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a) $\Delta p = 2.0 kg \cdot m/s$
b) $F=5.8 \times 10^2 N$

Giancoli 6th Edition, Chapter 7, Problem 15


Chapter 7, Problem 15 is solved.

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Transcript for this Giancoli solution

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.


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