Answer: 3. The puck B remains at the point of collision.
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.
The initial momentum is due only to puck B, as Puck A is at rest.
The final momentum is given by the sum of the momenta of both pucks, so we can write the following equation:
mA*viA = (mA * vfA) + (mB * vfB)
As mA = mB = m, we can simplify the former equation as follows:
viA = vfA + vfB (1)
Now, we also know, that the collision was an elastic collision, so total kinetic energy must be conserved too:
½ m viA²= ½ m vfA² + ½ m vfB²
Simplifying on both sides, we finally have:
viA² = vfA² + vfB² (2)
Now, if we square both sides of (1), we get:
viA² = (vfA + vfB)²= vfA² + 2* vfA * vfB +vfB² (3)
As the right side in (2) and (3) must be equal each other (as the left sides do), the only choice is that either vfA or vfB, be zero.
As we are told that puck A (initially stationary) after the collision, moves, the only possible choice is that puck B remain at rest in the point of collision, after the collision, exchanging his speed with puck A.
