Question

Blocks A (mass 3.00 kg ) and B (mass8.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block A.

A. Find the maximum energy stored in the spring bumpers.

B. Find the velocity of block

A when the energy stored in the spring bumpers is maximum.

C. Find the velocity of block

B when the energy stored in the spring bumpers is maximum.

D. Find the velocity of block

A after they have moved apart.

E. Find the velocity of

B after they have moved apart.

Answer 1

Frictionless movement means conservation of energy:

A) $E_{k}=E_{elMAX}\Leftrightarrow \frac{m_{A}v_{A}^{2}}{2}=E_{elMAX}=13.5\: J$

(the total kinetic energy-that of mass A-is completely transformed in potential elastic energy)

B) v_A=0 (all the kinetic energy of mass A in stranformed in potential elastic energy stored in the spring bumpers)

C) v_B=0 The same reason at that at point B).

D)+E)

$\left\{\begin{matrix} m_{A}v_{A}=m_{A}v_{fA}-m_{B}v_{fB}\\ \frac{m_{A}v_{A}^{2}}{2}=\frac{m_{A}v_{fA}^{2}}{2}+\frac{m_{B}v_{fB}^{2}}{2} \end{matrix}\right.$

Indices "f" means speeds after collision and "relaxation" of the springs (no elastic potenatial energy exists)

Solve the system and get:

$v_{fB}=\frac{2v_{A}}{\frac{m_{B}}{m_{A}}+1}\approx 1.64\: m/s$

$v_{fA}=-v_{A}+\frac{m_{B}}{m_{A}}v_{fB}\approx 1.36\: m/s$