To determine in which case the force is maximum, we need to calculate the force for each collision using the formula:
[tex]\[ F = \frac{\text{Impulse}}{\text{Time Interval}} \][/tex]
We are given the following values:
1. Collision A:
[tex]\[ \text{Impulse} = 10,000 \, \text{kgm/s} \][/tex]
[tex]\[ \text{Time Interval} = 10^3 \, \text{s} \][/tex]
2. Collision B:
[tex]\[ \text{Impulse} = 1,000 \, \text{kgm/s} \][/tex]
[tex]\[ \text{Time Interval} = 10^2 \, \text{s} \][/tex]
3. Collision C:
[tex]\[ \text{Impulse} = 100 \, \text{kgm/s} \][/tex]
[tex]\[ \text{Time Interval} = 10^1 \, \text{s} \][/tex]
4. Collision D:
[tex]\[ \text{Impulse} = 10 \, \text{kgm/s} \][/tex]
[tex]\[ \text{Time Interval} = 10^0 \, \text{s} \][/tex]
5. Collision E:
[tex]\[ \text{Impulse} = 1 \, \text{kgm/s} \][/tex]
[tex]\[ \text{Time Interval} = 10^{-2} \, \text{s} \][/tex]
Now, we calculate the force for each collision:
1. For Collision A:
[tex]\[ F_A = \frac{10,000}{10^3} \][/tex]
[tex]\[ F_A = \frac{10,000}{1,000} \][/tex]
[tex]\[ F_A = 10 \, \text{N} \][/tex]
2. For Collision B:
[tex]\[ F_B = \frac{1,000}{10^2} \][/tex]
[tex]\[ F_B = \frac{1,000}{100} \][/tex]
[tex]\[ F_B = 10 \, \text{N} \][/tex]
3. For Collision C:
[tex]\[ F_C = \frac{100}{10^1} \][/tex]
[tex]\[ F_C = \frac{100}{10} \][/tex]
[tex]\[ F_C = 10 \, \text{N} \][/tex]
4. For Collision D:
[tex]\[ F_D = \frac{10}{10^0} \][/tex]
[tex]\[ F_D = \frac{10}{1} \][/tex]
[tex]\[ F_D = 10 \, \text{N} \][/tex]
5. For Collision E:
[tex]\[ F_E = \frac{1}{10^{-2}} \][/tex]
[tex]\[ F_E = \frac{1}{0.01} \][/tex]
[tex]\[ F_E = 100 \, \text{N} \][/tex]
Among these forces, the maximum force is [tex]\( 100 \, \text{N} \)[/tex], which occurs in Collision E.
Thus, the correct answer is:
E. Collision E
