Question

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 73.2-kg man just before contact with the ground has a speed of 3.19 m/s. (a) In a stiff-legged landing he comes to a halt in 4.74 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.184 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 1.70 kg. The bullet has a speed of 645 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

Answer 1

ONE QUESTION AT A TIME AS PER THE RULES

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1)

(a)

we will use impusle - momentum theorem

force * time = change in momentum

F = 73.2 * 3.19 / 4.74e-3

F = 49263.3 N

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(b)

F = 73.2 * 3.19 / 0.184

F = 1269 N

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(c)

the force from ground acts upward and force of gravity acts downwards

so,

so,

force applied in part (a) from ground = 49263.3 + 73.2 * 9.8 = 49980.7 N

force applied in part (b) from ground = 1269 + 73.2 * 9.8 = 1986.42 N