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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 = m a as a special case. We can derive this form as follows. First, note that the change in momentum Δ p size 12{Δp} {} is given by

Δ p = Δ ( m v ) . size 12{Δp=Δ left (mv right )} {}

If the mass of the system is constant, then

Δ ( m v ) = m Δ v . size 12{Δ left (mv right )=mΔv} {}

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

F net = Δ p Δ t = m Δ v Δ t . size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}

Because Δ v Δ t = a size 12{ { {Δv} over {Δt} } =a} {} , we get the familiar equation

F net = m a

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.

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 = Δ p Δ t . 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

Δ p = m Δ v = m v f v i . 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 Δ p size 12{Δp} {} is calculated, F net = Δ p Δ t 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.

Δ p = m v f v i = 0.057 kg 58 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 F net = Δ p Δ t size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } } {} :

F net = Δ p Δ t = 3.306 kg m/s 5 . 0 × 10 3 s = 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’ 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 net = 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.

Section summary

  • Linear momentum ( momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.
  • In symbols, linear momentum p is defined to be
    p = m v , size 12{p=mv} {}
    where m size 12{m} {} is the mass of the system and v size 12{v} {} is its velocity.
  • The SI unit for momentum is kg · m/s size 12{"kg" cdot "m/s"} {} .
  • Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes.
  • In symbols, Newton’s second law of motion is defined to be
    F net = Δ p Δ t , size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}
    F net is the net external force, Δ p size 12{Δp} {} is the change in momentum, and Δ t size 12{Δt} {} is the change time.

Conceptual questions

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?

Problems&Exercises

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7 . 50 m/s size 12{7 "." "50"``"m/s"} {} . (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s size 12{"600"``"m/s"} {} . (c) What is the momentum of the 90.0-kg hunter running at 7 . 40 m/s size 12{7 "." "40"``"m/s"} {} after missing the elephant?

(a) 1.50 × 10 4 kg m/s size 12{1 "." "50" times "10" rSup { size 8{4} } `"kg" cdot "m/s"} {}

(b) 625 to 1

(c) 6 . 66 × 10 2 kg m/s size 12{6 "." "66" times "10" rSup { size 8{2} } `"kg" cdot "m/s"} {}

(a) What is the mass of a large ship that has a momentum of 1 . 60 × 10 9 kg · m/s , when the ship is moving at a speed of 48.0 km/h? size 12{"48" "." 0``"km/h?"} {} (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s size 12{"1200"``"m/s"} {} .

(a) At what speed would a 2 . 00 × 10 4 -kg size 12{2 "." "00" times "10" rSup { size 8{4} } "-kg"} {} airplane have to fly to have a momentum of 1 . 60 × 10 9 kg · m/s size 12{1 "." "60" times "10" rSup { size 8{9} } "kg" cdot "m/s"} {} (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 m/s size 12{"60" "." 0``"m/s"} {} ? (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.

(a) 8 . 00 × 10 4 m/s size 12{8 "." "00" times "10" rSup { size 8{4} } " m/s"} {}

(b) 1 . 20 × 10 6 kg · m/s size 12{1 "." "20" times "10" rSup { size 8{6} } " kg" cdot "m/s"} {}

(c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be 0 . 0100 m/s size 12{ - 0 "." "0100"`"m/s"} {} , which is probably not noticeable.

(a) What is the momentum of a garbage truck that is 1 . 20 × 10 4 kg size 12{1 "." "20" times "10" rSup { size 8{4} } " kg"} {} and is moving at 10 . 0 m/s size 12{10 "." "0 m/s"} {} ? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

A runaway train car that has a mass of 15,000 kg travels at a speed of 5 .4 m/s size 12{5 "." 4`"m/s"} {} down a track. Compute the time required for a force of 1500 N to bring the car to rest.

54 s

The mass of Earth is 5 . 972 × 10 24 kg size 12{5 "." "972" times 10 rSup { size 8{"24"} } " kg"} {} and its orbital radius is an average of 1 . 496 × 10 11 m size 12{1 "." "496" times 10 rSup { size 8{"11"} } " m"} {} . Calculate its linear momentum.

Questions & Answers

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ninjadapaul
20/(×-6^2)
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Abe advanced level physics. OpenStax CNX. Jul 11, 2013 Download for free at http://legacy.cnx.org/content/col11534/1.3
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