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 a fixed block of wood. The average resistance of wood during the penetration of nail is:
A. $ [\dfrac{M}{{m + M}}]\dfrac{{{u^2}}}{{2a}} $
B. $ [\dfrac{{{M^2}}}{{{{(m + M)}^2}}}]\dfrac{{{u^2}}}{{2a}} $
C. $ [\dfrac{{{M^2}}}{{m + M}}]\dfrac{{{u^2}}}{{2a}} $
D. $ [\dfrac{{M + m}}{m}]\dfrac{{{u^2}}}{{2a}} $

Answer 1

Hint: To solve this question, first we will find the velocity of the hammer when it strikes the nail. Then we will find the Kinetic energy after the collision between both of them. And finally we will conclude the work done. That’s how we can find the average resistance of the penetration.

Complete step by step solution:
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 = \dfrac{{M \times u + m \times 0}}{{M + m}} = \dfrac{{Mu}}{{M + m}} $
So, their kinetic energy after collision will become:
 $ \dfrac{1}{2}(M + m){(\dfrac{{Mu}}{{M + m}})^2} = \dfrac{{{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}| = \dfrac{{{M^2}{u^2}}}{{2(M + m)}} $
It penetrated a $ m $ .
While penetrating, assuming the resistive force $ f $ to be constant, work done by the force:
 $ W = fa $
So,
$ \therefore fa = \dfrac{{{M^2}{u^2}}}{{2(M + m)}} \\
   \Rightarrow f = [\dfrac{{{M^2}}}{{m + M}}]\dfrac{{{u^2}}}{{2a}} \\
 $
Hence, the correct option is C. $ [\dfrac{{{M^2}}}{{m + M}}]\dfrac{{{u^2}}}{{2a}} $ .

Note:
There is a relation between kinetic energy and momentum as both the properties are linked with velocity. Momentum gets expressed as a multiplication of velocity and mass, whereas kinetic energy is the product of the square of speed and half of the mass.