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Momentum, Impulse, and Conservation of Momentum

CSCA Momentum, Impulse, and Conservation of Momentum study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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International students preparing for CSCA Math, Physics, Chemistry, or Chinese exams.

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Momentum, Impulse, and Conservation of Momentum

1. Momentum & Impulse

#### A. Momentum

* **Definition**: A measure of the "inertia" of motion.

* **Formula**: $$\vec{p} = m\vec{v}$$

* **Nature**: **State Quantity**, **Vector** (same direction as $\vec{v}$).

#### B. Impulse

* **Definition**: The cumulative effect of force over time.

* **Formula**: $$\vec{I} = \vec{F} \cdot \Delta t$$ (Constant Force)

* **Nature**: **Process Quantity**, **Vector** (same direction as $\vec{F}$).

* **Graphical Meaning**: The **Area** under the Force-Time ($F-t$) graph represents Impulse.

2. Impulse-Momentum Theorem

The core tool for collision/impact problems.

* **Statement**: The impulse of the net force equals the change in momentum.

* **Formula**:

$$\vec{I}_{\text{net}} = \Delta \vec{p} = \vec{p}_2 - \vec{p}_1 = m\vec{v}_2 - m\vec{v}_1$$

**⚠️ CSCA Exam Trap (Vector Nature)**:

This is a vector equation. You MUST define a **positive direction**, especially for "rebound" problems.

Example: A ball hits a wall at $v$ and rebounds at $v$. The change is NOT 0. It is $\Delta p = m(v) - m(-v) = 2mv$ (assuming rebound is positive).

3. Conservation of Momentum

#### A. Conditions for Conservation

1. **No external forces** (Ideal).

2. **Net external force is zero** (Common, e.g., smooth surface).

3. **Internal forces >> External forces** (Approximate, e.g., explosions, collisions).

4. **Net force is zero in one direction** (Momentum is conserved in that direction).

#### B. Collision Models

* **Elastic Collision**: Momentum conserved + Kinetic Energy conserved.

* **Inelastic Collision**: Momentum conserved + Kinetic Energy lost.

* **Perfectly Inelastic**: Momentum conserved + Max Kinetic Energy lost (stick together, $v_1 = v_2$).

4. Classic Model: The Man-Boat Model (Recoil)

A standard "Recoil" problem in Chinese curricula, often appearing in CSCA.

**Scenario**: A boat (mass $M$) floats on still water. A person (mass $m$) walks from the bow to the stern (relative distance $L$). Ignore water resistance.

**Conclusion**: The center of mass of the system remains stationary. Displacement is inversely proportional to mass.

* **Displacement Relationship**:

$$m x_{\text{man}} = M x_{\text{boat}}$$

$$x_{\text{man}} + x_{\text{boat}} = L$$

* **Shortcut Formula (Boat's Displacement)**:

$$x_{\text{boat}} = \frac{m}{M+m} L$$

(Mnemonic: To find *this* object's displacement, put the *other* object's mass in the numerator)

5. Typical Examples

**Example 1: Rebound (Impulse Theorem)**

**Problem**: A $0.5\text{kg}$ ball hits a wall at $10\text{m/s}$ and rebounds at $8\text{m/s}$. Contact time $0.1\text{s}$. Find the average force.

**Solution**:

Define **Rebound Direction** as positive.

$v_1 = -10\text{m/s}, \quad v_2 = +8\text{m/s}$

$\Delta p = m v_2 - m v_1 = 0.5 \times [8 - (-10)] = 0.5 \times 18 = 9 \text{kg} \cdot \text{m/s}$

Using $F \Delta t = \Delta p$:

$F = \frac{9}{0.1} = 90\text{N}$, directed away from the wall.

**Example 2: Man-Boat Application**

**Problem**: A $40\text{kg}$ person stands on a $60\text{kg}$, $5\text{m}$ long boat. The person walks to the other end. How far does the boat move?

**Solution**: Use the shortcut formula.

$$x_{\text{boat}} = \frac{m_{\text{person}}}{M_{\text{boat}} + m_{\text{person}}} L = \frac{40}{60+40} \times 5 = \frac{40}{100} \times 5 = 2\text{m}$$