Given info : a pile of MASS 500kg is driven into ground by dropping hammer freely, having mass of 318 KG through height of 2.7m. If pile is driven into ground, by 0.15m.

To find : the average resistance of soil is..

solution : mass of hammer, m = 318 kg

it is freely dropping, so initial velocity of hammer is zero.

velocity of hammer just before striking the pile, v = √(2gh)

here, g = 9.81 m/s² , h = 2.7 m

so, v = √(2 × 9.81 × 2.7) ≈ 7.28 m/s

now after striking the pile, hammer and pile both move together into the ground.

so, common velocity of them, v'

using law of conservation of linear momentum,

mv + 0 = (m + M)v'

⇒(318 × 7.28) = (318 + 500) × v'

⇒v' = (31.8 × 7.28)/(31.8 × 500) = 2.83 m/s

now KINETIC energy of both and WORK done while falling through a distance 0.15 m in the soil, will be observed by the soil.

so, kinetic energy of both, K = 1/2 (m + M)v'²

= 1/2 × (318 + 500) × (2.83)²

= 3275.6 J

and work done = (m + M) × g × h'

= (318 + 500) × 9.81 × 0.15

= 1203.7 J

now total energy gained by soil = 3275.6 + 1203.7 = 4479.3 J

from conservation of energy THEOREM,

4479.3 J = resistance of soil × 0.15 m

resistance of soil = 4479.3/0.15 = 29862N