Conservation of Linear Momentum
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| conservation/ˌkɒnsəˈveɪʃn/ | bảo toàn |
| isolated system/ˈaɪsəleɪtɪd ˈsɪstəm/ | hệ cô lập |
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.)
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$0 = 1000 v + 5(200)$, 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.
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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.
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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 $\tfrac{\text{m}}{\text{s}}$?
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Before: $p = 2(3) = 6$. After: $3v = 6$, so $v = 2\ \tfrac{\text{m}}{\text{s}}$.
Select all situations where total momentum is conserved.
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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:
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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.
Write from a common speed of 2.5 m/s
- A 4 kg cart at 5 m/s hits a 4 kg cart at rest and they stick.
- Find the common speed.
A 4 kg cart at 5 m/s hits a stationary 4 kg cart and they stick together. The system is isolated. What is their common speed?
4 times 5 is 20, and 20 divided by 8 is 2.5.
What is the common speed?
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Momentum 20 shared by 8 kg is 2.5 m/s. Other options — 2 m/s (the lesson sticking speed); 5 m/s (the incoming speed unchanged).