Need help doing 5-7. Calculate the average decelerating impact force if a person

Question

Need help doing 5-7. Calculate the average decelerating impact force if a person falling with a terminal velocity of 62.5 m/sec is decelerated to zero velocity over a distance of 1 m. Assume that the person's mass is 70 kg and that she lands flat on her hack so that the area of impact is 0.3 m^2. Is this force below the level for serious injury? (For body tissue, this is about 5 times 10^6 dyn/cm^2.) A boxer punches a 50-kg bag. Just as his list hits the bag, it travels at a speed of 7 m/sec. As a result of hitting the bag, his hand comes to a complete stop. Assuming that the moving part of his hand weighs 5 kg, calculate the rebound velocity and kinetic energy of the bag. Is kinetic energy conserved in this example? Why? (Use conservation of momentum.)

Explanation / Answer

5-7 solution

In a closed system, momentum and total energy are always conserved.

a) First, let's use the law of conservation of momentum to solve for the rebound velocity of the bag.

Before the bag is punched, the total momentum is equal to the following:

= (mass of the boxer's hand)(initial velocity of the boxer's hand) + (mass of bag)(initial bag velocity)
= (5Kg)(7m/s) + (50Kg)(0m/s)
= 35Kgm/s

After the bag is punched, the total momentum is also equal to 35Kgm/s (remember: momentum is conserved, so you get

35Kgm/s = (mass of boxer's hand)(final velocity of boxer's hand) + (mass of bag)(final bag velocity)
35Kgm/s = (5Kg)(0m/s) + (50Kg)(final bag velocity)
35Kgm/s = 50Kg(final bag velocity)
(35Kg/50Kg)m/s = final bag velocity
0.7m/s = final bag velocity

b). Kinetic energy = 1/2 (m)(v^2)
KE = 1/2(50Kg) (0.7m/s X0.7m/s)
KE = 25Kg (0.49m^2/s^2)
KE = 12.25 Joules

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