Question

A pile driver of mass 300 kg is used to drive a pile of mass 500 kg vertically into the ground. The pile driver falls freely through a distance of 54.0 m, rebounding with a velocity relative to the pile and equal to the relative velocity immediately before impact. Determine: the velocity of the driver immediately before impact: the velocity of he pile immediately after the impact: the depth of penetration of the pile after impact given that the ground resisting force is constant and equal to 115 kN: (4 marks) the time taken for the penetration. (a) (b) (c) (d) (4 marks) (7 marks) (5 marks)​

Answer 1

a. The velocity of the driver immediately before impact is given by:

v = sqrt(2gh) ≈ 34.14 m/s

where g is the acceleration due to gravity, h is the falling distance of the driver, and v is the velocity of the driver.

b. The velocity of the pile immediately after impact can be found by applying the law of conservation of momentum:

m1v1 + m2v2 = (m1 + m2)v'

where m1 and v1 are the mass and velocity of the driver before impact, m2 and v2 are the mass and velocity of the pile before impact, and v' is the velocity of the pile and driver immediately after impact.

Since the driver falls freely, its initial velocity is equal to v. The velocity of the pile before impact is zero, since it is stationary. Therefore:

(300 kg)(34.14 m/s) + (500 kg)(0 m/s) = (300 kg + 500 kg)v'

Solving for v', we get:

v' ≈ 20.48 m/s

c. The depth of penetration of the pile after impact can be found by applying the work-energy principle:

W = ΔK

where W is the work done by the ground resisting force, ΔK is the change in kinetic energy of the pile and driver, and K = (1/2)(m1 + m2)v'^2 is the final kinetic energy of the system.

The work done by the ground resisting force is:

W = Fd = (115 kN)(d)

where d is the depth of penetration of the pile into the ground.

The change in kinetic energy of the system is:

ΔK = (1/2)(m1 + m2)(v^2 - v'^2)

Substituting the given values, we get:

(115 kN)(d) = (1/2)(300 kg + 500 kg)(34.14^2 - 20.48^2)

Solving for d, we get:

d ≈ 0.728 m

Therefore, the depth of penetration of the pile after impact is about 0.728 m.

d. The time taken for the penetration can be found by applying the equation of motion:

d = (1/2)gt^2

where d is the depth of penetration, g is the acceleration due to gravity, and t is the time taken for the penetration.

Substituting the given values, we

Answer 2

a. By conservation of energy, the potential energy lost by the pile driver is equal to the kinetic energy gained by the pile and the pile driver. Therefore, (1/2) * 300 * v^2 = (1/2) * (300 + 500) * u^2, where v is the velocity of the pile driver immediately before impact and u is the velocity of the pile and pile driver immediately after impact. Solving for v, we get v ≈ 28.9 m/s.

b. By conservation of momentum, the momentum of the pile and the pile driver before impact is equal to the momentum of the pile and pile driver after impact. Therefore, 300 * v = (300 + 500) * u, where v is the velocity of the pile driver immediately before impact and u is the velocity of the pile and pile driver immediately after impact. Solving for u, we get u ≈ 17.3 m/s.

c. The work done by the ground resisting force is equal to the change in kinetic energy of the pile and pile driver. Therefore, (1/2) * (300 + 500) * u^2 = 115000 * d, where u is the velocity of the pile and pile driver immediately after impact, and d is the depth of penetration of the pile after impact. Solving for d, we get d ≈ 0.575 m.

d. The time taken for the penetration is equal to the distance penetrated divided by the average velocity during penetration. The average velocity during penetration is equal to u/2, since the velocity decreases linearly from u to 0 during penetration. Therefore, the time taken for the penetration is equal to d / (u/2) = 0.575 / (17.3/2) ≈ 0.066 s.