Question

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?

Answer 1

a. -10.8kgm/s

b. 1542.86kgm/s^2 or 1.52kN

Step-by-step explanation:

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

`J=p_f-p_i\\\\=mv_f-mv_i\\\\=-10.8\ kg.m/s`

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:

`F_(avg)=(J)/(\bigtriangle t)\\\\={-10.8kgm/s}{0.007s}\\\\=-1542.86kgm/s^2`

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