In tae-kwon-do, a hand is slammed down onto a target at a speed of 14 m/s and comes to a stop during the 3.0 ms collision. Assume that during the impact the hand is independent of the arm and has a mass of 0.90 kg. What are the magnitudes of the (a) impulse and (b) average force on the hand from the target

A) The impulse is 11,7 kg.m/s .  B) The average force on the hand is 3900N

Explanation:

Givens:

Hand's mass (m)= 0,90kg

Hand's initial velocity (vi)= 14m/s

Time of collision (t)= 3ms

A) Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time. We use the formula:

/F/ = m. /Δv/

Where:

F= Force of Impulse

m= Mass

Δv= change in speed ( vf-vi)

Using that formula we get:

/F/= 0,90kg . / (0m/s-14m/s) /

/F/=0,90kg . 14m/s

/F/ = 11,7 kg. m/s

*note that (vf) is 0 because the hand stops in the action, so it's final

velocity = 0

B)The average force is equal to the change in the momentum over the change in time. We use the formula:

/F/= m. (Δv) : Δt

Where:

F= the average force from the target

m= the mass of the hand

Δv= hange in speed (vf/vi)

Δt= change in time

*if we look closely, we find the we alredy have half of the equation because we have alredy calculated m.(Δv) in point A.

We boil the equation down to:

/F/= Impulse : Δt

/F/= 11,7kg.m/s : 0,003s

/F/= 3900N

*we use 0,003s as our time because the given time was 3ms.

*the final result is expressed in Newtons because our final result ends up beeing kg.m/s² = N

i think its c if i'm wrong