens when both carts move in the same direction and cart 1 is slower?)
Set both velocities to their maximum magnitudes (v₁ = +10, v₂ = −10) with extreme mass ratios. Does the KE readout still confirm exact conservation even at the slider extremes?