The scientific definition of linear momentum is consistent with most people’s intuitive understanding of momentum: a large, fast-moving object has greater momentum than a smaller, slower object. Linear momentum is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum is expressed as
Momentum is directly proportional to the object’s mass and also its velocity. Thus the greater an object’s mass or the greater its velocity, the greater its momentum. Momentum $\vb{p}$ is a vector having the same direction as the velocity $\vb{v}$. The SI unit for momentum is $\kg · \ms$.
Linear momentum is defined as the product of a system’s mass multiplied by its velocity:
(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, $p$. (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
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.
Solution for (b)
To determine the momentum of the ball, substitute the known values for the ball’s mass and speed into the equation.
The ratio of the player’s momentum to that of the ball is
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.
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. Using symbols, this law is
where $\vb{F}\_{\text{net}}$ is the net external force, $\Delta \vb{p}$ is the change in momentum, and $\Delta t$ is the change in time.
The net external force equals the change in momentum of a system divided by the time over which it changes.
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 $\vb{F}_{\text{net}}=m\vb{a}$ as a special case. We can derive this form as follows. First, note that the change in momentum $\Delta \vb{p}$ is given by
If the mass of the system is constant, then
So that for constant mass, Newton’s second law of motion becomes
Because $\frac{ \Delta \vb{v}}{\Delta t}=\vb{a}$, we get the familiar equation
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.
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
As noted above, when mass is constant, the change in momentum is given by
In this example, the velocity just after impact and the change in time are given; thus, once $\Delta p$ is calculated, $F_{\text{net}}=\frac{\Delta p}{\Delta t}$ 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.
Now the magnitude of the net external force can determined by using $F_{\text{net}}=\frac{ \Delta p}{\Delta t}$:
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 $F_{\text{net}}=ma$, but one additional step would be required compared with the strategy used in this example.
where $m$ is the mass of the system and $\vb{v}$ is its velocity.
$\vb{F}_{\text{net}}$ is the net external force, $\Delta \vb{p}$ is the change in momentum, and $\Delta t$ is the change time.
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?
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?
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.
How can a small force impart the same momentum to an object as a large force?
(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of $7.50 \ms$.
(b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of $600 \ms$.
(c) What is the momentum of the 90.0-kg hunter running at $7.40 \ms$ after missing the elephant?
Strategy
This problem applies the definition of linear momentum $p = mv$ to calculate the momentum of each object. For part (b), we compare the momenta by finding their ratio.
Solution for (a)
Using the definition of momentum:
Solution for (b)
First, calculate the dart’s momentum:
Now find the ratio:
Solution for (c)
Discussion
Despite the dart traveling 80 times faster than the elephant, the elephant’s enormous mass (50,000 times the dart’s mass) results in a momentum 625 times larger. This illustrates why momentum depends on both mass and velocity. The hunter’s momentum is much smaller than the elephant’s, explaining why the hunter wisely chose to run!
Answer
(a) The elephant’s momentum is 1.50 × 10⁴ kg·m/s.
(b) The elephant’s momentum is 625 times greater than the dart’s momentum.
(c) The hunter’s momentum is 666 kg·m/s (or 6.66 × 10² kg·m/s).
(a) What is the mass of a large ship that has a momentum of $1.60 \times 10^{9}\kg · \ms$, when the ship is moving at a speed of $48.0 \text{km/h}$ ?
(b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of $1200 \ms$.
Strategy
For part (a), we use the definition of momentum $p = mv$ and solve for mass. We must first convert the speed from km/h to m/s. For part (b), we calculate the shell’s momentum and find the ratio.
Solution for (a)
First, convert the ship’s speed to SI units:
Using $p = mv$ and solving for mass:
Solution for (b)
Calculate the artillery shell’s momentum:
Find the ratio:
Discussion
Even though the artillery shell travels at supersonic speed (about Mach 3.5), the ship’s enormous mass gives it over 1000 times more momentum. This explains why large ships require great distances and time to stop or change direction—their massive momentum must be overcome by relatively small forces from their propulsion systems.
Answer
(a) The ship’s mass is 1.20 × 10⁸ kg (120 million kilograms, or about 120,000 metric tons).
(b) The ship’s momentum is approximately 1210 times (or 1.21 × 10³ times) greater than the artillery shell’s momentum.
(a) At what speed would a $2.00 \times 10^{4}\text{-kg}$ airplane have to fly to have a momentum of $1.60 \times 10^{9}\kg · \ms$ (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 $60.0 \ms$
?
(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.
Strategy
For part (a), we use $p = mv$ and solve for velocity. For part (b), we calculate the plane’s actual momentum during takeoff. For part (c), we apply conservation of momentum to find the ship’s recoil velocity.
Solution for (a)
Using $p = mv$ and solving for velocity:
Solution for (b)
Solution for (c)
By conservation of momentum, when the catapult launches the plane forward, the ship recoils backward. The ship’s mass from the previous problem is $1.20 \times 10^{8} \kg$.
Discussion
The airplane’s momentum at takeoff is about 1300 times smaller than the ship’s momentum. Because the ship is 6000 times more massive than the plane, the recoil effect is imperceptible. This demonstrates why aircraft carriers can launch planes without noticeable movement, making them stable platforms for flight operations.
Answer
(a) The airplane would need to fly at 8.00 × 10⁴ m/s (80 km/s), which is about 235 times the speed of sound—clearly impossible for conventional aircraft.
(b) The plane’s momentum at takeoff is 1.20 × 10⁶ kg·m/s.
(c) The ship recoils at only 0.0100 m/s (1 cm/s) backward, which is negligible and demonstrates why aircraft carriers remain stable platforms for flight operations.
(a) What is the momentum of a garbage truck that is $1.20 \times 10^{4}\kg$ and is moving at $10.0 \ms$ ?
(b) At what speed would an 8.00-kg trash can have the same momentum as the truck?
Strategy
For part (a), we apply the definition of momentum $p = mv$. For part (b), we set the trash can’s momentum equal to the truck’s momentum and solve for the required velocity.
Solution for (a)
Solution for (b)
For equal momenta: $p_{\text{can}} = p_{\text{truck}}$
Discussion
This speed is about 44 times the speed of sound and roughly half of Earth’s escape velocity! The mass ratio between the truck and trash can is 1500:1, so the trash can would need to travel 1500 times faster than the truck to have equal momentum. This dramatically illustrates how mass and velocity contribute equally to momentum—doubling either one doubles the momentum.
Answer
(a) The garbage truck’s momentum is 1.20 × 10⁵ kg·m/s.
(b) The trash can would need to travel at 1.50 × 10⁴ m/s (15 km/s, or about 54,000 km/h) to have the same momentum as the garbage truck.
A runaway train car that has a mass of 15 000 kg travels at a speed of $5.4 \ms$ down a track. Compute the time required for a force of 1500 N to bring the car to rest.
Strategy
We use Newton’s second law in terms of momentum: $F_{\text{net}} = \frac{\Delta p}{\Delta t}$. The train must lose all its momentum, so we calculate the initial momentum and then find the time required for the given force to produce this change in momentum.
Solution
First, calculate the initial momentum of the train car:
The final momentum is zero (the car comes to rest), so the change in momentum is:
Using Newton’s second law in terms of momentum:
The force opposing the motion is negative (in the direction opposite to motion), so:
Solving for time:
Discussion
Nearly a minute is needed to stop the train car, which makes sense given its large mass and modest braking force. The ratio of momentum to force gives time, showing why heavy objects require either larger forces or longer times to stop. This is why trains have long stopping distances—even with substantial braking forces, their enormous momentum takes considerable time to reduce to zero.
Answer
The time required to bring the train car to rest is 54 s (54 seconds).
The mass of Earth is $5.972 \times {10}^{24}\kg$ and its orbital radius is an average of $1.496 \times {10}^{11}\m$. Calculate its linear momentum.
Strategy
To find Earth’s linear momentum, we need its orbital velocity. Earth travels in a nearly circular orbit around the Sun, completing one orbit in one year. We can calculate the orbital velocity from the circumference of the orbit divided by the orbital period, then apply $p = mv$.
Solution
First, calculate the circumference of Earth’s orbit:
Convert one year to seconds:
Calculate Earth’s orbital velocity:
Now calculate Earth’s momentum:
Discussion
This is an enormous momentum—about $10^{24}$ times larger than the ship’s momentum from earlier problems. Earth’s orbital velocity of about 30 km/s (roughly 108,000 km/h) combined with its massive $6 \times 10^{24} \kg$ mass produces this immense momentum. Despite this huge momentum, Earth’s orbit is stable because the Sun’s gravitational force continuously provides the centripetal acceleration needed to change the direction of this momentum without changing its magnitude.
Answer
The linear momentum of Earth in its orbit around the Sun is 1.78 × 10²⁹ kg·m/s.