The importance of momentum, unlike the importance of energy, was recognized early in the development of classical physics. Momentum was deemed so important that it was called the “quantity of motion.” Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes.

Newton’s Second Law of Motion in Terms of Momentum

The net external force equals the change in momentum of a system divided by the time over which it changes.

\[F_{net} = \dfrac{\Delta p}{\Delta t}\]

where \(F_{net} \) is the net external force, \(\Delta p\) is the change in momentum, and \(\Delta t\) is the change in time.

#### Making Connections: Force and Momentum

Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics.

This statement of Newton’s second law of motion includes the more familiar \(F_{net} = ma\) as a special case. We can derive this form as follows. First, note that the change in momentum \(\Delta p\) is given by

\[\Delta p = \Delta (mv)\]

If the mass of the system is constant, then

\[\Delta (mv) = m\Delta v.\]

So that for constant mass, Newton’s second law of motion becomes

\[F_{net} = \dfrac{\Delta p}{\Delta t} = \dfrac{m \Delta v}{\Delta t}.\]

Because \(\frac{\Delta v}{\Delta t} = a, \) we get the familiar equation

\[F_{net} = ma\]

when the mass of the system is *constant*.

Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail**;** however, the relationship between momentum and force remains useful when mass is constant, such as in the following example.

Example \(\PageIndex{2}\): Calculating Force: Venus Williams’ Racquet

During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?

**Strategy**

This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as

\[F_{net} = \dfrac{\Delta p}{\Delta t} \nonumber\]

As noted above, when mass is constant, the change in momentum is given by

\[\Delta p = m\Delta v = m(v_f - v_i). \nonumber\]

In this example, the velocity just after impact and the change in time are given; thus, once \(\Delta p\) s calculated, \(F_{net} = \frac{\Delta p}{\Delta t}\) can be used to find the force.

**Solution**

To determine the change in momentum, substitute the values for the initial and final velocities into the equation above.

\[\begin{align*} \Delta p &= m(v_f - v_i) \\[5pt] &= (0.057 \, kg)(58 \, m/s - 0 \, m/s)\\[5pt] &= 3.306 \, kg \cdot m/s = 3.3 \, kg \cdot m/s \end{align*} \]

Now the magnitude of the net external force can determined by using \(F_{net} = \frac{\Delta p}{\Delta t}\)

\[\begin{align*} F_{net} &= \dfrac{\Delta p}{\Delta t} = \dfrac{3.306 \, kg}{5.0 \times 10^{-3}}\\[5pt] &= 661 \, N,\end{align*} \]

where we have retained only two significant figures in the final step.

**Discussion**

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using \(F = ma\) but one additional step would be required compared with the strategy used in this example.