Question

(Will give brainliest) The values of the impulse and the time interval for various collisions are listed in the table. In which case is the force maximum?
Collision Impulse (kilogram·meters/second) Time Interval (seconds)
A              10,000       10^+3
B              1,000           10^+2
C              100              10^+1
D              10                10^0
E               1              10^-2

A) a
B) b
C) c
D) d
E) e

Answer 1

Answer:

Maximum force is in case E i.e. 100 N

Explanation:

Impulse shows the impact of force. Mathematically, it can be written as :

$J=F\times t$

$F=\dfrac{J}{t}$

i.e. the force is directly proportional to the impulse and inversely proportional to the time taken.

(A) $F=\dfrac{10,000\ kg\ m/s}{10^3}=10\ N$

(B) $F=\dfrac{1000\ kg\ m/s}{10^2}=10\ N$

(C) $F=\dfrac{100\ kg\ m/s}{10}=10\ N$

(D) $F=\dfrac{10\ kg\ m/s}{1}=10\ N$

(E) $F=\dfrac{1\ kg\ m/s}{10^{-2}}=100\ N$

So, it is clear that the maximum force is exerted in case E. Hence, the correct option is (E).

Answer 2

Answer:

E) e

Explanation:

Impulse is equal to the product between the force applied (F) and the time interval of the collision ($\Delta t$):

$I=F \Delta t$

This means that we can calculate the force for each collision dividing the impulse by the time interval:

$F=\frac{I}{\Delta t}$

Let's apply the formula to each case:

A) $F=\frac{10,000 kg m/s}{10^3 s}=10 N$

B) $F=\frac{1,000 kg m/s}{10^2 s}=10 N$

C) $F=\frac{100 kg m/s}{10^1 s}=10 N$

D) $F=\frac{10 kg m/s}{10^0 s}=10 N$

E) $F=\frac{1 kg m/s}{10^{-2} s}=100 N$

Therefore, the force is the maximum for collision E).