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

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

p = m v . 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 p size 12{p} {} is a vector having the same direction as the velocity v size 12{v} {} . The SI unit for momentum is kg · 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:

p = m v . size 12{p=mv} {}

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, p 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

p = 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.

p player = 110 kg 8 . 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.

p ball = 0.410 kg 25.0 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

p player p ball = 880 10 . 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.

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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

F net = Δ p Δ t , size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } ,} {}

where F net size 12{F rSub { size 8{"net"} } } {} is the net external force, Δ p size 12{Δp} {} is the change in momentum, and Δ t size 12{Δ`t} {} 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.

F net = Δ p Δ t

Questions & Answers

what's the period of velocity 4cm/s at displacement 10cm
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1/v+1/u=1/f
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double slit experiment
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Up thrust is a force
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upthrust is a upward force that acts vertical in the ground surface.
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Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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