{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 9(Momentum Notes)

# Chapter 9(Momentum Notes) - Unit 9 Momentum Impulse and...

This preview shows pages 1–2. Sign up to view the full content.

Knowns: Unknowns: m = .150 kg part a) p v i = 20 m/s part b) F v f = -20 m/s t= .014 s p mv momentum = mass times velocity = Solution: F t m v so F = - 6.0 .014s = -430N m v t kgm s = = Note: When this equation is used to solve for force, the force is rarely constant, so you are solving for the average force. Unit 9: Momentum, Impulse and Conservation of Momentum Part 1: Impulse and momentum In this chapter we will be looking at further applications of Newton’s 2nd and 3rd Laws. First a physics definition of the term momentum. Early scientists thought that when one object collided with another, the damage it did was directly proportional to how much mass it had and how fast it was moving. It turns out they weren’t quite correct, but we will study that in the next chapter. Momentum is defined as the product of an object’s mass and its velocity. The equation: The unit of momentum is kg m s . Momentum is a vector quantity in that both its magnitude and direction are important. If an object changes its velocity, of course it also changes its momentum. The equation for change in momentum is: Whenever one object exerts an unbalanced force on another, we know the second object will accelerate according to the equation F = ma. We can rewrite this as F = m v t . If both sides are multiplied by t, we end up with F t m v = . This equation is called the impulse-momentum equation. The left side of the equation is a quantity we call the impulse. Impulse is defined as product of the force exerted on an object and the time interval over which the force is applied. The equation for impulse is: The unit of impulse is the N·s . The most important thing the impulse-momentum equation tells us is that the impulse on an object and the object’s change in momentum are always equal to each other. Sample problems : 1) A club is used to hit a 45 gram golf ball off a tee. If the ball is moving at 60.0 m/s after being hit, how much impulse did the club exert on the ball? Since we do not know the force exerted on the ball, or the time the force was exerted, we must use the fact that the impulse on the ball is equal to the change in momentum for the ball. 2) A 150 gram softball is thrown towards home plate at 20.0 m/s and is hit by the batter at 20.0 m/s in the opposite direction.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern