91Ó°ÊÓ

Given that block A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface.

Initially, block B is at rest and block A is moving towards it at 2.00 m/s. The blocks are equipped with ideal spring bumpers.

The collision is head-on, so all the motion before and after collision is along a straight line.

Mass of block A ${\mathrm{m}}_{\mathrm{A}}=2\mathrm{kg}$

Mass of block B ${\mathrm{m}}_{\mathrm{B}}=6\mathrm{kg}$

Initial speed of block A ${\mathrm{u}}_{\mathrm{A}}=2\mathrm{m}/\mathrm{s}$

Initial speed of block B ${\mathrm{u}}_{\mathrm{B}}=0$

Conservation of momentum states the if no external force is acting, then total momentum of the system is conserved.

When the springs are compressed fully, the two blocks have same velocity say V.

During collision, momentum is conserved.

So, Initial momentum of blocks = final momentum of blocks

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{V} \\ \left(2\mathrm{kg}\right)\left(2\mathrm{m}/\mathrm{s}\right)+\left(6\mathrm{kg}\right)\left(0\right)=\left(2\mathrm{kg}+6\mathrm{kg}\right)\mathrm{V} \\ 4=8\mathrm{V} \\ \mathrm{V}=0.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Let U be the energy stored in the spring bumpers.

According to conservation of energy

$$\begin{gathered} \frac{1}{2}{\mathrm{m}}_{\mathrm{A}}{{\mathrm{u}}_{\mathrm{A}}}^2=\mathrm{U}+\frac{1}{2}\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{V} \\ \frac{1}{2}\left(2\mathrm{kg}\right)\left(2\mathrm{m}/\mathrm{s}\right)=\mathrm{U}+\frac{1}{2}\left(2\mathrm{kg}+6\mathrm{kg}\right)\left(0.5\mathrm{m}/\mathrm{s}\right) \\ \mathrm{U}=3\mathrm{J} \end{gathered}$$

Hence energy stored in spring bumpers is 3J and velocity of each block at that time is 0.5 m/s in the direction of initial motion of Block A.

Velocity of block A after moving apart $v_A$ is

$$\begin{gathered} {\mathrm{v}}_{\mathrm{A}}=\left(\frac{{\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}\right) \\ =\left(\frac{2\mathrm{kg}-6\mathrm{kg}}{8\mathrm{kg}}\right)\left(2\mathrm{m}/\mathrm{s}\right) \\ =-1\mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity of block A is 1 m/s in opposite direction

Velocity of block B after moving apart ${\mathrm{v}}_{\mathrm{B}}$is

$$\begin{gathered} {\mathrm{v}}_{\mathrm{A}}=\left(\frac{2{\mathrm{m}}_{\mathrm{A}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}\right){\mathrm{u}}_{\mathrm{B}} \\ =\left(\frac{2\left(2\mathrm{kg}\right)}{8\mathrm{kg}}\right)\left(2\mathrm{m}/\mathrm{s}\right) \\ =1\mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity of block B is 1 m/s in initial direction as Block A