Conservation of Linear Momentum
Introduced| English | Meaning |
|---|---|
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ | A system that exchanges neither matter nor energy with its surroundings in thermodynamics, or experiences no net external influence in the relevant mechanical model. |
| conservation of linear momentum/ˌkɒnsəˈveɪʃn ɒv ˈlɪnɪə məʊˈmentəm/ | conservation of linear momentum |
| internal forces/ɪnˈtɜːnl ˈfɔːsɪz/ | internal forces |
Two skaters push apart
- Two skaters stand still on ice, then shove off each other.
- They glide away in opposite directions — with no outside help at all.
- Somehow their motions stay perfectly balanced.
- Behind it is one of the most powerful rules in physics.
Conservation of momentum
- If the net external force on a system is zero — an isolated system 孤立系统 — its total momentum stays constant.
- Whatever total momentum it starts with, it keeps.
- This is the conservation of linear momentum 动量守恒.
A head-on collision: total momentum before equals total momentum after
Momentum is conserved only when...
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An isolated system -- zero net external force -- conserves momentum. Internal forces are fine.
Internal forces cannot change it
- Internal forces 内力 come in third-law pairs that cancel across the system.
- So a push between the parts (like the skaters' shove) can never change the total.
- Only an external force can.
An internal explosion inside a floating system changes the system's total momentum.
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Explosion forces are internal (third-law pairs), so the total momentum is unchanged.
Forces between the parts of a system are called ____ forces and cannot change the total momentum.
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Internal forces come in canceling third-law pairs, so they leave the total momentum unchanged.
Collisions and explosions
- The law applies to collisions (objects hitting) and explosions (one thing flying apart) alike.
- In an explosion, the pieces fly off so their momenta still sum to the original total.
- Pick your system so the forces you care about are internal.
Conservation of momentum
In a collision with no outside forces, the total momentum before equals the total momentum after.
A $50\ \text{kg}$ skater at rest throws a $5\ \text{kg}$ ball at $8\ \tfrac{\text{m}}{\text{s}}$. The skater's recoil speed (in m/s)?
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$0 = 5(8) + 50 v$, so $v = -40/50 = -0.8\ \tfrac{\text{m}}{\text{s}}$ -- the skater recoils at $0.8\ \tfrac{\text{m}}{\text{s}}$.
A $4\ \text{kg}$ object at rest explodes into a $1\ \text{kg}$ piece moving right at $6\ \tfrac{\text{m}}{\text{s}}$ and a $3\ \text{kg}$ piece. The $3\ \text{kg}$ piece's speed (in m/s)?
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$0 = 1(6) + 3v$, so $v = -2\ \tfrac{\text{m}}{\text{s}}$ -- the $3\ \text{kg}$ piece moves left at $2\ \tfrac{\text{m}}{\text{s}}$.
Two dimensions
- Momentum is a vector, so it is conserved separately along each axis.
- Conserve the $x$-momenta and the $y$-momenta independently.
- Two equations let you solve two-dimensional collisions.
In a two-dimensional collision, momentum is conserved separately along each axis.
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Momentum is a vector, so its x- and y-components are each conserved.
A $60\ \text{kg}$ astronaut at rest throws a $2\ \text{kg}$ wrench at $10\ \tfrac{\text{m}}{\text{s}}$.
- Total momentum starts at zero, so it must stay zero: $0 = (2)(10) + (60)v$.
- The astronaut recoils at $v = -20/60 \approx -0.33\ \tfrac{\text{m}}{\text{s}}$ backward.
Momentum is conserved only when the net external force is zero. If a strong outside force acts (a wall, the ground, a rocket engine's exhaust leaving), you must either include it or choose a bigger system that makes it internal.
In an isolated system (zero net external force), total momentum is conserved — conservation of linear momentum. Internal forces cancel in pairs and cannot change the total; only external forces can. It holds for collisions and explosions, and in 2D you conserve each axis separately.
Write from a recoil speed of 2 m/s
- A 5 kg person on frictionless ice throws a 1 kg ball at 10 m/s.
- Find the person's recoil speed.
A 5 kg person on frictionless ice throws a 1 kg ball forward at 10 m/s. The person and ball start at rest. What is the person's recoil speed?
5 times 2 equals 10, which matches the ball's momentum.
What is the person's recoil speed?
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The person's momentum of 10 balances the ball, so the speed is 2 m/s. Other options — 0.8 m/s (the lesson recoil); 10 m/s (the same as the ball).