Question

A hammer of mass M strikes a nail of mass m with velocity of u m/s and drives it 'a' metres into fixed

block of wood. The average resistance of the wood to the penetration of nail is​

Answer 1

Answer: [M²u² / {2a(M+m)}]

Explanation:

Given data:

The mass of hammer = M

The mass of nail = m  

The initial velocity of hammer = u m/s

Distance =  a meters

To find: the average resistance of wood to the penetration of the nail

Solution:

Step 1:

As the hammer strikes the nail, the collision between them can be considered as inelastic.

Let the initial velocity of nail be "v".

After the collision, the velocity becomes,

v = [(M*u)+(m*0)] / [M+m]

⇒ v = [M*u] / [M+m]

Step 2:

The final kinetic energy after the collision

= ½ * [M+m] * [(M*u)/(M+m)]²

= [(Mu)²] / [2*(M+m)]

Since after the penetration, the nail and hammer will come to rest, therefore, so, their K.E. will become zero.

∴ Work done, W = ∆K = Kf – Ki = [(Mu)²] / [2*(M+m)] …. (i)

Step 3:

Let’s assume the average resistive force of the wood while penetrating be constant and denoted as “f”.

So,

Work done by the force = resistive force * distance = f * a …. (ii)

From (i) & (ii), we get

f * a = [(Mu)²]/[2*(M+m)]

⇒ f = [M²/(M+m)] * [u²/2a]

Thus, the average resistance of the wood to the penetration of nail is [M²u²/{2a(M+m)}].