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Aleks04 [339]
3 years ago
9

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 initiall

y 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 form.)?
Physics
1 answer:
Rina8888 [55]3 years ago
5 0
<span>(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</span>
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