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Linear Momentum and Collisions

52 Linear Momentum and Force

Learning Objectives

  • Explain the relationship between momentum and force.
  • State Newton’s second law of motion in terms of momentum.
  • Calculate momentum given mass and velocity.

Understanding Linear Momentum

In everyday life, we often describe athletes, vehicles, or movements as having “momentum.” Scientifically, this idea is captured by the concept of linear momentum, which quantifies how difficult it is to stop a moving object. For students in the biosciences or health sciences, understanding momentum can help in contexts like biomechanics, sports medicine, or even physical therapy.

A large person running quickly will be harder to stop than a smaller one walking slowly—that’s momentum in action. Linear momentum is particularly important in medical physics, injury mechanics, and rehabilitation science.

Definition of Linear Momentum

Linear momentum is defined as the product of an object’s mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction.

[latex]\mathbf{p} = m \mathbf{v}[/latex]

Where:

  • [latex]\mathbf{p}[/latex] is the momentum (in [latex]\text{kg} \cdot \text{m/s}[/latex])
  • [latex]m[/latex] is the mass (in kilograms)
  • [latex]\mathbf{v}[/latex] is the velocity (in meters per second)

Momentum increases linearly with both mass and velocity. So, a fast-moving small object and a slow-moving large object can have comparable momenta. This principle is crucial in understanding collisions, impacts, and motion in living systems—whether you’re analyzing gait in physical therapy or the dynamics of blood flow in cardiovascular research.

Example 52.1: Calculating Momentum: A Football Player and a Football

(a) Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player’s momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.

Strategy

No information is given regarding direction, and so we can calculate only the magnitude of the momentum, [latex]p[/latex]. (As usual, a symbol that is in italics is a magnitude, whereas one that is italicized, boldfaced, and has an arrow is a vector.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum given in the equation, which becomes

[latex]p=\text{mv}[/latex]

when only magnitudes are considered.

Solution for (a)

To determine the momentum of the player, substitute the known values for the player’s mass and speed into the equation.

[latex]{p}_{\text{player}}=\left(\text{110 kg}\right)\left(8\text{.}\text{00 m/s}\right)=\text{880 kg}·\text{m/s}[/latex]

Solution for (b)

To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.

[latex]{p}_{\text{ball}}=\left(\text{0.410 kg}\right)\left(\text{25.0 m/s}\right)=\text{10.3 kg}·\text{m/s}[/latex]

The ratio of the player’s momentum to that of the ball is

[latex]\frac{{p}_{\text{player}}}{{p}_{\text{ball}}}=\frac{\text{880}}{\text{10}\text{.}3}=\text{85}\text{.}9.[/latex]

Discussion

Although the ball has greater velocity, the player has a much greater mass. Thus the momentum of the player is much greater than the momentum of the football, as you might guess. As a result, the player’s motion is only slightly affected if he catches the ball. We shall quantify what happens in such collisions in terms of momentum in later sections.

Momentum and Newton’s Second Law

Momentum has long been recognized as a fundamental quantity in physics. Historically referred to as the “quantity of motion,” it played a central role in early mechanics. In fact, Newton’s second law of motion was originally stated in terms of momentum, rather than acceleration.

In its most general form, Newton’s second law states:

[latex]\mathbf{F}_{\text{net}} = \frac{\Delta \mathbf{p}}{\Delta t}[/latex]

Here:

  • [latex]\mathbf{F}_{\text{net}}[/latex] is the net external force acting on a system,
  • [latex]\Delta \mathbf{p}[/latex] is the change in linear momentum,
  • [latex]\Delta t[/latex] is the time interval over which the momentum changes.

Newton’s Second Law of Motion in Terms of Momentum

The net external force acting on a system equals the rate of change of momentum:

[latex]\mathbf{F}_{\text{net}} = \frac{\Delta \mathbf{p}}{\Delta t}[/latex]

Making Connections: Force and Momentum

Force and momentum are fundamentally linked. A net force acting over time causes a change in momentum. This version of Newton’s second law applies broadly—beyond cases of constant mass—and remains a critical concept in biomechanics, cellular motion, and molecular collisions.

Let’s now derive the more familiar version of Newton’s second law from the momentum form. Begin with the definition of momentum:

[latex]\Delta \mathbf{p} = \Delta (m \mathbf{v})[/latex]

If the mass [latex]m[/latex] remains constant, we can factor it out:

[latex]\Delta (m \mathbf{v}) = m \Delta \mathbf{v}[/latex]

Substituting into Newton’s law gives:

[latex]\mathbf{F}_{\text{net}} = \frac{\Delta \mathbf{p}}{\Delta t} = \frac{m \Delta \mathbf{v}}{\Delta t}[/latex]

Recognizing that [latex]\Delta \mathbf{v} / \Delta t = \mathbf{a}[/latex], we recover the familiar equation:

[latex]\mathbf{F}_{\text{net}} = m \mathbf{a}[/latex]

This form is valid only when the mass of the system remains constant. However, Newton’s second law stated in terms of momentum applies even when mass is changing—such as in biological systems involving fluid flow, metabolism, or transport, and in rocket motion, which we’ll explore later.

Example 52.2: 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

[latex]{\mathbf{F}}_{\text{net}}=\frac{\Delta \mathbf{p}}{\Delta t}\text{.}[/latex]

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

[latex]\Delta p=m\Delta v=m\left({v}_{f}-{v}_{i}\right).[/latex]

In this example, the velocity just after impact and the change in time are given; thus, once [latex]\Delta p[/latex] is calculated, [latex]{F}_{\text{net}}=\frac{\Delta p}{\Delta t}[/latex] 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.

[latex]\begin{array}{lll}\Delta p& =& m\left({v}_{f}–{v}_{i}\right)\\ & =& \left(\text{0.057 kg}\right)\left(\text{58 m/s}–0 m/s\right)\\ & =& 3\text{.306 kg}·\text{m/s}\approx \text{3.3 kg}·\text{m/s}\end{array}[/latex]

Now the magnitude of the net external force can determined by using [latex]{F}_{\text{net}}=\frac{\Delta p}{\Delta t}[/latex]:

[latex]\begin{array}{lll}{F}_{\text{net}}& =& \frac{\Delta p}{\Delta t}=\frac{\text{3.306 kg}\cdot \text{m/s}}{5\text{.}0×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}s}\\ & =& \text{661 N}\approx \text{660 N,}\end{array}[/latex]

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 [latex]{F}_{\text{net}}\phantom{\rule{0.15em}{0ex}}\text{=}\phantom{\rule{0.15em}{0ex}}\text{ma}[/latex], but one additional step would be required compared with the strategy used in this example.

Section Summary

  • Linear momentum (or simply momentum) is a vector quantity defined as the product of an object’s mass and its velocity.
  • In equation form:
    [latex]\mathbf{p} = m\mathbf{v}[/latex]

    where:

    • [latex]m[/latex] is the mass of the object or system
    • [latex]\mathbf{v}[/latex] is its velocity vector
  • The SI unit of momentum is [latex]\text{kg} \cdot \text{m/s}[/latex].
  • Newton’s second law of motion can be stated in terms of momentum: the net external force on a system is equal to the rate of change of its momentum.
  • In symbolic form:
    [latex]\mathbf{F}_{\text{net}} = \frac{\Delta \mathbf{p}}{\Delta t}[/latex]

    where:

    • [latex]\mathbf{F}_{\text{net}}[/latex] is the net external force
    • [latex]\Delta \mathbf{p}[/latex] is the change in momentum
    • [latex]\Delta t[/latex] is the time interval during which the change occurs

Conceptual Questions

  1. An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?
  2. An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum?
  3. Professional Application Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground.
  4. How can a small force impart the same momentum to an object as a large force?

Problems & Exercises

  1. (a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of [latex]7\text{.}\text{50 m/s}[/latex]. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of [latex]\text{600 m/s}[/latex]. (c) What is the momentum of the 90.0-kg hunter running at [latex]7\text{.}\text{40 m/s}[/latex] after missing the elephant?
  2. (a) What is the mass of a large ship that has a momentum of [latex]1\text{.}\text{60}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{kg}·\text{m/s}[/latex], when the ship is moving at a speed of [latex]\text{48.0 km/h?}[/latex] (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of [latex]\text{1200 m/s}[/latex].
  3. (a) At what speed would a [latex]2\text{.}\text{00}×{\text{10}}^{4}\text{-kg}[/latex] airplane have to fly to have a momentum of [latex]1\text{.}\text{60}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{kg}·\text{m/s}[/latex] (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of [latex]\text{60.0 m/s}[/latex]? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.
  4. (a) What is the momentum of a garbage truck that is [latex]1\text{.}\text{20}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and is moving at [latex]10\text{.}\text{0 m/s}[/latex]? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?
  5. A runaway train car that has a mass of 15,000 kg travels at a speed of [latex]5\text{.4 m/s}[/latex] down a track. Compute the time required for a force of 1500 N to bring the car to rest.
  6. The mass of Earth is [latex]5\text{.}\text{972}×{10}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and its orbital radius is an average of [latex]1\text{.}\text{496}×{10}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{m}[/latex]. Calculate its linear momentum.

Glossary

linear momentum
the product of mass and velocity
second law of motion
physical law that states that the net external force equals the change in momentum of a system divided by the time over which it changes
definition

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