Chapter 1: Science Skills Coach Kelsoe Physical Science Pages 2-25.
Chapter 6: Momentum and Collisions Coach Kelsoe Physics Pages 196–227.
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Transcript of Chapter 6: Momentum and Collisions Coach Kelsoe Physics Pages 196–227.
Chapter 6:Momentum and Collisions
Coach KelsoePhysicsPages 196–227
Section 6–1:Momentum and Impulse
Coach KelsoePhysicsPages 198–204
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.
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.
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
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.
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?
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
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.
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.
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.
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
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.
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?
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
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!
Scrambled Eggs, Anyone?