A karate master strikes a board with an initial velocity of 12.0 m/s,decreasing to 3.0 m/s as his hand passes through the board.If the time of contact with the board is 0.002 s,and the mass of the coordinated hand and arm is 1.0 kg,what is the force exerted on the board?
A)7.5E+3 N
B)1.8E-2 N
C)3.0E-2 N
D)4.5E+3 N
E)6.0E+3 N

Question

Quizplus · Accepted Answer

We can use the impulse-momentum theorem to solve this problem. The impulse on the board is equal to the change in momentum of the karate master's hand and arm:

Impulse = Δp = mΔv

where m is the mass of the hand and arm, and Δv is the change in velocity.

From the given data, we have:

m = 1.0 kg
Δv = 12.0 m/s - 3.0 m/s = 9.0 m/s

The time of contact is also given as t = 0.002 s.

The impulse on the board is therefore:

Impulse = mΔv = (1.0 kg)(9.0 m/s) = 9.0 N s

The average force exerted on the board during the time of contact is:

F = Impulse / t = 9.0 N s / 0.002 s = 4500 N

Therefore, the force exerted on the board is 4.5 × 10^3 N, which corresponds to choice D.

A Karate Master Strikes a Board with an Initial Velocity