Question

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

1. The magnitude of the impulse on the hand from the target,$\left|\mathrm{J}\right|=9.1\mathrm{N}·\mathrm{s}$.
2. The magnitude of the average force on the hand from the target, ${\mathrm{F}}_{\mathrm{avg}}=1.8\times {10}^3\mathrm{N}$.

Step-by-step solution

Understanding the given information

The mass of hand,$\mathrm{m}=0.70\mathrm{kg}$.

The speed of hand is,${\mathrm{v}}_1=13\mathrm{m}/\mathrm{s}$.

The time of contact, $\mathrm{t}=5.0\mathrm{ms}\left(\frac{1\times {10}^{-3}\mathrm{s}}{1\mathrm{ms}}\right)=5.0\times {10}^{-3}\mathrm{s}$.

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.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ ∆\mathrm{P}=\mathrm{m}\left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}.\mathrm{t} \end{gathered}$$

(a) Calculation for the impulse

The equation of the impulse-momentum theorem is,

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ =\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ =0.70\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-13\mathrm{m}/\mathrm{s}\right) \\ =-9.1\mathrm{N}·\mathrm{s} \end{gathered}$$

So, the magnitude of impulse is,

$\left|\mathrm{J}\right|=9.1\mathrm{N}·\mathrm{s}$

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

We have,

$\mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times ∆\mathrm{t}$

So,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{∆\mathrm{t}} \\ =\frac{9.1\mathrm{N}·\mathrm{s}}{5\times {10}^{-3}\mathrm{s}} \\ =1.8\times {10}^3\mathrm{N} \end{gathered}$$