Question

A person slaps her leg with her hand, bringing her hand to rest in 2.50 milliseconds from an initial speed of 4.00 m/s. (a) What is the average force exerted on the leg, taking the effective mass of the hand and forearm to be 1.50 kg? (b) Would the force be any different if the woman clapped her hands together at the same speed and brought them to rest in the same time? Explain why or why not.

Answer 1

Answer:

(a) 2400N

(b) different

Explanation:

2.5 milliseconds = 0.0025 seconds

a) The momentum of the moving arm-hand of 1.5 kg and at 4m/s speed is the product of the mass and the speed

$p = mv = 1.5 * 4 = 6 kgm/s$

After the slap, the arm-hand comes to rest, making the change in momentum of 6 - 0 = 6 kgm/s. This change in momentum is due to the impulse generated by the force within t = 0.0025 seconds.

So the average force is:

$F = \frac{\Delta p}{\Delta t} = \frac{6}{0.0025} = 2400 N$

b) If there are 2 hands clapping together at same speed and mass, their total momentum must be double of that. This makes the average force 2 times as before.