Conservation of Linear Momentum
Introduced| English |
|---|
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ |
| conservation of linear momentum/ˌkɒnsəˈveɪʃn ɒv ˈlɪnɪə məʊˈmentəm/ |
| internal forces/ɪnˈtɜːnl ˈfɔːsɪz/ |
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.
Momentum is conserved only when...
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.
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.
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)?
$0 = 5(8) + 50 v$, 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)?
$0 = 1(6) + 3v$, 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.
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.