Question

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

Boxer 2 stops the punch with a force of 45,000 N.
How much time did it take for boxer 1's hand to stop moving?

Answer 1

0.001 s

Step-by-step explanation:

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

`F=(\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=(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=(m(v-u))/(F)=((5)(0-9))/(-45,000)=0.001 s`