Question

A block C of mass m is moving with velocity $v_0$ and collides elastically with block A of mass m and connected to another block B of mass 2m through spring with spring constant K. What is the value of K, if $x_0$ is compression of spring, when velocity of A and B is same?

• A. $\frac{m{v}_0^2}{x_0^2}$
• B. $\frac{m{v}_0^2}{2x_0^2}$
• C. $\frac{3m{v}_0^2}{2x_0^2}$
• D. $\frac{2m{v}_0^2}{3x_0^2}$

Answer 1

The correct option is D $\frac{2m{v}_0^2}{3x_0^2}$

Conservation of momentum
$p_i=p_{f}m{v}_0=mv+2mvv=\frac{v_0}{3}$
Applying conservation of energy
$\frac{1}{2}m{v}_2^0=\frac{1}{2}K{x}_0^2+\frac{1}{2}\left(3m\right){\left(\frac{v_0}{3}\right)}^2\frac{1}{2}m{v}_0^2=\frac{K{x}_0^2}{2}+\frac{1}{2}\frac{v_0}{3}m{v}_0^2=K{x}_0^2+\frac{v_0^2}{3}K=\frac{2m{v}_0^2}{3x_0^2}$