Question

To drive a vertical nail of mass \(10\,\text{g}\) into wood through \(9\,\text{cm}\), an iron block of mass \(990\,\text{g}\) is dropped onto it freely from a height of \(10\,\text{m}\). If the collision between the nail and the block is perfectly inelastic, then the force of resistance offered by wood is \[ (g=10\,\text{m s}^{-2}) \]

• A. \(898\,\text{N}\)
• B. \(989\,\text{N}\)
• C. \(1089\,\text{N}\)
• D. \(1198\,\text{N}\)

Hint:

For a perfectly inelastic collision: \[ \boxed{ m_1u_1+m_2u_2=(m_1+m_2)V } \] Then use the work--energy theorem during penetration.

Answer 1

The Correct Option is C

Step 1: Find the speed of the iron block before collision. \[ v=\sqrt{2gh} =\sqrt{2\times10\times10} =10\sqrt2\ \text{m s}^{-1}. \]

Step 2: Use conservation of momentum. Since the collision is perfectly inelastic, \[ 0.99(10\sqrt2) =(0.99+0.01)V. \] Thus, \[ V=9.9\sqrt2\ \text{m s}^{-1}. \]

Step 3: Apply work--energy theorem. Combined mass, \[ m=1\,\text{kg}. \] Initial kinetic energy, \[ K=\frac12(1)(9.9\sqrt2)^2 =98.01\ \text{J}. \] During penetration of \(s=0.09\,\text{m}\), \[ (F-mg)s=K. \] Hence, \[ F=\frac{98.01}{0.09}+10 =1089\ \text{N}. \] Therefore, \[ \boxed{F=1089\ \text{N}} \] Thus, \[ \boxed{(C)} \] is the correct answer.