Question

two blocks $A$ and $B$ of masses $m$ and $2m$, respectively are connected by a spring of force constant $k$. Them masses are moving to the right with uniform velocity $\nu$ each, the heavier mass leading the lighter one. The spring is in the natural length during this motion. Block $B$ collides head-on with a third block $C$ of mass $2m$, at rest, the collision is completely inelastic. Calculate the maximum compression of the spring.

Answer 1

Using conservation of momentum

${MV}_0+2{MV}_0=4{MV}_1+{MV}_0$ Velocity here won't change immediately

$3{MV}_0-{MV}_0=4{MV}_1$

$2{MV}_0=4{MV}_1$

$V_1=\frac{V_0}{2}$

Using conservation of momentum before collision.

${MV}_0+2{MV}_0=5M\times V$

${3MV}_0=5MV$

$V_0=\frac{5V}{3}$

$V=\frac{3V_0}{5}$

applying same law alter collision.

$\frac{1}{2}{MV}_0^2+\frac{1}{2}\times 4M\times V_0^2=\frac{1}{2}\times 5M\times V^2+\frac{1}{2}{Kx}^2$

$\frac{1}{2}{MV}_0^2+\frac{1}{2}\times 4M\times V_0^2=\frac{1}{2}\times 5M\times \frac{9}{25}{V}_0^2+\frac{1}{2}{Kx}^2$

$\frac{{MV}_0^2}{2}(1+1-\frac{45}{25})=\frac{1}{2}{Kx}^2$

$\frac{{MV}_0^2}{2}\times \frac{1}{5}=\frac{1}{2}{Kx}^2$

${MV}_0^2=5{Kx}^2$

$x^2=\frac{{MV}^2}{5K}$

$x=\sqrt{\frac{{MV}_0^2}{5K}}$