Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
7
Linear Momentum
Change chapter

7-1 and 7-2: Momentum and Its Conservation
7-3: Collisions and Impulse
7-4 and 7-5: Elastic Collisions
7-6: Inelastic Collisions
7-7: Collisions in Two Dimensions
7-8: Center of Mass
7-9: CM for the Human Body
7-10: CM and Translational Motion

Problem 15
A
a) Δp=2.0kgm/s\Delta p = 2.0 kg \cdot m/s
b) F=5.8×102NF=5.8 \times 10^2 N
Giancoli 6th Edition, Chapter 7, Problem 15 solution video poster
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VIDEO TRANSCRIPT

This golf ball experiences a change in momentum equal to the final momentum minus the initial momentum. The initial momentum is zero because it started at rest when it was on the tee. Change in momentum is then ‘m’ times ‘vf’. We are given the final speed and the mass, the mass being zero point zero four five kilograms and the speed being forty five meters per second. The change in momentum then is: two point zero kilogram meters per second. For part b we want to know the force which is equal to the change in momentum divided by the change in time it took for that change in momentum to occur. That ratio is the force and it’s two point zero two five kilogram meters per second and we divide that by the amount of time that the golf club was in contact with the golf ball, three point five times ten raised to power minus three seconds. And the 6th Edition answer is five point eight times ten raised to power two newtons. For the 5th Edition the golf club is assumed to be in contact with the ball for five milliseconds so for the 5th Edition the answer will instead be four point zero times ten to the two Newtons.

COMMENTS
By tete on Mon, 2/27/2012 - 5:29 AM

Thanks for the feedback LuisJCR, hope these pages continue to help

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