Question

A block C of mass $m$ is moving with velocity $V_S$ and collides elastically with block A of mass $m$ which is connected to another block B of mass $2m$ through a spring of spring constant k. What is k if $x_0$ is the compression of spring when velocity of A and B is same?

• A. $\frac{m{V}_S^2}{x_0^2}$
• B. $\frac{m{V}_S^2}{2x_0^2}$
• C. $\frac{3}{2}\frac{m{V}_S^2}{x_0^2}$
• D. $\frac{2}{3}\frac{m{V}_S^2}{x_0^2}$

Answer 1

The correct option is D $\frac{2}{3}\frac{m{V}_S^2}{x_0^2}$

After collision $V_c=0$
By momentum conservation,
$m{V}_S=(m+2m)uu=\frac{1}{3}{V}_S$
Applying energy conservation,
$\frac{1}{2}m{V_S}^2=\frac{1}{2}(m+2m)u^2+\frac{1}{2}k{x_0}^2\frac{1}{2}m{V_S}^2=\frac{3m}{2}(\frac{1}{3}{V}_s{)}^2+\frac{1}{2}k{x_0}^2\frac{1}{2}(m{V_S}^2-\frac{{m{V}_S}^2}{3})=\frac{1}{2}k{x_0}^{2}k=\frac{2m{V_S}^2}{3{x_0}^2}$