Interactive physics simulator
Conservation of Momentum (P_initial = P_final)
The Law of Conservation of Momentum states that the total momentum of a closed system remains constant if no external forces act on it. Explore elastic bounces, inelastic couplings, and deep space recoil separations in our live interactive physics laboratory.
Momentum Conservation Simulator
Verify that the total momentum (green telemetry) of the system remains constant before, during, and after collisions or recoil pushes.
Live Telemetry
- Cart 1 Mass
- 1.0 kg
- Cart 2 Mass
- 1.0 kg
- Cart 1 Speed
- 12.0 m/s
- Cart 2 Speed
- -8.0 m/s
- Cart 1 Momentum (p1)
- 12.0 kg·m/s
- Cart 2 Momentum (p2)
- -8.0 kg·m/s
- Total Momentum (P)
- 4.0 kg·m/s
What is the Law of Conservation of Momentum?
The Law of Conservation of Momentum states that within a closed, isolated system of interacting bodies, the total linear momentum remains completely constant over time. This means that momentum cannot be created or destroyed, only transferred from one object to another during events like collisions or push separations.
Mathematically, for a system consisting of two colliding bodies, the law is written as:
Expanding the individual momenta terms using mass (m) and velocity (v):
Derivation from Newton's Third Law
Conservation of momentum is a direct consequence of Action-Reaction pairs:
F_12 = -F_21
- During collision, the contact duration Δt is identical for both objects.
- Therefore, their impulse values are equal and opposite: F_12 · Δt = -F_21 · Δt.
- Since Impulse equals change in momentum, this translates to: Δp_2 = -Δp_1.
- Adding them gives Δp_1 + Δp_2 = 0, proving the system's net momentum change is exactly zero.
Elastic vs. Inelastic Collisions
While total momentum is always conserved, kinetic energy behavior varies:
- Elastic Collisions (e = 1.0): Objects bounce perfectly. Both total momentum and total kinetic energy are conserved.
- Inelastic Collisions (e < 1.0): Momentum is conserved, but kinetic energy is lost. It transforms into thermal energy, sound, and structural deformation.
- Completely Inelastic (e = 0.0): The objects stick together post-collision, moving with a single shared velocity: vf = (m1·v1i + m2·v2i) / (m1 + m2).
Deep Space Recoil Mechanics
Recoil occurs when components of a system start at rest and push apart:
- Since initial velocities are zero, initial system momentum is zero: P_initial = 0.
- When Astronaut 1 pushes Astronaut 2, they move in opposite directions: m1·v1f = -m2·v2f.
- This ensures that the final total momentum remains exactly zero: P_final = 0.
- Heavy components move slower, while light components recoil faster: v2f = -v1f · (m1 / m2).
Graphical Proof of Conservation
Using real-time telemetry plots visually demonstrates the law:
- Individual Momentum vs. Time: Shows two curves (red and blue). At the moment of impact or push, their lines undergo sharp, equal, and opposite shifts.
- Total Momentum vs. Time: Sums the two individual curves. It forms a perfectly flat, horizontal line (green) showing zero change, proving conservation.
- Even under partial elasticity, the green summation line remains completely flat.
Solved Examples
A red cart with a mass of 2.0 kg moving East at +6.0 m/s collides elastically (e = 1.0) with a blue cart of mass 1.0 kg moving West at -3.0 m/s. Find the final velocity of each cart after the collision.
- First, calculate the total initial momentum of the system: P_initial = m1 * v1i + m2 * v2i = (2.0 kg * 6.0 m/s) + (1.0 kg * -3.0 m/s) = 12.0 - 3.0 = +9.0 kg·m/s.
- Since the collision is perfectly elastic (e = 1.0), both momentum and kinetic energy are conserved.
- Using the restitution equations: v1f = [m1*v1i + m2*v2i + m2*e*(v2i - v1i)] / (m1 + m2).
- Substitute the values: v1f = [9.0 + 1.0 * 1.0 * (-3.0 - 6.0)] / (2.0 + 1.0) = [9.0 - 9.0] / 3.0 = 0.0 m/s.
- Solve for Cart 2: v2f = [m1*v1i + m2*v2i + m1*e*(v1i - v2i)] / (m1 + m2).
- Substitute the values: v2f = [9.0 + 2.0 * 1.0 * (6.0 - (-3.0))] / (2.0 + 1.0) = [9.0 + 18.0] / 3.0 = +9.0 m/s.
- The final velocity of Cart 1 is 0.0 m/s (stops) and Cart 2 is +9.0 m/s (moving East).
Answer: Cart 1: 0.0 m/s | Cart 2: +9.0 m/s (East)
A 3.0 kg bumper cart traveling at +4.0 m/s collides with a 2.0 kg bumper cart moving in the opposite direction at -1.0 m/s. If the collision is completely inelastic (e = 0.0) and they stick together, find their combined velocity.
- Calculate the initial momentum of each cart: p1i = 3.0 kg * 4.0 m/s = +12.0 kg·m/s. p2i = 2.0 kg * -1.0 m/s = -2.0 kg·m/s.
- Total initial momentum: P_initial = 12.0 - 2.0 = +10.0 kg·m/s.
- In a completely inelastic collision (e = 0.0), the carts stick together and move with a single final velocity (vf).
- By the Law of Conservation of Momentum: P_initial = P_final = (m1 + m2) * vf.
- Solve for vf: vf = P_initial / (m1 + m2) = 10.0 kg·m/s / (3.0 kg + 2.0 kg) = 10.0 / 5.0 = +2.0 m/s.
- The stuck carts move together at +2.0 m/s East.
Answer: Combined Velocity: +2.0 m/s (East)
A 60 kg astronaut at rest in space pushes off from a 120 kg satellite (also at rest). If the astronaut recoils in the negative direction at a velocity of -2.0 m/s, find the recoil velocity of the satellite.
- Since both objects start at rest, the total initial momentum of the system is zero: P_initial = 0 kg·m/s.
- By the Law of Conservation of Momentum: P_final = m_ast * v_ast_f + m_sat * v_sat_f = 0.
- Rearrange the equation to solve for the satellite's velocity: v_sat_f = - (m_ast * v_ast_f) / m_sat.
- Substitute values: v_sat_f = - (60 kg * -2.0 m/s) / 120 kg = - (-120 kg·m/s) / 120 kg = +1.0 m/s.
- The satellite recoils at +1.0 m/s in the opposite direction.
Answer: Satellite Velocity: +1.0 m/s (East)
Common Misconceptions
- "Momentum is only conserved in elastic collisions": False. Momentum is conserved in all collisions, whether elastic, partially elastic, or completely inelastic. Only kinetic energy is conserved strictly in elastic collisions.
- "Recoil creates momentum out of nothing": False. The total momentum before recoil is zero, and it remains zero afterward. The forward momentum of one object is balanced by the backward momentum of the other.
- "External forces like friction don't affect conservation": False. Momentum is only conserved in isolated systems. External forces like friction or gravity change the system\'s total momentum.
Quick Summary
- The Law of Conservation of Momentum states: P_initial = P_final.
- It applies strictly to closed, isolated systems (no external net forces).
- Derived directly from Newton\'s Third Law: Δp1 = -Δp2.
- In elastic collisions, kinetic energy is also conserved. In inelastic, it is lost.
- Recoil velocities are inversely proportional to masses: v2f = -v1f · (m1 / m2).
- The system\'s Total Momentum graph is a completely flat horizontal line.
Practice Questions
1. What is the Law of Conservation of Momentum, and under what conditions does it hold true?
The Law of Conservation of Momentum states that the total momentum of a closed system remains constant over time. It holds true under two conditions: 1) the system is closed (no mass enters or leaves) and 2) the system is isolated (the net external force acting on the system is zero).
2. How does Newton's Third Law of Motion explain the conservation of momentum in a collision?
During a collision, the force exerted by Cart 1 on Cart 2 (F12) is equal in magnitude and opposite in direction to the force exerted by Cart 2 on Cart 1 (F21 = -F12). Since the contact time (Δt) is the same for both, the impulses are equal and opposite: F12·Δt = -F21·Δt. Since impulse equals change in momentum, Δp2 = -Δp1, which means Δp1 + Δp2 = 0. The total change in momentum is zero, so momentum is conserved.
3. Why does a rocket accelerate in the vacuum of space where there is no air to push against?
A rocket accelerates due to the conservation of momentum. When the rocket fuel burns, gas is expelled backward at high velocity, gaining massive backward momentum. To conserve total momentum, the rocket body must gain an equal and opposite forward momentum. Thus, the rocket pushes against its own exhaust, not the surrounding air.
4. If momentum is always conserved in collisions, why does a bouncing ball eventually come to a stop on the ground?
Momentum is only conserved in a closed, isolated system. A ball bouncing on the ground is not isolated because it interacts with the Earth (external gravitational force) and air resistance. When it hits the floor, it transfers momentum to the entire Earth, whose massive inertia makes the resulting change in the Earth's speed completely undetectable.
FAQ
Frequently Asked Questions
What is the Law of Conservation of Momentum?
It is a fundamental physics principle stating that the total linear momentum of a closed, isolated system remains constant in magnitude and direction, regardless of any internal interactions or collisions.
What is a closed, isolated system in physics?
A closed system has a constant mass (no matter enters or leaves). An isolated system has a net external force of zero. Both conditions are required for momentum to be conserved.
Is momentum conserved in everyday car crashes?
Yes. During the brief moments of a crash, the internal impact forces between the vehicles are vastly larger than external forces like friction. Therefore, the total momentum immediately before and after the collision is highly conserved.
What is the difference between elastic and inelastic collisions?
In an elastic collision, both total momentum and total kinetic energy are conserved. In an inelastic collision, momentum is conserved, but kinetic energy is lost (turned into heat, sound, or physical damage).
Can a system have zero total momentum but contain moving parts?
Yes. If two objects push off each other from rest in opposite directions, one has positive momentum and the other has equal negative momentum. The sum of their momenta remains exactly zero.
How does the coefficient of restitution (e) affect collisions?
The coefficient of restitution (e) measures elasticity. e = 1.0 means perfectly elastic (full bounce), e = 0.0 means completely inelastic (stick together), and values in between represent realistic partially elastic bounces.
Why does a cannon jump backward when firing a shell?
Initially, the cannon and shell are at rest (momentum = 0). Firing expels the shell forward with positive momentum. To conserve momentum, the heavy cannon must recoil backward with equal and opposite negative momentum.
Does gravity break the law of conservation of momentum?
No. Gravity is an external force, so it changes a falling object's momentum. However, if you expand the system to include both the object and the Earth, their mutual gravitational pull is an internal force, and total momentum remains conserved.
How is momentum conserved in a rocket launch?
The rocket expels hot combustion gases downward at high speeds. To balance this downward momentum, the rocket body is propelled upward with equal and opposite momentum.
What does the flat line on the Total Momentum graph represent?
The flat line represents the system's total momentum. It remains completely horizontal and constant, showing that while individual cart momenta fluctuate during impacts, the sum is perfectly conserved.
How do I run a recoil experiment in the simulator?
Switch the simulator mode to Space Recoil. Set the astronaut masses, push force, and push duration. Click "Play / Launch" to watch them recoil in opposite directions while total momentum remains exactly zero.