Question

Object A and Object B are both in motion when they collide with each other. They then continue in a new direction unaffected by any other forces.

Answer 1

What you described is an isolated two-body collision. The fundamental law that governs what happens is conservation of linear momentum (vector form). Whether kinetic energy is conserved depends on the type of collision:

- If no external forces act on the pair during the collision, total linear momentum is conserved:
m1 v1_i + m2 v2_i = m1 v1_f + m2 v2_f
(vectors if the motion is in 2 or 3 dimensions).

- If the collision is perfectly elastic, kinetic energy is also conserved:
1/2 m1 |v1_i|^2 + 1/2 m2 |v2_i|^2 = 1/2 m1 |v1_f|^2 + 1/2 m2 |v2_f|^2.

- If the collision is partially inelastic, kinetic energy is not conserved; a convenient parameter is the coefficient of restitution e (0 ≤ e ≤ 1), which measures the ratio of relative speeds along the line of impact:
(relative separation speed) = e × (relative approach speed).
In 1D: v2_f − v1_f = e (v1_i − v2_i).

Common useful results
- Perfectly inelastic (they stick together): final common velocity
v_f = (m1 v1_i + m2 v2_i) / (m1 + m2).

- 1D elastic collision (e = 1) — closed-form final velocities:
v1_f = [(m1 − m2) v1_i + 2 m2 v2_i] / (m1 + m2)
v2_f = [2 m1 v1_i + (m2 − m1) v2_i] / (m1 + m2).

- 1D with restitution e (general):
v1_f = [ (m1 − m2 e) v1_i + m2 (1 + e) v2_i ] / (m1 + m2)
v2_f = [ m1 (1 + e) v1_i + (m2 − m1 e) v2_i ] / (m1 + m2).

2D (or 3D) collisions
- Treat momentum as a vector (conserve both x- and y-components).
- If collision is central (line-of-centers), 1D formulas apply along that line.
- For off-center collisions, decompose each velocity into components normal and tangential to the contact plane:
- Tangential components are unchanged if there is no friction or tangential impulse.
- Normal components are changed according to momentum conservation and restitution (or energy conservation if elastic).
- Working in the center-of-mass frame often simplifies elastic-collision analysis: total momentum is zero and in an elastic collision each particle’s speed relative to the CM is unchanged (only directions change).

Other notes
- If bodies can rotate or the impact is off-center, angular momentum and torques matter and some translational kinetic energy can convert to rotational energy.
- If you give masses, initial velocities, and whether the collision is elastic/inelastic (or a value for e), I can compute the final velocities for you (1D or 2D).