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Chapter 6: Momentum and Collisions

Chapter 6: Momentum and Collisions. Coach Kelsoe Physics Pages 196–227. Section 6–1: Momentum and Impulse. Coach Kelsoe Physics Pages 198–204. Objectives. Compare the momentum of different moving objects. Compare the momentum of the same object moving with different velocities.

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Chapter 6: Momentum and Collisions

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  1. Chapter 6:Momentum and Collisions Coach Kelsoe Physics Pages 196–227

  2. Section 6–1:Momentum and Impulse Coach Kelsoe Physics Pages 198–204

  3. Objectives • Compare the momentum of different moving objects. • Compare the momentum of the same object moving with different velocities. • Identify examples of change in the momentum of an object. • Describe changes in momentum in terms of force and time.

  4. Linear Momentum • Consider a soccer player heading a kick. • So far the quantities and kinematic equations we’ve introduced predict the motion of an object, such as a soccer ball, before and after it is struck. • What we will consider in this chapter is how the force and duration of the collision of an object affects the motion of the ball.

  5. Linear Momentum • The linear momentum of an object of mass m moving with a velocity v is defined as the product of the mass and velocity. • Momentum is represented by the symbol “p,” which was given by German mathematician Gottfried Leibniz, who used the term “progress.” • In formula terms: p = mv

  6. Linear Momentum • Momentum is a vector quantity, with its direction matching that of the velocity. • The unit for momentum is kilogram-meters per second (kg·m/s), NOT a Newton, which are kilogram-meters per square seconds (kg·m/s2). • The physics definition for momentum conveys a similar meaning to the everyday definition for momentum.

  7. Sample Problem A • A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

  8. Sample Problem Solution • Identify givens and unknowns: • m = 2250 kg • v = 25 m/s east • Choose the correct formula: • p = mv • Plug values into formula: • p = (2250 kg)(25 m/s east) • p = 56,000 kg·m/s to the east

  9. Changes in Momentum • A change in momentum is closely related to force. You know this from experience – it takes more force to stop something with a lot of momentum than with little momentum. • When Newton expressed his second law of motion, he didn’t say that F = ma, but instead, he expressed it as F = Δp/Δt. • We can rearrange this formula to find the change in momentum by saying Δp = FΔt.

  10. Impulse-Momentum Theorem • The expression Δp = FΔt is called the impulse-momentum theorem. • Another form of this equation that can be used is Δp = FΔt = mvf – mvi. • The impulse component of the equation is the FΔt. This idea also helps us understand why proper technique is important in sports.

  11. Sample Problem B • A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the force exerted on the car during the collision.

  12. Sample Problem Solution • Identify your givens and unknowns: • m = 1400 kg vi = 15 m/s west or -15 m/s • Δt = 0.30 s vf = 0 m/s • F = ? • Choose the correct equation: • FΔt = mvf – mvi F = mvf – mvi/Δt • Plug values into equation: • F = mvf – mvi/Δt • F = (1400 kg)(0 m/s) – (1400 kg)(-15 m/s)/0.30 s • F = 70,000 N to the east

  13. Impulse-Momentum Theorem • Highway safety engineers use the impulse-momentum theorem to determine stopping distances and safe following distances for cars and trucks. • For instance, if a truck was loaded down with twice its mass, it would have twice as much momentum and would take longer to stop.

  14. Sample Problem C • A 2240 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8410 N to the east? How far does the car travel during the deceleration?

  15. Sample Problem Solution • Identify givens and unknowns: • m = 2240 kg vi = 20.0 m/s west or -20.0 m/s • vf = 5.00 m/s west or -5.00 m/s • F = 8410 N east or +8410 N • Δt = ? Δx = ? • Choose the correct formulas: • FΔt = Δp  Δt = Δp/F • Δt = mvf - mvi/F • Plug values into formula: • Δt = (2240 kg)(-5.00 m/s) – (2240 kg)(-20.0 m/s)/8410 N • Δt = 4.00 s

  16. Reducing Force • We can reduce the force an object experiences by increasing the stop time. • It’s the same idea behind catching a water balloon or an egg. The change is momentum is the same. • The reason this works is that time and force are indirectly proportional. As one increases, the other decreases. • It’s the difference between a bunt or a homerun!

  17. Scrambled Eggs, Anyone?

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