Question

From what minimum height h must the system be released when a spring is unstretched so that after perfectly inelastic collision (e=0) with ground, B may be lifted off the ground (spring constant =k)

• A. mg/(4k)
• B. 4 mg/k
• C. mg/(2k)
• D. None

Answer 1

The correct option is B

4 mg/k

$u=\sqrt{2}gh$

To lift the block B, the elongation x must be 2 mg/k.

Applying work - energy theorem between A and B, we get

$0-\frac{1}{2}m{u}^2=-\frac{1}{2}mgx-\frac{1}{2}k{x}^2$

$\Rightarrow \frac{1}{2}m{u}^{2}mg\frac{2mg}{k}+\frac{1}{2}k\frac{4m^2{g}^2}{k}$

$\therefore u^2=\frac{4m{g}^2}{k}=2gh\Rightarrow h=\frac{4mg}{k}$