Q.A girl riding a bicycle along a straight road with a speed of $5$ m s$^{-1}$ throws a stone of mass 0.5 kg which has a speed of $15$ m s$^{-1}$ with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

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:

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

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

Or in terms of masses and velocities:

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:

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.

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 …