Question

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10.5 m/s and comes to a stop during the 2.13 ms collision. Assume that during the impact the hand is independent of the arm and has a mass of 0.400 kg. What is the magnitude of the impulse?

Answer 1

Answer:

4.2 kg.m/s

Explanation:

The impulse is the vector greatness that relates force and time, and the momentum is the vector greatness that relates the mass and the velocity of a body. The impulse theorem relates the impulse with the momentum (Q) by the expression:

I = ΔQ

The momentum is Q = mxv, where m is the mass and v the velocity. How the hand stops, the final velocity must be 0, so:

I = mxvf - mxvi

I = 0.4x0 - 0.4x10.5

I = - 4.2 kg.m/s

The minus signal indicates the direction, so the magnitude is 4.2 kg.m/s