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?

a. -10.8kgm/s

b.  1542.86kgm/s^2 or 1.52kN

Explanation:

a. Given  initial velocity , final velocity , duration of collision as 0.007s and the mass as , the impulse on the hand can be calculated as:

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:

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

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where are the vectors ?