First, we need to understand the concept of force as it relates to impulse and time. The force exerted during a collision can be calculated using the formula:
[tex]\[ F = \frac{\text{Impulse}}{\text{Time Interval}} \][/tex]
We'll apply this formula to each collision case in the provided table.
### Collision A
- Impulse: [tex]\(10,000\)[/tex] kg·m/s
- Time Interval: [tex]\(10^3\)[/tex] s
So, the force for Collision A is:
[tex]\[ F_A = \frac{10{,}000}{10^3} = 10 \text{ N} \][/tex]
### Collision B
- Impulse: [tex]\(1,000\)[/tex] kg·m/s
- Time Interval: [tex]\(10^2\)[/tex] s
So, the force for Collision B is:
[tex]\[ F_B = \frac{1{,}000}{10^2} = 10 \text{ N} \][/tex]
### Collision C
- Impulse: [tex]\(100\)[/tex] kg·m/s
- Time Interval: [tex]\(10^1\)[/tex] s
So, the force for Collision C is:
[tex]\[ F_C = \frac{100}{10^1} = 10 \text{ N} \][/tex]
### Collision D
- Impulse: [tex]\(10\)[/tex] kg·m/s
- Time Interval: [tex]\(10^0\)[/tex] s
So, the force for Collision D is:
[tex]\[ F_D = \frac{10}{10^0} = 10 \text{ N} \][/tex]
### Collision E
- Impulse: [tex]\(1\)[/tex] kg·m/s
- Time Interval: [tex]\(10^{-2}\)[/tex] s
So, the force for Collision E is:
[tex]\[ F_E = \frac{1}{10^{-2}} = 100 \text{ N} \][/tex]
Now, we compare the calculated forces:
- Force for Collision A: [tex]\(10 \text{ N}\)[/tex]
- Force for Collision B: [tex]\(10 \text{ N}\)[/tex]
- Force for Collision C: [tex]\(10 \text{ N}\)[/tex]
- Force for Collision D: [tex]\(10 \text{ N}\)[/tex]
- Force for Collision E: [tex]\(100 \text{ N}\)[/tex]
The maximum force is [tex]\(100 \text{ N}\)[/tex], which occurs in Collision E.
Therefore, the force is maximum in Collision E.
