Newton's second law: force, momentum and impulse

Momentum first. Linear momentum bundles mass and velocity into one vector,

where: $m$ is mass in kg, $\vec v$ is velocity in m/s, and $\vec p$ is momentum in kg·m/s. It is momentum, not velocity alone, that a force actually changes — a heavy truck and a light scooter at the same speed do not respond to the same push in the same way.

The law itself. Newton's second law states that the net force on a body equals the rate at which its momentum changes, and points in the same direction as that change:

The newton is defined precisely so that no extra constant of proportionality is needed here — a force of $1\ \text{N}$ produces a momentum change of $1\ \text{kg}\,\cdot\,\text{m/s}$ every second.

Constant mass gives $F = ma$. Almost every school problem holds mass fixed, so the derivative falls only on velocity:

where: $\vec a$ is acceleration in m/s². This is the familiar working form, and it applies independently along each axis, $F_x = ma_x$, $F_y = ma_y$, $F_z = ma_z$ — a force along $y$ never shows up in the $x$-equation.

Impulse links force and time. Rearranging the general law over a finite interval gives

The quantity $\vec F\,\Delta t$ is called impulse; it is numerically equal to the momentum change the force produces, whether the force is steady or, as in a bat striking a ball, a brief and varying spike. (The syllabus notes only as background that relativistic momentum is $\vec p = m_0\vec v/\sqrt{1-v^2/c^2}$; $\vec F = m\vec a$ is its low-speed limit, and no relativistic numericals are set here.)

Deriving the impulse–momentum relation from the second law, and using it to show that stretching the contact time of a collision lowers the peak force needed for a fixed momentum change, is the single most useful piece of algebra this topic offers, since it explains cricket gloves, airbags and bent-knee landings with the same three lines.

Step 1. Start from the general form of the second law, $\vec F = d\vec p/dt$, which holds whether or not the force is constant during the interval — this generality matters because a bat's push on a ball is not steady, it rises and falls within milliseconds.

Step 2. Multiply both sides by the small time interval $dt$ and integrate from the start of contact ($t_1$, momentum $\vec p_1$) to the end ($t_2$, momentum $\vec p_2$): $\int_{t_1}^{t_2}\vec F\,dt = \int_{\vec p_1}^{\vec p_2} d\vec p = \vec p_2 - \vec p_1 = \Delta\vec p$. The left side, $\int \vec F\,dt$, is defined as the impulse; for a school-level treatment we replace the varying force by its time-average $\vec F_{\text{avg}}$ over the interval $\Delta t = t_2 - t_1$, giving the working relation $\vec F_{\text{avg}}\,\Delta t = \Delta\vec p$.

Step 3. Read off the consequence for a fixed momentum change. If $\Delta p$ is set by the situation — say, a ball's velocity must reverse — then $F_{\text{avg}} = \Delta p/\Delta t$ shows $F_{\text{avg}}$ and $\Delta t$ are inversely related: doubling the contact time halves the average force needed to produce the same $\Delta p$.

Step 4. Apply this to padding and cushioning. A cricket glove, an airbag, or bending your knees on landing all work by increasing $\Delta t$ for the same $\Delta p$, which lowers $F_{\text{avg}}$ and reduces the chance of injury — the physics is entirely contained in $F_{\text{avg}}\,\Delta t = \Delta\vec p$, no new law is needed.

A cricket ball of mass $0.16\ \text{kg}$ moving at $20\ \text{m/s}$ is struck by a bat and rebounds along the same line at $18\ \text{m/s}$, with bat–ball contact lasting $0.01\ \text{s}$. (a) Find the average force on the ball. (b) If a padded practice net instead stretched the contact time to $0.04\ \text{s}$ for the same speed change, find the new average force.

Step 1. Take the ball's initial direction as positive. Initial momentum $p_i = mu = 0.16 \times 20 = 3.2\ \text{kg}\,\cdot\,\text{m/s}$. Final momentum, with the ball reversing direction, $p_f = mv = 0.16 \times (-18) = -2.88\ \text{kg}\,\cdot\,\text{m/s}$.

Step 2. Change in momentum: $\Delta p = p_f - p_i = -2.88 - 3.2 = -6.08\ \text{kg}\,\cdot\,\text{m/s}$. This is the same $\Delta p$ for both parts (a) and (b), since the ball's speed change is identical in each — only the contact time differs.

Step 3. Part (a): using $F\,\Delta t = \Delta p$ with $\Delta t = 0.01\ \text{s}$, $F = \Delta p/\Delta t = -6.08/0.01 = -608\ \text{N}$, magnitude $608\ \text{N}$, directed back toward the bowler.

Step 4. Part (b): the same $\Delta p = -6.08\ \text{kg}\,\cdot\,\text{m/s}$ now acts over $\Delta t = 0.04\ \text{s}$, four times longer, so $F = -6.08/0.04 = -152\ \text{N}$ — exactly one quarter of the rigid-bat force, confirming that quadrupling contact time quarters the average force for the same momentum change.

The average force is $608\ \text{N}$ for the $0.01\ \text{s}$ contact and $152\ \text{N}$ for the $0.04\ \text{s}$ contact.

• Treating $F = ma$ as the definition of force, forgetting it is only the constant-mass special case of $\vec F = d\vec p/dt$.
• Dropping the sign when a body reverses direction: a ball bouncing back needs a negative velocity relative to its initial direction, not "same speed, so no change in momentum."
• Confusing impulse (a single event, unit N·s = kg·m/s, numerically equal to $\Delta p$) with force itself (unit N) — they are related but not the same quantity.
• Applying $F_x = ma_x$ with a force or acceleration component from the wrong axis mixed in — each axis equation is independent.

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