Please provide a detailed, handwritten answer. Thank you. Billiard balls A and B, of equal mass, move at right angles and meet at the origin ofan xy coordinate system as shown in Fig. 7-36. Initially ball A is moving along the y axisat +2.0 m/s, and ball B is moving to the right along the x axis with speed +3.7 m/s.After the collision (assumed elastic), ball B is moving along the positive y axis (Fig. 7-36)with velocity v. What is the final direction of ball A, and what are the speeds of the twoballs?Figure 7-36BVB = 3.7 m/sProblem 46. (Ball A after the collision is not shown.)+yVBB0[VA = 2.0 m/sA+x

Think of this one as a conceptual question: If these balls are equal mass and there is a collision there will be a vector conservation of momentum. Rephrase the question as: "Ball A is going north at a speed of 'x'. Ball B is going east at a speed of 'y'. A and B have the same mass." We are told that after the collision Ball B is now going north. So, since they are the same mass, and since Ball B is now moving in the exact same direction that Ball A originally was, we can think of Ball B as absorbing all of the vector momentum that was originally in ball A. We know that ball B is now moving at A's original speed ('x'). The remaining system momentum (the original momentum in ball B) is now transferred in ball A. So ball A now has all of ball B's old momentum (that is, heading east at speed 'y'). ,So back to the question you asked, Ball A is now going east (+x direction at an angle of zero). Ball A has a speed of 3.7 and Ball B has a speed of 2.Momentum of a body is given by the product of the mass of the body and the velocity of the body. Mathematically, P = m×v Now when a system undergoes a process in which there is no net force on the system is acting in a particular direction then the total momentum of the system in that direction is conserved. If a system of two bodies is going under a collision under the influence of no any external force then the total momentum will be conserved.