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

Answer 1

The correct option is B $\frac{m^{2}gh}{(M+m)d}+(M+m)g$

Using law of conservation of total momentum, we get

$(M+m)v=M\sqrt{2gh}$

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

Let opposition to penetration be F then

Total work done $=$ change in K.E.

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

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