Newton's Second Law of Motion

Momentum first. A speeding truck is harder to stop than a speeding scooter, and a fast scooter is harder to stop than a slow one — mass and velocity both matter to "how much motion" a body carries. Momentum bundles the two together:

where: $p$ is momentum in $\text{kg}\,\cdot\,\text{m/s}$, $m$ is mass in kg, and $v$ is velocity in m/s. Momentum is a vector — it points the way the body is moving — and a body at rest carries zero momentum no matter how large its mass.

The law itself. Newton's Second Law states: the rate of change of momentum of a body is directly proportional to the applied unbalanced force, and the change takes place in the direction of the force. Two separate claims sit inside that one sentence — a bigger unbalanced force produces a bigger, faster change in momentum, and the momentum changes exactly along the line of the applied force, never sideways to it. Momentum can change three ways in principle — a change in mass, a change in speed, or a change in direction of velocity — but Class 9 always treats mass as fixed, so every change in momentum here traces back to acceleration, positive (speeding up) or negative (slowing down).

From momentum to $F = ma$. Writing the proportionality as an equation and working through it with mass constant (the full algebraic derivation is the subject of the companion "Mathematical formulation" article) collapses to the everyday form:

where: $F$ is the net, unbalanced force in newtons (N), $m$ is mass in kg, and $a$ is acceleration in $\text{m/s}^2$. This single line is why the newton is defined as the force that gives a $1\ \text{kg}$ mass an acceleration of $1\ \text{m/s}^2$: $1\ \text{N} = 1\ \text{kg}\,\cdot\,\text{m/s}^2$. Everything that follows — every "find the force" or "find the acceleration" numerical in this chapter — is one rearrangement of $F = ma$ away from the momentum statement it came from.

A trolley of mass $4\ \text{kg}$ is moving at $2\ \text{m/s}$. A constant unbalanced force pushes it for $5\ \text{s}$, after which its velocity becomes $6\ \text{m/s}$ in the same direction. Find the force applied, and verify the answer using the momentum definition of the Second Law.

Step 1. List the known values: $m = 4\ \text{kg}$, $u = 2\ \text{m/s}$, $v = 6\ \text{m/s}$, $t = 5\ \text{s}$.

Step 2. Find the acceleration using $a = \dfrac{v-u}{t}$:

Step 3. Apply $F = ma$:

Step 4. Verify with the momentum route. The change in momentum is $\Delta p = m(v-u) = 4 \times (6-2) = 16\ \text{kg}\,\cdot\,\text{m/s}$, produced over $t = 5\ \text{s}$, so the rate of change of momentum is $16/5 = 3.2\ \text{kg}\,\cdot\,\text{m/s}^2 = 3.2\ \text{N}$ — the same answer, confirming that $F=ma$ is just the momentum statement written in a shortcut form. The unbalanced force acting on the trolley, in the direction of its motion, is $3.2\ \text{N}$.

Competitive stretch

If the same $3.2\ \text{N}$ force instead acted on a lighter trolley of mass $2\ \text{kg}$ starting from rest, its acceleration would be $a = F/m = 3.2/2 = 1.6\ \text{m/s}^2$ — double the acceleration for half the mass, at exactly the same force, which is the practical face of $a = F/m$: acceleration and mass vary inversely for a fixed force.

• Using one of several forces acting on a body instead of the net (unbalanced) force — if friction or gravity also act, $F$ in $F=ma$ must be the resultant of all forces, not just the applied push.
• Confusing momentum ($p=mv$, units $\text{kg}\,\cdot\,\text{m/s}$) with force ($F=ma$, units N) — they are related but not the same physical quantity and cannot be substituted for each other.
• Assuming a larger force always gives a larger final velocity while ignoring mass — the same force gives different accelerations to different masses, since $a = F/m$.
• Forgetting that the change in momentum happens in the direction of the applied unbalanced force, which is why a force applied opposite to motion produces a negative (decelerating) change in momentum.

Ask the tutor: Want to practise picking out $u$, $v$, $a$, $t$, and $m$ from a tricky word problem before applying $F=ma$? Bring one to the AI tutor and work through the setup step by step.