Physics Problem 4

Question

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 7.0 ms collision. Assume that during the impact the hand is independent of the arm and has a mass of 0.60 kg. What are the magnitudes of the (a) impulse and (b) average force on the hand from the target?

Answer 1

The momentum of the hand as it hits the target is 14 m/s * 0.6 kg = 8.4 kg-m/s.  After the collision the momentum is 0.  Impulse is change in momentum, so the impulse is -8.4 kg-m/s.

 

That impulse occurs during a period of 7 x 10^-3 seconds.

 

Force is impulse / time, so the force is -8.4 kg-m/s / 7 x 10^-3 s = -1.2 x 10³ N.

 

The negative sign means that the force is exerted in a direction opposite to the motion of the hand.