Newton's second law

Newton's first law tells you that a body's motion stays put until an unbalanced force intervenes — but it says nothing about how much the motion changes. That gap is exactly what Newton's second law fills, and it is the single most-used equation in this entire chapter of ICSE Class 9 Physics. Push twice as hard and a body accelerates twice as fast; push the same force on twice the mass and the acceleration halves. Buried in this everyday intuition is a precise, testable formula, $F = ma$ , that connects three quantities you can measure independently: the net force applied, the mass being pushed, and the acceleration produced. This article builds that formula up from the more fundamental idea of momentum — "quantity of motion" — shows exactly how $F=ma$ falls out of the rate of change of momentum, and works through fully solved numericals of the kind you will meet in exams, where $F=ma$ is combined with the equations of uniformly accelerated motion.

The core insight

The net unbalanced force on a body equals the rate of change of its momentum, which for constant mass simplifies to $F = ma$ .

The core idea

Momentum first. Before force, physicists define linear momentum as a measure of how much motion a moving body carries — the product of its mass and its velocity:

p = mv

where: $p$ is momentum (kg·m/s), $m$ is mass (kg), and $v$ is velocity (m/s). Momentum captures something speed alone cannot: a heavy truck rolling slowly can carry as much "quantity of motion" as a light car speeding along, because momentum weighs mass and velocity together.

From momentum to force. If a body of constant mass $m$ has its velocity change from $u$ to $v$ over a time $\Delta t$ , its momentum changes by:

\Delta p = \Delta(mv) = m\Delta v

Mass is treated as constant here, so it factors straight out of the change — only velocity is changing. Newton's second law is the statement that the net (unbalanced) force on a body equals the rate of change of its momentum, and that this force acts in the same direction the motion changes:

F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}

Since $\Delta v/\Delta t$ is just the acceleration $a$ , this collapses to the working equation you will use constantly:

F = ma

where: $F$ is the net force (newtons, N), $m$ is mass (kg), and $a$ is the resulting acceleration ( $\text{m/s}^2$ ). One newton is defined as the force that gives a 1 kg mass an acceleration of $1\ \text{m/s}^2$ . In the older CGS system, force is measured in dyne, where $1\ \text{dyne} = 1\ \text{g} \times 1\ \text{cm/s}^2$ and $1\ \text{N} = 10^5\ \text{dyne}$ — a conversion factor worth memorising because unit mix-ups are the most common way marks are lost on this topic. The figure below shows why the same force produces different outcomes depending on mass — the entire content of $F=ma$ in one picture.

Figure

Two carts pushed by the same force showing the light cart accelerating fast and the heavy cart accelerating slowly, illustrating that acceleration equals force divided by mass

The rate of change of momentum, in full

The goal of this derivation is to show, step by step, that Newton's second law — the net unbalanced force on a body equals the rate of change of its momentum — reduces exactly to the familiar equation $F = ma$ whenever the body's mass stays constant, which is the case for almost every numerical you will meet in this chapter.

Step 1. Start from the definition of momentum, $p = mv$ , and consider a body whose velocity changes from $u$ to $v$ over a time interval $\Delta t$ while its mass $m$ stays fixed. Its momentum changes from $p_i = mu$ to $p_f = mv$ .

Step 2. Write the change in momentum. $\Delta p = p_f - p_i = mv - mu = m(v-u) = m\Delta v$ , where $\Delta v = v - u$ is the change in velocity. Because $m$ is the same before and after, it can be factored out of the subtraction.

Step 3. Apply Newton's second law statement — force equals rate of change of momentum. $F = \dfrac{\Delta p}{\Delta t} = \dfrac{m\Delta v}{\Delta t}$ . Group the terms as $m \times \dfrac{\Delta v}{\Delta t}$ .

Step 4. Recognise the acceleration. By definition, acceleration is the rate of change of velocity, $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v-u}{t}$ . Substituting gives the final working equation, $F = ma$ , valid whenever mass is constant — the form you will use in every numerical in this chapter, and which connects directly to $v=u+at$ , $S=ut+\tfrac12at^2$ and $v^2=u^2+2aS$ from the motion chapter once you know $a$ .

Worked example

A car of mass 1000 kg, initially at rest, is pushed by a constant net force and reaches a velocity of 5 m/s in 4 s. Find (a) the acceleration, (b) the force applied, and (c) the change in momentum of the car, and (d) verify $F = \Delta p/\Delta t$ .

Step 1 — find the acceleration. Given $u = 0$ , $v = 5\ \text{m/s}$ , $t = 4\ \text{s}$ :

a = \frac{v-u}{t} = \frac{5-0}{4} = 1.25\ \text{m/s}^2

Step 2 — apply Newton's second law.

F = ma = 1000 \times 1.25 = 1250\ \text{N}

Step 3 — find the change in momentum directly.

\Delta p = m\Delta v = 1000 \times (5-0) = 5000\ \text{kg}\,\cdot\,\text{m/s}

Step 4 — check via rate of change of momentum.

\frac{\Delta p}{\Delta t} = \frac{5000}{4} = 1250\ \text{N}

This matches Step 2 exactly, confirming $F = \Delta p/\Delta t = ma$ are two routes to the same answer.

Competitive stretch

A ball of mass 0.2 kg moving at 15 m/s is brought to rest by a resisting force in 0.05 s. Find the change in momentum and the average force.

Step 1. $u = 15\ \text{m/s}$ , $v = 0$ , $\Delta t = 0.05\ \text{s}$ . Change in momentum: $\Delta p = m\Delta v = 0.2 \times (0-15) = -3\ \text{kg}\,\cdot\,\text{m/s}$ ; the negative sign shows momentum decreased.

Step 2. Average force: $F = \dfrac{\Delta p}{\Delta t} = \dfrac{-3}{0.05} = -60\ \text{N}$ . The average retarding force has magnitude 60 N, directed opposite to the ball's original motion.

Common traps

Mixing units — using mass in grams with acceleration in $\text{cm/s}^2$ and expecting an SI answer in newtons. Stay consistently in one system (SI or CGS) through a whole calculation.
Using an $a$ that is not the net acceleration — $F=ma$ needs the resultant acceleration caused by the unbalanced (net) force, not any single one of several forces acting on the body.
Dropping the sign on a decelerating force — a negative $\Delta v$ correctly gives a negative (retarding) force; report "60 N, opposing motion," not just "60 N."
Confusing momentum with force — momentum is $p=mv$ , a snapshot quantity; force is the rate of change of momentum, $\Delta p/\Delta t$ , a different physical quantity entirely.

Ask the tutor: Want to practise combining Newton's second law with the equations of motion on a few more numericals?

Why does F = ma only work when mass is constant?

Newton's second law is really $F = \Delta p/\Delta t$ , the rate of change of momentum, and $p = mv$ has two parts that could change — mass and velocity. When mass stays fixed, only velocity changes, so $\Delta p = m\Delta v$ and the mass factors out cleanly to give $F=ma$ . If mass itself is changing, as in a rocket burning fuel, you must go back to the full momentum form; $F=ma$ alone is not enough. At ICSE Class 9 level, every numerical assumes constant mass, so $F=ma$ is always the tool to reach for.

How is momentum different from force if they use the same F = ma type numbers?

They are not the same kind of quantity at all. Momentum $p=mv$ describes a body's motion at one instant — how much "quantity of motion" it currently has. Force is what changes that momentum over time, given by the rate $\Delta p/\Delta t$ . A car moving at constant velocity has momentum but zero net force acting on it, because its momentum is not changing. Only when momentum is changing — speeding up, slowing down, or turning — is there a net force present.

Why does the negative sign matter when a force stops something?

The sign tells you the direction of the force relative to the body's original motion, and examiners specifically check for it. When a moving ball is brought to rest, $\Delta v = v - u$ is negative because the final velocity is less than the initial one, so $F = m\Delta v/\Delta t$ comes out negative too. That minus sign says the force acted opposite to the direction of motion — it was a retarding or resisting force, not a driving one. Reporting only "60 N" without this direction is treated as an incomplete answer.