Question

A hammer of mass $M$ strikes a nail of mass $m$ with a velocity of $u$ $m{s}^{-1}$ and drives it $a$ meter into fixed block of wood. The average resistance of wood during the penetration of nail is :

• A. $\left[\frac{M}{m+M}\right]\frac{u^2}{2a}$
• B. $\left[\frac{M^2}{(m+M{)}^2}\right]\frac{u^2}{2a}$
• C. $\left[\frac{M^2}{m+M}\right]\frac{u^2}{2a}$
• D. $\left[\frac{M+m}{m}\right]\frac{u^2}{2a}$

Answer 1

Answer: (C)

$\left[\frac{M^2}{m+M}\right]\frac{u^2}{2a}$

When the hammer strikes the nail, both of them will go together. So, the collision can be treated as inelastic.
So just after the hammer strikes the nail (which is initially at rest), their velocity will become

$v=\frac{M\times u+m\times 0}{M+m}=\frac{Mu}{M+m}$

So their kinetic energy after collision will become $\frac{1}{2}(M+m)(\frac{Mu}{M+m}{)}^2=\frac{M^2{u}^2}{2(M+m)}$

After the penetration, the nail and the hammer come to rest. So their kinetic energy is zero.

So the work done $W=|\Delta K|=|K_f-K_i|=\frac{M^2{u}^2}{2(M+m)}$
It penetrated $a$ m.

While penetration, assuming the resistive force $f$ to be constant, work done by the force $W=fa$

$\therefore fa=\frac{M^2{u}^2}{2(M+m)}$

$\therefore f=\left[\frac{M^2}{m+M}\right]\frac{u^2}{2a}$