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
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