(6.0x10^-22, -1.40x10^-21, 0) kg*m/s Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction. (3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s After the interaction (2.6x10^-21, 1.40x10^-21, 0) kg*m/s and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be 3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of (6.0x10^-22, -1.40x10^-21, 0) kg*m/s