Dynamics

Impulse–Momentum & 1-D Collisions

Impulse mode: F·Δt = m·Δv for one body — solves the missing one of {final velocity, impulse, force, duration}. Collision mode: direct central impact of two bodies with a coefficient of restitution — post-impact velocities, impulse between the bodies, and kinetic energy lost. Use for: hammer blows, jet impingement timing, cars/carts colliding, ballistic pendulum first stage. Not for: oblique (2-D) impacts.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

modeimpulse = one body under a force/impulse; collision = two-body direct central impactmass (optional)Body mass (impulse mode) [dimension: mass]initial velocity (optional)SIGNED initial velocity along +x (impulse mode) [dimension: velocity]final velocity (optional)SIGNED final velocity (impulse mode; omit to solve for it) [dimension: velocity]force (optional)Average force on the body (impulse mode) [dimension: force]duration (optional)Duration the force acts (impulse mode) [dimension: time]impulse (optional)Impulse magnitude in force·seconds — give value with unit N or lbf and it is read as N·s / lbf·s (impulse mode, alternative to force+duration) [dimension: force]mass1 (optional)Mass of body 1 (collision mode) [dimension: mass]velocity1 (optional)SIGNED velocity of body 1 before impact: +x positive, so two approaching bodies have OPPOSITE signs [dimension: velocity]mass2 (optional)Mass of body 2 (collision mode) [dimension: mass]velocity2 (optional)SIGNED velocity of body 2 before impact (collision mode) [dimension: velocity]restitution (optional)Coefficient of restitution e, 0 (perfectly plastic, bodies stick) to 1 (elastic)

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

elastic equal-mass swap: 2 kg at +4 hits 2 kg at −2, e=1

Given

mode: collision
mass1: 2 kg
velocity1: 4 m/s
mass2: 2 kg
velocity2: -2 m/s
restitution: 1

Assumptions

Direct central impact along one line; the impact is short enough that external forces are negligible during contact.

Solution steps

Conserve momentum and apply restitution
Momentum is conserved through the impact; restitution relates the separation speed to the approach speed.
m1·v1 + m2·v2 = m1·v1' + m2·v2'; e = (v2' − v1')/(v1 − v2)
Total momentum = 4 kg·m/s; v1' = -2 m/s; v2' = 4 m/s
Impulse and energy loss
The internal impulse changes each body's momentum equally and oppositely; the lost kinetic energy depends on (1 − e²).
ΔE = (m1·m2/(m1+m2))·(1 − e²)·(v1 − v2)²/2
Impulse on body one = -12 N·s; energy lost = 0 J
= 0 J

Results

Velocity of body 1 after impact

-2m/s

Velocity of body 2 after impact

4m/s

Impulse between the bodies (N·s, signed on body 1)

-12

Kinetic energy lost in the impact

0kJ

Where this answer was checked

source: Classical result: equal masses in elastic impact exchange velocities — e=1, m1=m2
verified by: hand-recomputed
derivation: p = 2·4 + 2·(−2) = 4. v1' = (4 + 2·1·(−6))/4 = −2; v2' = (4 + 2·1·6)/4 = +4 — exact swap. ΔE = 0.

Example 2

perfectly plastic: 3 kg at 5 m/s hits 2 kg at rest, e=0

Given

mode: collision
mass1: 3 kg
velocity1: 5 m/s
mass2: 2 kg
velocity2: 0 m/s
restitution: 0

Assumptions

Direct central impact along one line; the impact is short enough that external forces are negligible during contact.

Solution steps

Conserve momentum and apply restitution
Momentum is conserved through the impact; restitution relates the separation speed to the approach speed.
m1·v1 + m2·v2 = m1·v1' + m2·v2'; e = (v2' − v1')/(v1 − v2)
Total momentum = 15 kg·m/s; v1' = 3 m/s; v2' = 3 m/s
Impulse and energy loss
The internal impulse changes each body's momentum equally and oppositely; the lost kinetic energy depends on (1 − e²).
ΔE = (m1·m2/(m1+m2))·(1 − e²)·(v1 − v2)²/2
Impulse on body one = -6 N·s; energy lost = 15 J
= 15 J

Results

Velocity of body 1 after impact

3m/s

Velocity of body 2 after impact

3m/s

Impulse between the bodies (N·s, signed on body 1)

-6

Kinetic energy lost in the impact

0.015kJ

Where this answer was checked

source: Common velocity = momentum-weighted mean; ΔE via reduced mass — e=0 sticking impact
verified by: hand-recomputed
derivation: v' = 15/5 = 3 m/s both. KE1 = 0.5·3·25 = 37.5 J; KEpost = 0.5·5·9 = 22.5 J; ΔE = 15 J. Reduced-mass check: 0.5·(6/5)·1·25 = 15 ✓.

Example 3

US impulse: 1 slug at 10 ft/s pushed by 20 lbf for 2 s

Given

mode: impulse
mass: 1 slug
initial velocity: 10 ft/s
force: 20 lbf
duration: 2 s

Assumptions

Straight-line motion; the average force acts along +x; SIGNED velocities.

Solution steps

Apply the impulse–momentum theorem
The impulse on the body equals its change in momentum.
F·Δt = m·(v2 − v1)
J = 177.9 N·s; v2 = 50 ft/s
= 50 ft/s

Results

Final velocity v2

50ft/s

Impulse (N·s)

177.9

Average force

0.02kip

Where this answer was checked

source: F·Δt = m·Δv in consistent US units (slug·ft/s² = lbf) — impulse from force and duration
verified by: hand-recomputed
derivation: J = 20·2 = 40 lbf·s = 40 slug·ft/s. Δv = 40/1 = 40 ft/s; v2 = 50 ft/s.