Skip to content
CivilSolve

Dynamics

Impulse–Momentum & 1-D Collisions

Solve force-time impulse problems or direct 1D collisions. Find velocities, impulse, and kinetic energy loss with conservation equations.

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

Inputs

Load a sample problem:

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

  1. 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

  2. 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

  1. 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

  2. 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

  1. 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.