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

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

(b) State and prove parallel axes theorem.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2019Subjective· 5mImportance★★★★★

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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 …

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The law of conservation of momentum follows from Newton's third law; applied to a gun firing a bullet (starting from rest), it gives recoil velocity V = -mv/M.

This question offers an OR alternative (parallel axes theorem); this solution answers the primary part (a) as instructed.

PROOF OF THE LAW OF CONSERVATION OF MOMENTUM:

Consider two bodies A and B of masses m1 and m2, moving with initial velocities u1 and u2 respectively, which collide (or interact) for a short time t and then move apart with final velocities v1 and v2.

During the collision, by Newton's Third Law, the force F_AB that A exerts on B is equal and opposite to the force F_BA that B exerts on A at every instant:

F_AB = - F_BA

By Newton's Second Law, force equals rate of change of momentum, so for body B:

F_AB = m2(v2 - u2)/t

and for body A:

F_BA = m1(v1 - u1)/t

Substituting into F_AB = -F_BA:

m2(v2 - u2)/t = - m1(v1 - u1)/t

Cancelling t and rearranging:

m2 v2 - m2 u2 = - m1 v1 + m1 u1

m1 v1 + m2 v2 = m1 u1 + m2 u2

This says: total momentum AFTER the interaction equals total momentum BEFORE the interaction, provided no external force acts on the system. This is the law of conservation of momentum: the total linear momentum of an isolated system remains constant.

APPLICATION -- Recoil velocity of a gun:

Before firing, both the gun (mass M) and the bullet (mass m) are at rest, so the total initial momentum of the (gun + bullet) system is zero.

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

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