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n tae-kwon-do, a hand is slammed down onto a target at a speed of 12 m/s and comes to a stop during the 7.0 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?

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opudodennis

Answer:

a. -10.8kgm/s

b.  1542.86kgm/s^2 or 1.52kN

Explanation:

a. Given  initial velocity [tex]v_i=12m/s[/tex] , final velocity [tex]v_f=0\ m/s[/tex] , duration of collision as 0.007s and the mass as [tex]m=0.90kg[/tex], the impulse on the hand can be calculated as:

[tex]J=p_f-p_i\\\\=mv_f-mv_i\\\\=-10.8\ kg.m/s[/tex]

Hence the impulse on the hand is -10.8 kgm/s

b.  From a above we know the impulse on the hand as -10.8 kgm/s and the duration of the impulse as 0.007s:

-We use these values to calculate the average force on the hand:

[tex]F_{avg}=\frac{J}{\bigtriangle t}\\\\={-10.8kgm/s}{0.007s}\\\\=-1542.86kgm/s^2[/tex]

Hence the average force on the hand is  1542.86kgm/s^2 or 1.52kN