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

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Presentation on theme: "Chapter 6: Momentum and Collisions Coach Kelsoe Physics Pages 196–227."— Presentation transcript:

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/s 2 ). 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 = mv f – mv i. 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 kgv i = 15 m/s west or -15 m/s –Δt = 0.30 sv f = 0 m/s –F = ? Choose the correct equation: –FΔt = mv f – mv i  F = mv f – mv i /Δt Plug values into equation: –F = mv f – mv i /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 kgv i = 20.0 m/s west or -20.0 m/s –v f = 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 = mv f - mv i /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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