For this problem on the topic of momentum and collisions, we are told that two blocks A and B are forced together, which compresses a spring that is laying between them. The spring has a spring constant of 2500 newtons per meter, and the spring is compressed by three centimeters from its equilibrium length. We want to know the speeds of block A. And B after the blocks move off and the spring returns to its original length. Now they are essentially on a frictionless surface, and we'll use the conservation of momentum and the conservation of energy here to find the speed at which the blocks move off. Now the conservation of momentum tells us that the initial momentum of the system is equal to the final momentum, and the conservation of energy tells us that the initial energy is equal to the final energy. Now the initial momentum is zero since the two blocks are addressed and the initial kinetic energy E. I. Is equal to the energy stored in the spring, which is the elastic potential energy a half K data X squared. Now, since P I is equal to P F is equal to zero, we have M A V. A plus M B. The B is equal to zero, which means that M A V. A. The massive block a tendered speed is equal to minus the mass of block, B times its speed after it moves off. Now from the conservation of energy, the initial energy they put the final energy, which means that the initial elastic potential energy half K. Delta X squared. Where delta X. Is. The compression of the spring is converted all into kinetic energy. A half M. The A squared goes to block A plus a half M B. The b squared goes to block B. So simplifying this, we get okay, DELTA X squared is equal to M B squared over M A times V B squared less M B Vb squid. And so if we simplify further, we get okay. Delta X squared is equal to M B squared over M A plus MB all into V B squared, which means that we can rearrange and we can solve for Phoebe. And so we get Vb to be the square root of K. Delta X squared divided by M B into one plus M B of M A. And we can write an expression for V A. V A is simply minus M B of uh M A times B B. So we can calculate VB first by substituting the relevant values into that equation. And we get VB is equal to the square root of 2500 newtons commit to which is K. And the compression of the spring three centimeters or three times 10 to the minus two m squared, divided by three kg, which is the mass of B into one plus and be over M. A, which is three kg divided by the massive block A one kg. And so we get VB to be 0.43 the zero or 0.433 meters a second. Now that we have VB, we can find V. A. And V A. Is equal to minus and be three kgs over M. A, which is one kg times zero point 4330 meters per second, which gives us the speed of A. After the collision to be minus one 0.3 meters the second.