Answer:
E) e
Explanation:
Impulse is equal to the product between the force applied (F) and the time interval of the collision ([tex]\Delta t[/tex]):
[tex]I=F \Delta t[/tex]
This means that we can calculate the force for each collision dividing the impulse by the time interval:
[tex]F=\frac{I}{\Delta t}[/tex]
Let's apply the formula to each case:
A) [tex]F=\frac{10,000 kg m/s}{10^3 s}=10 N[/tex]
B) [tex]F=\frac{1,000 kg m/s}{10^2 s}=10 N[/tex]
C) [tex]F=\frac{100 kg m/s}{10^1 s}=10 N[/tex]
D) [tex]F=\frac{10 kg m/s}{10^0 s}=10 N[/tex]
E) [tex]F=\frac{1 kg m/s}{10^{-2} s}=100 N[/tex]
Therefore, the force is the maximum for collision E).
Answer:
Maximum force is in case E i.e. 100 N
Explanation:
Impulse shows the impact of force. Mathematically, it can be written as :
[tex]J=F\times t[/tex]
[tex]F=\dfrac{J}{t}[/tex]
i.e. the force is directly proportional to the impulse and inversely proportional to the time taken.
(A) [tex]F=\dfrac{10,000\ kg\ m/s}{10^3}=10\ N[/tex]
(B) [tex]F=\dfrac{1000\ kg\ m/s}{10^2}=10\ N[/tex]
(C) [tex]F=\dfrac{100\ kg\ m/s}{10}=10\ N[/tex]
(D) [tex]F=\dfrac{10\ kg\ m/s}{1}=10\ N[/tex]
(E) [tex]F=\dfrac{1\ kg\ m/s}{10^{-2}}=100\ N[/tex]
So, it is clear that the maximum force is exerted in case E. Hence, the correct option is (E).
