Question

Estimate the force with which a karate master strike a board, assuming the hand’s speed at the moment of Impact is $10.0m{s^{ - 1}}$ , decreasing to $1.00m{s^{ - 1}}$ during a $0.002s$ time internal of contact between the hand and the board. The mass of his hand and arm is $1.00kg$

Answer 1

we will use the concept of impulse-momentum theorem which states that for an object the change in momentum will be equal to the impulse applied. This will give us the relation between the force applied for a duration of time, velocity, and mass. Using this relation we will find the force with which a karate master strikes a board.
$Ft = mv$
Where, $F$ is the force applied, $t$ denotes time, $m$ is the mass of the body , $v$ is the speed of impact.

Complete Step By Step Answer:
momentum of an object is the product of mass and velocity. It tells us the measurement of mass in motion.
$\rho = mv$ where $m$ is the mass of the body , $v$ is the speed of the body.
For change of momentum we write $\Delta \rho = m\Delta v$ where $\Delta v$ is change in speed.
Impulse is the quantity of the total force acting on a body in a period of time.
$J = F\Delta t$ where $F$ is the force applied, $t$ is the time.
We can create a direct relationship between how a force operates on an object over time and the object's motion according to the impulse-momentum theorem.
$Ft = m\left( {{v_f} - {v_i}} \right)$ where ${v_f}$ and ${v_i}$ are the final and initial velocities.
$\Rightarrow F\left( {0.002s} \right) = \left( {1.00kg} \right)\left( {1.00m{s^{ - 1}} - 10.0m{s^{ - 1}}} \right)$
$\Rightarrow F = - 4500N$
Hence, the force with which a karate master strikes a board is $F = 4500N$ .

Note:
The impulse-momentum theorem is equivalent to Newton's second law of motion.
According to the impulse-momentum theorem $J = \Delta \rho$
For constant mass $F\Delta t = m\Delta v$
For variable mass $Fdt = vdv + vdm$
SI unit of impulse is $Newton\sec$
SI unit of momentum is $\dfrac{{kg\left( {meter} \right)}}{{\sec }}$