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sergij07 [2.7K]
4 years ago
11

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. Assum

e 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
Physics
1 answer:
GrogVix [38]4 years ago
8 0

Answer:

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

<u>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:</u>

<em />

<em>/F/ = m. /Δv/</em>

<u>Where:</u>

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

<em>/F/ = 11,7 kg. m/s</em>

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

velocity = 0

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

<em>/F/= m. (Δv) : Δt</em>

<u>Where:</u>

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

<em>/F/= 3900N</em>

*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 <em>kg.m/s²</em> = N

<u />

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