A person slaps her leg with her hand, which results in her hand coming to rest i

Question

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.65 milliseconds from an initial speed of 3.75 m/s. What is the magnitude of the a force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.65 kg? contact Number Would the contact force be any different if the woman clapped her hands together, each with an initial speed of 3 75 m / s, if they come to rest in the same time interval as above O No, because the change in momenturn for the first hand will be the same. It depends on the coeffcient of friction between the two hands. O No, because the second hand has zero momentum. O Yes, because the initial momentum of the system ill be different due to the second hand O Yes, because the change in momentum is different

Explanation / Answer

here,

time taken , t = 0.00265 s

initial speed , u = 3.75 m/s

mass , m = 1.65 kg

let the accelration be a

v = u + a * t

0 = 3.75 + a * 0.00265

a = 1415 m/s^2

the force exerted , F = m * a

F = 1.65 * 1415 N

F = 2334.9 N

the magnitude of force is 2334.9 N

when the clapped her hands

the initial momentum of the system will be different due to the seccond hand and the change in momentum is also different

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