Question

Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.

Answer 1

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

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

where

F is the force applied

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be written as

$\Delta p=m(v-u)$

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

$F=\frac{m(v-u)}{\Delta t}$

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

$\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s$