Conservation of Linear Momentum
Introduced| English |
|---|
| conservation/ˌkɒnsəˈveɪʃn/ |
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ |
A cannon kicks back — and the books balance
- Fire a cannon and it lurches backward as the ball flies forward.
- The forward momentum of the ball is matched by the backward momentum of the cannon.
- Before firing, nothing moved: total momentum was zero. After firing, it is still zero.
- Momentum was not created — it was shared out, and the total was conserved 守恒.
The conservation law
- In an isolated system 孤立系统 (no external net force), total momentum stays constant.
- Total momentum before an interaction equals total momentum after.
- $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ (using $u$ for before, $v$ for after).
- It follows directly from Newton's third law: the internal forces cancel in pairs.
Momentum is conserved
Set the masses and speeds, run the collision, and check the total momentum before equals after.
A $1000\ \text{kg}$ cannon fires a $5\ \text{kg}$ ball at $200\ \tfrac{\text{m}}{\text{s}}$. What is the cannon's recoil speed, in $\tfrac{\text{m}}{\text{s}}$? (Total momentum starts at zero.)
$0 = 1000 v + 5(200)$, so $v = -1000/1000 = -1\ \tfrac{\text{m}}{\text{s}}$ — speed $1\ \tfrac{\text{m}}{\text{s}}$ backward.
In an isolated collision, the total momentum after equals the total momentum before.
That is exactly conservation of momentum: $\sum p_{\text{before}} = \sum p_{\text{after}}$.
Why the internal forces cancel
- During a collision, the two objects push on each other equally and oppositely.
- These internal impulses are equal and opposite, so they cancel in the total.
- With no external force, nothing changes the system's total momentum.
- Each object's momentum can change wildly — but their sum does not.
Conservation of momentum follows directly from Newton's ____ law.
The equal-and-opposite internal forces (third law) cancel, leaving the total momentum unchanged.
Solving with conservation
- Write down the total momentum before, set it equal to the total after, and solve.
- Remember to include signs — momentum is a vector.
- Recoil (guns, rockets), explosions, and every collision obey this law.
- It works even when you have no idea about the messy forces during contact.
A $2\ \text{kg}$ trolley at $3\ \tfrac{\text{m}}{\text{s}}$ hits a stationary $1\ \text{kg}$ trolley and they stick. What is their common speed after, in $\tfrac{\text{m}}{\text{s}}$?
Before: $p = 2(3) = 6$. After: $3v = 6$, so $v = 2\ \tfrac{\text{m}}{\text{s}}$.
Select all situations where total momentum is conserved.
The first three have no external net force. Strong external friction adds outside impulse, so momentum is not conserved there.
Momentum is only conserved for an isolated system — one with no external net force. Friction, gravity or a wall can add outside impulse and change the total. Choose your system so the big forces are internal, then conservation applies.
Momentum is conserved only when the system is:
With no external net force, the internal forces cancel and total momentum stays constant.
A $2\ \text{kg}$ trolley at $3\ \tfrac{\text{m}}{\text{s}}$ hits a stationary $1\ \text{kg}$ trolley and they stick together.
- Before: $p = 2(3) + 1(0) = 6\ \text{kg}\cdot\tfrac{\text{m}}{\text{s}}$.
- After: combined mass $3\ \text{kg}$ moves at $v$, so $3v = 6 \Rightarrow v = 2\ \tfrac{\text{m}}{\text{s}}$.
Conservation of momentum: in an isolated system, total momentum is unchanged, so $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$. It comes from Newton's third law (internal forces cancel). Keep track of signs, and it solves recoil, explosions and collisions.