Proof for the Conservation of Linear Momentum: A diagram showing two objects A and B colliding, with labeled masses (m₁, m₂), velocities (v₁, v₂), and forces (F₁₂, F₂₁) before, during, and after the collision.

The image you've uploaded contains a handwritten physics problem asking for the proof of the conservation of linear momentum during a collision between two objects. Let's go through it step-by-step, based on the diagrams and text provided.

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🔍 Problem Summary:

We are given:
- Two objects, A and B, with masses \( m_1 \) and \( m_2 \).
- They are moving along the same straight line with initial velocities \( u_1 \) and \( u_2 \), where \( u_1 > u_2 \).
- The diagram shows:
- Before collision: Object A (mass \( m_1 \)) moving faster than object B (mass \( m_2 \)), so they will collide.
- During collision: Forces act between them — \( F_{AB} \) (force by A on B), and \( F_{BA} \) (force by B on A).
- After collision: They move with new velocities \( v_1 \) and \( v_2 \).

We are to prove the conservation of linear momentum, i.e., show that total momentum before collision = total momentum after collision.

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✔ Step-by-Step Proof

We'll use Newton’s Third Law and Newton’s Second Law to derive the conservation of momentum.

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#### Step 1: Newton’s Second Law
For each object, the rate of change of momentum is equal to the net force acting on it.

Let:
- \( \vec{F}_{AB} \): Force exerted by A on B
- \( \vec{F}_{BA} \): Force exerted by B on A

From Newton’s second law:

For object B:
\[
\frac{d}{dt}(m_2 u_2) = \vec{F}_{AB}
\]

For object A:
\[
\frac{d}{dt}(m_1 u_1) = \vec{F}_{BA}
\]

But from Newton’s Third Law:
\[
\vec{F}_{AB} = -\vec{F}_{BA}
\]

So:
\[
\frac{d}{dt}(m_1 u_1) + \frac{d}{dt}(m_2 u_2) = \vec{F}_{BA} + \vec{F}_{AB} = \vec{F}_{BA} - \vec{F}_{BA} = 0
\]

Thus:
\[
\frac{d}{dt}(m_1 u_1 + m_2 u_2) = 0
\]

This means the total momentum of the system is constant over time.

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#### Step 2: Integrate Over Time

Since the derivative of total momentum is zero, the total momentum does not change:

\[
m_1 u_1 + m_2 u_2 = \text{constant}
\]

So, before collision:
\[
\text{Total momentum} = m_1 u_1 + m_2 u_2
\]

After collision, let their velocities be \( v_1 \) and \( v_2 \). Then:
\[
\text{Total momentum} = m_1 v_1 + m_2 v_2
\]

By conservation:
\[
m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
\]

✔ This proves the conservation of linear momentum in the absence of external forces.

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📌 Key Assumptions:

- No external forces act on the system (e.g., friction, gravity neglected or balanced).
- The only forces are internal (between A and B).
- The collision occurs over a short time interval, but we consider the net effect.

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🧠 Conclusion:

> The total linear momentum of an isolated system remains constant if no external forces act on it.

In this case, since the internal forces between A and B are equal and opposite (Newton’s Third Law), the changes in momentum cancel out, and total momentum is conserved.

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✔ Final Answer:

\[
\boxed{m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2}
\]

This equation represents the conservation of linear momentum during the collision.

Let me know if you'd like to see a numerical example or apply this to elastic/inelastic collisions!