Linear momentum ratio when charged particles approach

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2 replies · 2K views
missrikku
Hello again!

I am having trouble with the following linear momentum problem:

A has a mass mA = 9.11x10^31 kg
B has a mass mB = 1.67x10^-27 kg

They are attracted to each other by some electrical force.

Say they are released from rest with a distance between them of Di = 3.0x10^-6 m

When their distance is decreased to D = 1.0x10^-6 m, what is the ratio of

a) their linear momentums

b) speeds

c) kinetic energies

-----

i know:

p = mv

so for a)

I am looking for: Pe/Pp
mAvA/mBvB

I have mA and mB, but not vA or vB

I know that they are released from rest, but how do you find vA and vB?

Can I use V^2 = Vi^2 + 2a(D-Di) ? If so, how do I find the a?

Thanks. I just need to be pointed in the right direction.
 

Discussion

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Actually, the phrasing of the questions is a major hint. Notice that the question does not ask for the momenta or velocities of the separate particles but only for the ratios. Also the problem asks first for the ratio of the momenta- if you find that first, the others are easy.
Since the force attracting the two particles to one another is "internal", momentum is conserved- that is, MAvA+ MBvB is a constant.. Since the two particles were initially motionless, what was the initial momentum of each and what was the total momentum (the "constant")?
 


Hi there,

To solve this problem, we can use the conservation of momentum principle, which states that the total momentum of a closed system remains constant. In this case, the system is closed since there are no external forces acting on A and B.

a) To find the ratio of their linear momentums, we can use the equation p = mv. Since they are released from rest, their initial momentums are both zero. When their distance is decreased to D = 1.0x10^-6 m, their final momentums can be calculated using the equation p = mv again. So the ratio of their momentums will simply be the ratio of their masses, since their velocities are the same. Therefore, Pe/Pp = mA/mB = 9.11x10^31/ 1.67x10^-27 = 5.46x10^58.

b) To find the ratio of their speeds, we can use the equation v = sqrt(2K/m), where K is the kinetic energy and m is the mass. Since the initial kinetic energy is zero, we can only calculate the final kinetic energy using the equation K = (1/2)mv^2. Plugging in the values, we get K = (1/2)(1.67x10^-27)(v^2). To find the speed, we can rearrange the equation to v = sqrt(2K/m). So the ratio of their speeds will be vA/vB = sqrt(mB/mA) = sqrt(1.67x10^-27/9.11x10^31) = 1.38x10^-16.

c) Finally, to find the ratio of their kinetic energies, we can use the equation K = (1/2)mv^2. We already have the values for m and v from the previous calculations, so we can plug them in to get the kinetic energies for A and B. The ratio of their kinetic energies will be Kp/Ke = (1/2)mA*vA^2/(1/2)mB*vB^2 = (mA*vA^2)/(mB*vB^2) = (mA/mB)*(vA/vB)^2 = 9.11x10^31/1.67x10^-27 * (1.38x10^-16)^2 = 2.84x10^34.

I hope