Question

A hammer of mass M strikes a nail of mass m with velocity of $um{s}^{-1}$ and drives it a metre in to fixed block of wood. The average resistance of the wood to the penetration of the nail is

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

Answer 1

The correct option is D $\left(\frac{M^2}{M+m}\right)\frac{u^2}{2}$

When the hammer, $M$ strikes the nail, $m$ both move with velocity $v$ after the collision.

By applying momentum conservation we get

$v=\frac{Mu}{M+m}$

Now this phenomena can be treated as a $system$ of mass $m+M$ is moving with $initial$ speed $u$ and is being stopped

during a travel of $1meter$

So we need calculate the $retardation$ first which is $a=\frac{v^2}{2s}=\frac{v^2}{2\times 1}=\frac{v^2}{2}$

Now the resistive force is $mass\times retardation=(m+M)v^2/2$

putting value of $v$ we get the force as $\frac{M^2{u}^2}{2(M+m)}$