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Chapter 9: Q28P (page 249)

In tae-kwon-do, a hand is slammed down onto a target at a speed of13 m/sand comes to a stop during the 5.0 mscollision. Assume that during the impact the hand is independent of the arm and has a mass of0.70 kg . What are the magnitudes of the (a) Impulse and (b) Average force on the hand from the target?

Short Answer

Expert verified
  1. The magnitude of the impulse on the hand from the target,J=9.1N路s.
  2. The magnitude of the average force on the hand from the target, Favg=1.8脳103N.

Step by step solution

01

Understanding the given information

The mass of hand,m=0.70kg.

The speed of hand is,v1=13m/s.

The time of contact, t=5.0ms1脳10-3s1ms=5.0脳10-3s.

02

Concept and formula used in the given question

We can use the equation of impulse-momentum theorem to calculate the impulse, and then we can use this value of impulse in the equation of average force to calculate its value.

J=鈭�P鈭�P=mv2-v1J=Favg.t

03

(a) Calculation for the impulse

The equation of the impulse-momentum theorem is,

J=鈭�P=m脳v2-v1=0.70kg脳0m/s-13m/s=-9.1N路s

So, the magnitude of impulse is,

J=9.1N路s

04

(b) Calculation for the Average force on the hand from the target

We have,

J=Favg脳鈭�t

So,

Favg=J鈭�t=9.1N路s5脳10-3s=1.8脳103N

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