Momentum and Impulse: Collisions and Explosions in One Dimension

1) Linear Momentum and Impulse

Linear momentum

In one dimension (motion along a line), the linear momentum of an object is defined as

p = m v

where m is mass and v is velocity along the chosen axis. Momentum is a signed quantity in 1D: if you choose +x to the right, then an object moving left has negative velocity and therefore negative momentum.

Impulse

Collisions and explosions involve large forces acting over short times. The quantity that connects force and time is impulse, defined (in 1D) by

J = ∫ F(t) dt

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If the force is roughly constant over the interaction time interval Δt, then

J ≈ F_avg Δt

The key relationship is the impulse–momentum theorem:

J = Δp = p_f − p_i

This is extremely practical: if you can estimate the average force and the interaction time, you can predict the change in momentum (and therefore the change in velocity).

Why “bigger force” does not automatically mean “bigger impulse”

Impulse depends on both force and time. A very large force acting for a very short time can produce the same impulse as a smaller force acting longer.

• Example: F = 1000 N for 0.01 s gives J = 10 N·s.
• Example: F = 100 N for 0.10 s also gives J = 10 N·s.

Both cause the same momentum change Δp = 10 kg·m/s.

2) Momentum Conservation: When and How It Works

The conservation statement

For a chosen system, the total momentum changes according to the net external impulse:

Δp_system = J_external

If the net external impulse is approximately zero during the interaction, then

p_system,f ≈ p_system,i

This is the practical criterion for momentum conservation in collisions/explosions:

• Choose a system (often the two colliding objects together).
• Check external forces during the short interaction time.
• If external forces exist but act over a very short time, their impulse can be negligible compared with the internal collision forces.

System selection (what to include)

Momentum conservation is easiest when you include all objects that exert large forces on each other during the event.

• Two carts colliding: choose “cart A + cart B” as the system. The collision forces are internal and cancel in pairs (action–reaction), so they do not change total system momentum.
• Recoil of a gun: choose “gun + bullet” (and sometimes “gun + bullet + gases” for better realism). If you choose only the bullet, the gun’s force on it is external to that system, so the bullet’s momentum is not conserved by itself.

When external impulse is not negligible

If the interaction lasts long enough (or external forces are large), you must include external impulse:

p_f = p_i + J_external

Example situations: a car collision with significant braking during impact, or a block colliding while attached to a spring that pulls strongly during the collision.

3) Collisions in One Dimension

Types of collisions (momentum always, energy sometimes)

In an isolated system (net external impulse ≈ 0), momentum is conserved in all collisions. What changes between collision types is what happens to kinetic energy.

• Elastic collision: momentum conserved and kinetic energy conserved.
• Inelastic collision: momentum conserved but kinetic energy decreases (some becomes thermal energy, sound, deformation).
• Perfectly inelastic collision: the objects stick together and move with a common final velocity; kinetic energy loss is typically maximal for given masses and initial velocities.

Coefficient of restitution (conceptual)

The coefficient of restitution e describes how “bouncy” a collision is in 1D by comparing relative speeds along the line of impact:

e = (speed of separation) / (speed of approach)

With sign conventions handled carefully, a common 1D form is

e = (v2f − v1f) / (v1i − v2i)

where object 1 approaches object 2 initially. Conceptually:

• e = 1 elastic (no kinetic energy loss).
• 0 < e < 1 inelastic (some kinetic energy loss).
• e = 0 perfectly inelastic (they leave with the same velocity).

Perfectly inelastic collision formula (stick together)

If two objects stick, they share a final velocity v_f. Momentum conservation gives

m1 v1i + m2 v2i = (m1 + m2) v_f

so

v_f = (m1 v1i + m2 v2i) / (m1 + m2)

Energy loss interpretation (without assuming energy conservation)

In inelastic collisions, kinetic energy is not conserved because internal forces do work that changes internal energy (deformation, heat). Momentum conservation still holds (if external impulse ≈ 0) because internal forces come in equal-and-opposite pairs and do not change total momentum.

A common misconception is: “If kinetic energy is not conserved, momentum is not conserved.” In fact, momentum conservation depends on external impulse, not on whether kinetic energy is conserved.

4) Step-by-Step Problems (with Sign Conventions and Checks)

Problem A: Two carts collide and stick (perfectly inelastic)

Situation. Cart A (m1 = 0.50 kg) moves right at v1i = +2.0 m/s. Cart B (m2 = 1.0 kg) moves left at v2i = −1.0 m/s. They collide and stick. Find the final velocity.

Step 1: Choose sign convention. Take right as positive. (Already built into the given velocities.)

Step 2: Choose system and check criterion. System = both carts. During the short collision, external impulse from track friction is negligible, so momentum is conserved.

Step 3: Apply momentum conservation.

m1 v1i + m2 v2i = (m1 + m2) v_f

(0.50)(+2.0) + (1.0)(−1.0) = (1.50) v_f

1.0 − 1.0 = 1.50 v_f

0 = 1.50 v_f

v_f = 0 m/s

Step 4: Check reasonableness. Initial total momentum is zero, so final total momentum must be zero. Sticking implies a shared velocity; the only shared velocity giving zero momentum is 0. Also note: kinetic energy decreases (they stick), but momentum can still be conserved.

Problem B: Recoil (gun and bullet) in 1D

Situation. A 3.0 kg launcher fires a 0.020 kg projectile. Initially, both are at rest on a frictionless track. The projectile leaves to the right at +300 m/s (relative to the ground). Find the launcher’s recoil velocity.

Step 1: Choose system. System = launcher + projectile. External horizontal impulse ≈ 0 (frictionless track), so momentum is conserved.

Step 2: Write initial momentum. Initially at rest:

p_i = 0

Step 3: Write final momentum and solve.

m_L v_Lf + m_p v_pf = 0

3.0 v_Lf + 0.020(300) = 0

3.0 v_Lf + 6.0 = 0

v_Lf = −2.0 m/s

Interpretation. The negative sign means recoil to the left. The launcher’s speed is much smaller than the projectile’s because its mass is much larger.

Limit check. If m_L were extremely large, v_Lf would approach 0, matching intuition.

Problem C: “Explosion” separation from rest (two carts push apart)

Situation. Two carts are initially at rest, touching, on a frictionless track. A spring mechanism releases, pushing them apart. Cart 1 has m1 = 0.40 kg, cart 2 has m2 = 0.60 kg. After release, cart 1 moves right at v1f = +3.0 m/s. Find v2f.

Step 1: System and criterion. System = both carts. External impulse ≈ 0, so total momentum is conserved.

Step 2: Initial momentum. Initially at rest:

p_i = 0

Step 3: Apply momentum conservation.

m1 v1f + m2 v2f = 0

0.40(+3.0) + 0.60 v2f = 0

1.2 + 0.60 v2f = 0

v2f = −2.0 m/s

Step 4: Check with mass ratio. From m1 v1f = −m2 v2f, speeds satisfy |v2f| = (m1/m2)|v1f| = (0.40/0.60)·3.0 = 2.0. The heavier cart moves slower, which fits intuition.

Problem D: Impulse from a force-time interaction (average force matters)

Situation. A ball of mass 0.20 kg moves left at v_i = −10 m/s. It hits a wall and rebounds right at v_f = +6 m/s. The contact time is Δt = 0.020 s. Find (a) the impulse on the ball and (b) the average force on the ball.

Step 1: Compute momentum change.

p_i = m v_i = 0.20(−10) = −2.0 kg·m/s

p_f = m v_f = 0.20(+6) = +1.2 kg·m/s

Δp = p_f − p_i = 1.2 − (−2.0) = 3.2 N·s

So (a) impulse: J = +3.2 N·s (to the right).

Step 2: Use average force.

F_avg = J / Δt = 3.2 / 0.020 = 160 N

So (b) average force: +160 N (to the right).

Misconception check. A “harder hit” can mean larger average force, longer contact time, or both. Without Δt, you cannot determine impulse from force magnitude alone.

Common Misconceptions and How to Avoid Them

Misconception 1: “Momentum is conserved only if kinetic energy is conserved”

Momentum conservation depends on external impulse, not on whether kinetic energy changes. In a perfectly inelastic collision, kinetic energy decreases, but momentum can still be conserved because the net external impulse is approximately zero.

Misconception 2: “The object that feels the larger force gets the larger impulse”

In an interaction between two objects, the forces are equal in magnitude and opposite in direction at each instant (action–reaction). Therefore, the impulses they exert on each other are equal in magnitude and opposite in direction (assuming the same interaction time). What differs is the resulting change in velocity, because Δv = Δp / m.

Misconception 3: “If something stops, its momentum is destroyed”

Momentum is not “destroyed”; it is transferred to other parts of the system (other objects, Earth, etc.). If you choose a system that does not include the object receiving the momentum, it will look like momentum is not conserved—because external impulse is acting on your chosen system.