An average force of 4800 N acts on a golf ball for a time interval of 0.003 s. Determine the change in the momentum of the golf ball. The change in the momentum of the golf ball is kg-m/s.
An average force of 4800 N acts on a golf ball for a time interval of 0.003 s. Determine the change in the momentum of the golf ball. The change in the momentum of the golf ball is kg-m/s.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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![**Problem Statement:**
An average force of 4800 N acts on a golf ball for a time interval of 0.003 s.
**Task:**
Determine the change in the momentum of the golf ball.
**Question:**
The change in the momentum of the golf ball is ____ kg·m/s.
---
**Explanation:**
To find the change in momentum (\(\Delta p\)) of the golf ball, use the equation derived from Newton's second law:
\[
\Delta p = F \times \Delta t
\]
where:
- \(F\) is the force applied (4800 N),
- \(\Delta t\) is the time interval (0.003 s).
Substitute the given values into the equation:
\[
\Delta p = 4800 \, \text{N} \times 0.003 \, \text{s} = 14.4 \, \text{kg·m/s}
\]
Thus, the change in momentum of the golf ball is 14.4 kg·m/s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0e8ca4f-0f8c-42f5-b521-ced309fbfcc3%2F22753a10-f669-43b5-a6b0-8a05bd154f3f%2Fu0r6ds_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
An average force of 4800 N acts on a golf ball for a time interval of 0.003 s.
**Task:**
Determine the change in the momentum of the golf ball.
**Question:**
The change in the momentum of the golf ball is ____ kg·m/s.
---
**Explanation:**
To find the change in momentum (\(\Delta p\)) of the golf ball, use the equation derived from Newton's second law:
\[
\Delta p = F \times \Delta t
\]
where:
- \(F\) is the force applied (4800 N),
- \(\Delta t\) is the time interval (0.003 s).
Substitute the given values into the equation:
\[
\Delta p = 4800 \, \text{N} \times 0.003 \, \text{s} = 14.4 \, \text{kg·m/s}
\]
Thus, the change in momentum of the golf ball is 14.4 kg·m/s.
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