# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

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

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.D.1.1 The student is able to justify the selection of data needed to determine the relationship between the direction of the force acting on an object and the change in momentum caused by that force. (S.P. 4.1)

# Linear Momentum

### Linear Momentum

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

8.1 $p=mv.p=mv. size 12{p=mv} {}$

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, $pp size 12{p} {}$, is a vector, having the same direction as the velocity, $vv size 12{v} {}$. The SI unit for momentum is $kg·m/skg·m/s size 12{"kg" cdot "m/s"} {}$.

### Linear Momentum

Linear momentum is defined as the product of a system’s mass multiplied by its velocity.

8.2 $p=mvp=mv size 12{p=mv} {}$

### Example 8.1Calculating 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 only calculate the magnitude of the momentum, $pp size 12{p} {}$. 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

8.3 $p=mvp=mv size 12{p= ital "mv"} {}$

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.

8.4 $pplayer=110 kg8.00 m/s=880 kg·m/s.pplayer=110 kg8.00 m/s=880 kg·m/s. size 12{p rSub { size 8{"player"} } = left ("110"" kg" right ) left (8 "." "00"" m/s" right )="880"" kg" cdot "m/s"} {}$

Solution for (b)

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

8.5 $pball=0.410 kg25.0 m/s=10.3 kg·m/s.pball=0.410 kg25.0 m/s=10.3 kg·m/s. size 12{p rSub { size 8{"ball"} } = left (0 "." "410"" kg" right ) left ("25" "." 0" m/s" right )="10" "." 3" kg" cdot "m/s"} {}$

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

8.6 $pplayerpball=88010.3=85.9.pplayerpball=88010.3=85.9. size 12{ { {p rSub { size 8{"player"} } } over {p rSub { size 8{"ball"} } } } = { {"880"} over {"10" "." 3} } ="85" "." 9} {}$

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 and Newton’s Second Law

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

8.7 $Fnet=ΔpΔt.Fnet=ΔpΔt. size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } ,} {}$

where $FnetFnet size 12{F rSub { size 8{"net"} } } {}$ is the net external force, $ΔpΔp size 12{Δp} {}$ is the change in momentum, and $ΔtΔt size 12{Δt} {}$ is the change in time.

### 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.

8.8 $Fnet=ΔpΔtFnet=ΔpΔt$

### 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 $Fnet = maFnet = ma$ as a special case. We can derive this form as follows. First, note that the change in momentum, $ΔpΔp size 12{Δp} {}$, is given by

8.9 $Δp=Δ(mv).Δp=Δ(mv). size 12{Δp=Δ left (mv right )} {}$

If the mass of the system is constant, then

8.10 $Δ(mv)=mΔv.Δ(mv)=mΔv. size 12{Δ left (mv right )=mΔv} {}$

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

8.11 $Fnet=ΔpΔt=mΔvΔt.Fnet=ΔpΔt=mΔvΔt. size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}$

Because $ΔvΔt=a,ΔvΔt=a, size 12{ { {Δv} over {Δt} } =a} {}$ we get the familiar equation

8.12 $Fnet=maFnet=ma$

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 8.2Calculating Force: Venus Williams’s 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’s 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?

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

8.13 $Fnet=ΔpΔt.Fnet=ΔpΔt. size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}$

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

8.14 $Δp=mΔv=mvf−vi.Δp=mΔv=mvf−vi. size 12{Δp=mΔv=m left (v rSub { size 8{f} } - v rSub { size 8{i} } right )} {}$

In this example, the velocity just after impact and the change in time are given; thus, once $ΔpΔp size 12{Δp} {}$ is calculated, $Fnet=ΔpΔtFnet=ΔpΔt size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } } {}$ 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.

8.15 $Δp=mvf–vi=0.057 kg58 m/s–0 m/s=3.306 kg·m/s≈3.3 kg·m/s.Δp=mvf–vi=0.057 kg58 m/s–0 m/s=3.306 kg·m/s≈3.3 kg·m/s.$

Now the magnitude of the net external force can determined by using $Fnet=ΔpΔtFnet=ΔpΔt size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } } {}$:

8.16 Fnet=ΔpΔt=3.306 kg⋅m/s5.0×10−3s=661 N≈660 N,Fnet=ΔpΔt=3.306 kg⋅m/s5.0×10−3s=661 N≈660 N,alignl { stack { size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {3 "." "306""kg" cdot "m/s"} over {5 "." 0 times "10" rSup { size 8{ - 3} } s} } } {} # " "="661 N" approx "660""N," {} } } {}

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

Discussion

This quantity was the average force exerted by Venus Williams’s 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 $Fnet=ma,Fnet=ma, size 12{F rSub { size 8{"net"} } " = " ital "ma"} {}$ but one additional step would be required compared with the strategy used in this example.

### Making Connections: Illustrative Example

Figure 8.2 A puck has an elastic, glancing collision with the edge of an air hockey table.

In Figure 8.2, a puck is shown colliding with the edge of an air hockey table at a glancing angle. During the collision, the edge of the table exerts a force F on the puck, and the velocity of the puck changes as a result of the collision. The change in momentum is found by the equation

8.17 $Δp=mΔv=mv'-mv=m[v'+(-v)].Δp=mΔv=mv'-mv=m[v'+(-v)].$

As shown, the direction of the change in velocity is the same as the direction of the change in momentum, which, in turn, is in the same direction as the force exerted by the edge of the table. Note that there is only a horizontal change in velocity. There is no difference in the vertical components of the initial and final velocity vectors; therefore, there is no vertical component to the change in velocity vector or the change in momentum vector. This is consistent with the fact that the force exerted by the edge of the table is purely in the horizontal direction.