During a snowball fight, a 0.30-kilogram snowball traveling at a speed of 17.0 m/s hits a student in the back of the ad. If the contact time is 0.10 s, what is the magnitude of the average force on the student's head? N ssf60
During a snowball fight, a 0.30-kilogram snowball traveling at a speed of 17.0 m/s hits a student in the back of the ad. If the contact time is 0.10 s, what is the magnitude of the average force on the student's head? N ssf60
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![**Physics Problem: Calculating Average Force During Collision**
**Problem Statement:**
During a snowball fight, a 0.30-kilogram snowball traveling at a speed of 17.0 m/s hits a student in the back of the head. If the contact time is 0.10 s, what is the magnitude of the average force on the student’s head?
**Answer Box:**
\[ \_\_\_\_\_\_ \text{ N} \]
**Explanation:**
In this problem, you are asked to calculate the magnitude of the average force exerted on the student's head by the snowball during the collision.
To solve this problem, you can use the concept of impulse and momentum. The change in momentum (impulse) is equal to the force exerted multiplied by the contact time.
The formula to use is:
\[ \text{Force} (F) = \frac{\Delta p}{\Delta t} \]
Where:
- \(\Delta p\) is the change in momentum (mass \(\times\) change in velocity),
- \(\Delta t\) is the contact time.
First, calculate the change in momentum \(\Delta p\):
\[
\Delta p = \text{mass} \times \text{change in velocity}
= 0.30 \, \text{kg} \times 17.0 \, \text{m/s}
= 5.1 \, \text{kg} \cdot \text{m/s}
\]
Next, plug the values into the formula:
\[
F = \frac{5.1 \, \text{kg} \cdot \text{m/s}}{0.10 \, \text{s}}
= 51 \, \text{N}
\]
So, the magnitude of the average force on the student’s head is \(\boxed{51 \, \text{N}}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c6fb9be-cb3b-43fd-a6cc-90b4ac965f54%2Fbd214703-86cd-4b15-8853-b971535d6ce7%2F0p8dgk8_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Calculating Average Force During Collision**
**Problem Statement:**
During a snowball fight, a 0.30-kilogram snowball traveling at a speed of 17.0 m/s hits a student in the back of the head. If the contact time is 0.10 s, what is the magnitude of the average force on the student’s head?
**Answer Box:**
\[ \_\_\_\_\_\_ \text{ N} \]
**Explanation:**
In this problem, you are asked to calculate the magnitude of the average force exerted on the student's head by the snowball during the collision.
To solve this problem, you can use the concept of impulse and momentum. The change in momentum (impulse) is equal to the force exerted multiplied by the contact time.
The formula to use is:
\[ \text{Force} (F) = \frac{\Delta p}{\Delta t} \]
Where:
- \(\Delta p\) is the change in momentum (mass \(\times\) change in velocity),
- \(\Delta t\) is the contact time.
First, calculate the change in momentum \(\Delta p\):
\[
\Delta p = \text{mass} \times \text{change in velocity}
= 0.30 \, \text{kg} \times 17.0 \, \text{m/s}
= 5.1 \, \text{kg} \cdot \text{m/s}
\]
Next, plug the values into the formula:
\[
F = \frac{5.1 \, \text{kg} \cdot \text{m/s}}{0.10 \, \text{s}}
= 51 \, \text{N}
\]
So, the magnitude of the average force on the student’s head is \(\boxed{51 \, \text{N}}\).
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