Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is

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Q. Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.

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Concept understanding — Conservation of Momentum

Conservation of Momentum: From Push to Principle

Imagine you're standing on perfectly smooth ice, wearing skates. You're completely still. Now, you push a heavy medicine ball away from you. What happens? You roll backward. The harder you push the ball, the faster you roll back.

That's the core intuition: you can't push something away without being pushed back yourself. The push you give the ball is matched by an equal push on you, in the opposite direction. This isn't a special property of ice or skates — it's a fundamental rule of how forces work in the universe.

The Hidden Quantity That Never Changes

Physicists call the "amount of motion" an object has its momentum. For everyday speeds, momentum is simple:

\vec{p} = m \vec{v}

Where $m$ is mass (how much stuff) and $\vec{v}$ is velocity (speed with direction). Momentum is a vector — it cares about which way you're going.

A truck creeping forward has huge momentum (big mass, small speed). A bullet zipping through air has moderate momentum (tiny mass, huge speed). A parked car has zero momentum (speed is zero).

Now here's the key: in any isolated system (no outside forces), total momentum stays the same. Always. Before, during, and after any interaction.

The Precise Statement

Important

Law of Conservation of Momentum:

In a closed, isolated system (no external forces), the total vector momentum of the system remains constant over time.

Mathematically, for two objects that interact (collide, push apart, explode):

\vec{p}_{1,\text{initial}} + \vec{p}_{2,\text{initial}} = \vec{p}_{1,\text{final}} + \vec{p}_{2,\text{final}}

Or in terms of masses and velocities:

m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2

Where $\vec{u}$ means initial velocity and $\vec{v}$ means final velocity.

Why This Works: Newton's Third Law in Disguise

When you push the medicine ball, your hand exerts a force $\vec{F}$ on the ball. By Newton's Third Law, the ball exerts an equal and opposite force $-\vec{F}$ back on your hand. These forces act for the same time $\Delta t$ .

Force times time equals impulse, which equals change in momentum:

\vec{F} \Delta t = \Delta \vec{p}

For you and the ball:

Ball's momentum change: $+\vec{F} \Delta t$ (ball goes forward)
Your momentum change: $-\vec{F} \Delta t$ (you go backward)

Add them: $+\vec{F} \Delta t + (-\vec{F} \Delta t) = 0$

Total change is zero. Momentum is conserved because forces always come in equal-and-opposite pairs.

Note

This is why a rocket works in the vacuum of space. It throws exhaust backward (one momentum change), and the rocket itself moves forward (equal opposite momentum change). No air needed — just Newton's Third Law and conservation of momentum.

What This Law Does NOT Mean

It does NOT mean individual objects keep constant momentum. Only the total of all objects in the system stays constant. Individual momenta can change wildly.
It does NOT apply if external forces act. If friction, gravity from outside, or a wall stops something, momentum is not conserved for that system. (You can expand the system to include the Earth or the wall, and then momentum is conserved again.)
It does NOT require collisions to be elastic. Even in a messy, sticky, energy-losing collision, momentum is still perfectly conserved. Energy can be lost to heat or deformation, but momentum never disappears.

A Quick Example

A 2 kg cart moving at 3 m/s right collides with a stationary 1 kg cart. After collision, they stick together. Find their speed.

Before:

Cart 1: $p_1 = (2)(3) = 6\ \text{kg m/s}$ right

Cart 2: $p_2 = (1)(0) = 0$

Total: $6\ \text{kg m/s}$ right

After:

Combined mass: $2 + 1 = 3\ \text{kg}$

Let $v$ be their common velocity.

Total momentum: $3v$

Conservation: $3v = 6 \implies v = 2\ \text{m/s}$ right

The carts slow down because mass increased, but total momentum stayed the same.

The Big Picture

Conservation of momentum is one of the most reliable laws in physics. It holds true from subatomic particles colliding in accelerators to galaxies merging in space. It's a symmetry of the universe — a consequence of the fact that the laws of physics are the same everywhere (Noether's theorem, if you ever study deeper).

For now, remember the ice-skater pushing the ball. That feeling of being pushed back — that's conservation of momentum, live and in person.

Many students search for "Conservation of Momentum class 11 physics" or "Conservation of Momentum: Definition, Formula & Real-World Examples" while revising for boards, and Conservation of Momentum is drawn directly from the Laws of Motion / System of Particles and Rotational Motion coverage of the NCERT/CBSE Class 11 Physics syllabus and recurs often in JEE Main and NEET papers. Working through the worked examples above alongside the official NCERT Physics textbook is the most reliable way to turn this understanding into exam-ready recall.

Step 1. Consider two particles that interact only with each other (an isolated system, no external forces). By Newton's third law, the force particle 1 exerts on particle 2, $\vec F_{21}$ , and the force particle 2 exerts on particle 1, $\vec F_{12}$ , satisfy $\vec F_{12}=-\vec F_{21}$ .

Step 2. By Newton's second law, $\vec F_{12}=\dfrac{d\vec p_1}{dt}$ and $\vec F_{21}=\dfrac{d\vec p_2}{dt}$ .

Step 3. Substituting into the third-law relation: $\dfrac{d\vec p_1}{dt}=-\dfrac{d\vec p_2}{dt}$ , i.e. $\dfrac{d}{dt}(\vec p_1+\vec p_2)=0$ .

Step 4. So $\vec p_1+\vec p_2=\vec p_{tot}=\text{constant vector}$ — the total linear momentum of an isolated system is conserved, even though $\vec p_1$ and $\vec p_2$ individually may change (as long as their sum stays fixed).

Step 5. Application — recoil of a gun: before firing, gun+bullet are both at rest, so $\vec p_{tot}=0$ . After firing, the bullet has momentum $\vec p_1'=m_{bullet}\vec v_{bullet}$ (forward) and the gun has momentum $\vec p_2'=m_{gun}\vec v_{recoil}$ . Conservation demands $\vec p_1'+\vec p_2'=0$ , so $m_{gun}\vec v_{recoil}=-m_{bullet}\vec v_{bullet}$ , giving $\vec v_{recoil}=-\dfrac{m_{bullet}}{m_{gun}}\vec v_{bullet}$ — opposite to the bullet, and much smaller in magnitude since $m_{gun}\gg m_{bullet}$ .

✓Final answer

Total momentum of an isolated system is conserved because internal (third-law) forces cannot change it; for a gun of mass M firing a bullet of mass m at speed v from rest, the recoil speed is V = mv/M, opposite to the bullet.

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Q.Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.

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