Q.State Newton's second law of motion in terms of the rate of change of momentum, and show how the familiar equation $F = ma$ follows from it for a body of constant mass.

Newton's second law states that the net external force on a body equals the rate of change of its momentum: $F = \dfrac{dp}{dt}$. Momentum is defined as $p = mv$. If the mass $m$ of the body stays constant while its velocity changes, then $F = \frac{d(mv)}{dt} = m\frac{dv}{dt}$ since $m$ can be pulled outside the derivative. But $\dfrac{dv}{dt}$ is, by definition, the body's acceleration $a$, so this becomes $F = ma$ This shows that $F = ma$ is not an independent law in its own right -- it is simply what the momentum form of the second law reduces to whenever the mass of the body does not change. In situations where the mass itself is changing with time (a rocket burning fuel, for example), $F = ma$ alone is not sufficient, and the full momentum form $F = dp/dt$ must be used instead.\n\n> [!ANSWER]\n> $F = ma$ follows from $F = dp/dt$ by treating $m$ as constant, so $F = d(mv)/dt = m(dv/dt) = ma$.

Differentiate $p = mv$ with respect to time, treating $m$ as a constant so it can be pulled out of the derivative, leaving $F = m(dv/dt) = ma$.

• Assuming $F = ma$ is itself the fundamental law, rather than a special case of the more general momentum form $F = dp/dt$.
• Forgetting that this derivation requires mass to be constant -- it silently breaks down for a variable-mass system such as a rocket.