Question

A hammer of mass M falls from a height h repeatedly to drive a pile of mass m into the ground. The hammer makes the pile penetrate in the ground to a distance d in single blow. Opposition to penetration is given by

• A.
• B. + (M + m)g
• C.
• D. $\frac { m ^ { 2 } g h } { ( m + M ) d } - ( M + m ) g$

Answer 1

Answer: (B)

+ (M + m)g

Answer 2

Using law of conservation of total momentum, we get

(M + m)v = M $\sqrt { 2 g h }$ s

or v = $\frac { M \sqrt { 2 g h } } { M + m }$

Let opposition to penetration be F then

Total work done = change in K.E.

i.e., F = $\frac { 1 } { 2 }$ (M + m) $\frac { \left( \mathrm { M } ^ { 2 } + 2 \mathrm { gh } \right) } { \mathrm { d } ( \mathrm { M } + \mathrm { m } ) ^ { 2 } }$ + (M + m)g

= $\frac { \mathrm { M } ^ { 2 } \mathrm { gh } } { ( \mathrm { M } + \mathrm { m } ) \mathrm { d } } + ( \mathrm { M } + \mathrm { m } ) \mathrm { g }$