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SCIENCE · INVESTIGATE AND EXPLAIN

Carry position and velocity across a contact-loss event

A support begins at rest with upward coordinate ys=−k t³/6, so its acceleration is−kt. A2 kg point initially rests on it. If contact holds, the point must share this acceleration. Newton's law gives N−mg=−mkt, or N=m(g−kt). A contact surface can push upward but cannot pull the point down.

Learning outcome: Locate separation analytically, compare a pre-event, exact-boundary and post-event state, and retain continuous position/velocity plus the growing support gap.

Starting recap

Differentiate/integrate with explicit initial conditions; identify each force recipient and the inertial frame. Preserve force dimensions, geometric constraints and the distinction between model and solver error.

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Try

For k=10, separation occurs at1 s. At t=2, point height is−35/3≈−11.6667 m while support height is−40/3≈−13.3333 m; gap5/3 m.

Open the experiment

Predict

Once the required normal becomes negative, the model must:

Experiment

Locate separation analytically, compare a pre-event, exact-boundary and post-event state, and retain continuous position/velocity plus the growing support gap.

Switch equations at a unilateral-contact event

Worked teaching model · synthetic values · Virtual2 kg point initially rests on a support with ysupport=−k t³/6, upward positive. Contact can push upward but not pull. Before separation it shares support motion; afterward only gravity−10 m/s² acts. No collisions or recontact occur on this bounded downward-accelerating branch.

Reset keeps your prediction, captures and practice history.

Contact trajectory and rejected continuation

height (m)time (s)020-8

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: actual point (solid).2: support (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Contact trajectory and rejected continuation

actual point

time (s)height (m)
00
0.025-1.56250e-5
0.05-0.000125
0.075-0.000421875
0.1-0.001
0.125-0.001953125
0.15-0.003375
0.175-0.005359375
0.2-0.008
0.225-0.01139063
0.25-0.015625
0.275-0.02079688
0.3-0.027
0.325-0.03432813
0.35-0.042875
0.375-0.05273438
0.4-0.064
0.425-0.07676562
0.45-0.091125
0.475-0.1071719
0.5-0.125
0.525-0.1447031
0.55-0.166375
0.575-0.1901094
0.6-0.216
0.625-0.2441406
0.65-0.274625
0.675-0.3075469
0.7-0.343
0.725-0.3810781
0.75-0.421875
0.775-0.4654844
0.8-0.512
0.825-0.5615156
0.85-0.614125
0.875-0.6699219
0.9-0.729
0.925-0.7914531
0.95-0.857375
0.975-0.9268594
1-1
1.025-1.076891
1.05-1.157625
1.075-1.242297
1.1-1.331
1.125-1.423828
1.15-1.520875
1.175-1.622234
1.2-1.728
1.225-1.838266
1.25-1.953125
1.275-2.072672
1.3-2.197
1.325-2.326203
1.35-2.460375
1.375-2.599609
1.4-2.744
1.425-2.893641
1.45-3.048625
1.475-3.209047
1.5-3.375
1.525-3.546578
1.55-3.723875
1.575-3.906984
1.6-4.096
1.625-4.291016
1.65-4.492125
1.675-4.699421
1.7-4.912963
1.725-5.132755
1.75-5.358796
1.775-5.591088
1.8-5.82963
1.825-6.074421
1.85-6.325463
1.875-6.582755
1.9-6.846296
1.925-7.116088
1.95-7.39213
1.975-7.674421
2-7.962963

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

support

time (s)height (m)
00
0.025-1.56250e-5
0.05-0.000125
0.075-0.000421875
0.1-0.001
0.125-0.001953125
0.15-0.003375
0.175-0.005359375
0.2-0.008
0.225-0.01139063
0.25-0.015625
0.275-0.02079688
0.3-0.027
0.325-0.03432813
0.35-0.042875
0.375-0.05273438
0.4-0.064
0.425-0.07676562
0.45-0.091125
0.475-0.1071719
0.5-0.125
0.525-0.1447031
0.55-0.166375
0.575-0.1901094
0.6-0.216
0.625-0.2441406
0.65-0.274625
0.675-0.3075469
0.7-0.343
0.725-0.3810781
0.75-0.421875
0.775-0.4654844
0.8-0.512
0.825-0.5615156
0.85-0.614125
0.875-0.6699219
0.9-0.729
0.925-0.7914531
0.95-0.857375
0.975-0.9268594
1-1
1.025-1.076891
1.05-1.157625
1.075-1.242297
1.1-1.331
1.125-1.423828
1.15-1.520875
1.175-1.622234
1.2-1.728
1.225-1.838266
1.25-1.953125
1.275-2.072672
1.3-2.197
1.325-2.326203
1.35-2.460375
1.375-2.599609
1.4-2.744
1.425-2.893641
1.45-3.048625
1.475-3.209047
1.5-3.375
1.525-3.546578
1.55-3.723875
1.575-3.906984
1.6-4.096
1.625-4.291016
1.65-4.492125
1.675-4.699422
1.7-4.913
1.725-5.132953
1.75-5.359375
1.775-5.592359
1.8-5.832
1.825-6.078391
1.85-6.331625
1.875-6.591797
1.9-6.859
1.925-7.133328
1.95-7.414875
1.975-7.703734
2-8

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
First separation time1.666667s
Actual point height-7.962963m
Support height-8m
Point above support gap0.03703704m
Actual vertical velocity-11.66667m/s
Actual upward normal force0N
Actual vertical acceleration-10m/s²

The required pulling normal is impossible; the point follows a freefall branch after separation.

While contact holds, N−mg=m(−k t), so N=m(g−k t). Its first zero is g/k. Carry continuous position and velocity from that event into freefall rather than resetting either to zero. At the exact boundary acceleration agrees on both sides; afterward the support falls away faster and the gap grows. A negative algebraic normal is a failed contact assumption, not a physical pulling surface.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The required pulling normal is impossible; the point follows a freefall branch after separation.

While contact holds, N−mg=m(−k t), so N=m(g−k t). Its first zero is g/k. Carry continuous position and velocity from that event into freefall rather than resetting either to zero. At the exact boundary acceleration agrees on both sides; afterward the support falls away faster and the gap grows. A negative algebraic normal is a failed contact assumption, not a physical pulling surface.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Carry position and velocity across a contact-loss event

A support begins at rest with upward coordinate ys=−k t³/6, so its acceleration is−kt. A2 kg point initially rests on it. If contact holds, the point must share this acceleration. Newton’s law gives N−mg=−mkt, or N=m(g−kt). A contact surface can push upward but cannot pull the point down.

The first zero normal occurs at ts=g/k. This is a branch boundary, not a reason to set the point’s position or velocity to zero. At separation, ys=−k ts³/6 and vs=−k ts²/2. Carry both values into freefall: y=ys+vs(t−ts)−g(t−ts)²/2 after the event.

For k=6 and g10, ts=5/3 s. At t=2 the support is at−8 m but the point is at−215/27≈−7.96296 m, leaving a1/27≈.037037 m gap. Actual normal is zero. Continuing the contact equation would demand a negative normal and a different trajectory.

At the exact event the accelerations match at−g. Afterward the support accelerates downward more strongly, so it falls away and this bounded model has no recontact. The graph compares an actual allowed branch with the support path; a rejected negative-force continuation must not be labeled an observed physical force.

Changed case: predict before checking

For k=10, separation occurs at1 s. At t=2, point height is−35/3≈−11.6667 m while support height is−40/3≈−13.3333 m; gap5/3 m.

Retained investigation

Locate separation analytically, compare a pre-event, exact-boundary and post-event state, and retain continuous position/velocity plus the growing support gap. Why is clipping a force insufficient if the trajectory is not recomputed on the new branch?

Sources and boundaries

Primary conceptual reference checked30 October2026: OpenStax University Physics Volume1. Original mathematical cases and synthetic records; no performed experiments, practical certification or actual equipment safety claims. Numerical method/error relations are derived explicitly for the supplied equations. Preserve independent checks, conditional regimes and first-response evidence. No physical release, launch, loading or pulley procedure is requested.

Apply

Once the required normal becomes negative, the model must:

Public practice checks a specific claim. It does not establish independent mastery.

Why is clipping a force insufficient if the trajectory is not recomputed on the new branch?

Continue

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