A drop hammer of mass 120 kg falls 2.5 m onto a pile of mass 250 kg and drives it 70 mm into the ground. Calculate:

a.The velocity with which the hammer strikes the pile (use the conservation of energy principle)

b.The velocity immediately after the impact(use the conservation of momentum principle,
assuming the hammer does not rebound on impact)

c. The average ground resistance (use work-energy method)

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Sure! Let's go through each part step by step.

a. To calculate the velocity with which the hammer strikes the pile, we can use the principle of conservation of energy. This principle states that the total energy before the collision is equal to the total energy after the collision, assuming no energy is lost due to friction or other factors.

The total energy before the collision is the potential energy of the hammer at its starting height. The potential energy (PE) is given by the formula:

PE = mgh

where m is the mass of the hammer, g is the acceleration due to gravity (usually taken as 9.8 m/s^2), and h is the height from which the hammer falls.

In this case, the mass of the hammer is 120 kg, and the height is 2.5 m. Plugging these values into the formula, we get:

PE = 120 kg × 9.8 m/s^2 × 2.5 m = 2940 J

The total energy after the collision is the kinetic energy of the moving hammer. The kinetic energy (KE) is given by the formula:

KE = (1/2)mv^2

where m is the mass of the hammer and v is the velocity of the hammer.

To find the velocity, we can equate the initial potential energy to the final kinetic energy:

PE = KE

2940 J = (1/2) × 120 kg × v^2

Simplifying the equation, we get:

2940 J = 60 kg × v^2

v^2 = 2940 J / 60 kg

v^2 = 49 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 7 m/s

Therefore, the velocity with which the hammer strikes the pile is approximately 7 m/s.

b. To calculate the velocity immediately after the impact, we can use the principle of conservation of momentum. This principle states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.

The momentum (p) of an object is given by the formula:

p = mv

where m is the mass of the object and v is its velocity.

Before the impact, the hammer has a momentum of 120 kg × 7 m/s = 840 kg·m/s. The pile has a momentum of 250 kg × 0 m/s = 0 kg·m/s since it is at rest.

After the impact, the hammer and the pile become one system, so their total momentum is equal to the momentum of this system. Let's assume the velocity of the combined system is v'.

Using the conservation of momentum, we have:

Total momentum before impact = Total momentum after impact

(120 kg × 7 m/s) + (250 kg × 0 m/s) = (120 kg + 250 kg) × v'

840 kg·m/s = 370 kg × v'

Dividing both sides by 370 kg, we get:

v' = 840 kg·m/s / 370 kg

v' ≈ 2.27 m/s

Therefore, the velocity immediately after the impact is approximately 2.27 m/s.

c. To calculate the average ground resistance, we can use the work-energy method. The work done by the ground resistance (W) is equal to the change in kinetic energy of the moving hammer.

The change in kinetic energy is given by:

ΔKE = KEf - KEi

where KEf is the final kinetic energy after impact and KEi is the initial kinetic energy before impact.

The initial kinetic energy (KEi) is given by:

KEi = (1/2)mv^2

where m is the mass of the hammer and v is its velocity.

Substituting the values, we have:

KEi = (1/2) × 120 kg × (7 m/s)^2
= 2,940 J

The final kinetic energy (KEf) is given by:

KEf = (1/2)mv'^2

where m is the combined mass of the hammer and the pile, and v' is the velocity immediately after impact.

The combined mass is the sum of the masses of the hammer and the pile:

m = 120 kg + 250 kg
= 370 kg

Substituting the values, we have:

KEf = (1/2) × 370 kg × (2.27 m/s)^2
≈ 963 J

Now we can calculate the change in kinetic energy:

ΔKE = KEf - KEi
= 963 J - 2940 J
≈ -1977 J

The negative sign indicates that energy has been lost due to the work done by the ground resistance.

The work done by the ground resistance is equal to the negative change in kinetic energy. So, we have:

W = -ΔKE
≈ 1977 J

Finally, the average ground resistance (R) is equal to the work done divided by the displacement:

R = W / d

where W is the work done and d is the displacement. In this case, the displacement is given as 70 mm, which is equal to 0.07 m.

Substituting the values, we have:

R ≈ 1977 J / 0.07 m
≈ 28242.86 N

Therefore, the average ground resistance is approximately 28242.86 Newtons.