When they are far apart, the momentum of a proton is 3.2 ✕ 10−21, 0, 0 kg · m/s as it approaches another proton that is initially at rest. the two protons repel each other electrically, without coming close enough to touch. when they are once again far apart, one of the protons now has momentum 2.6 ✕ 10−21, 1.40 ✕ 10−21, 0 kg · m/s. at that instant, what is the momentum of the other proton? (express your answer in vector

(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