Find the initial separation between the particles

[nckaytee]

One particle has a mass of 3.00 10-3 kg and a charge of +7.80 µC. A second particle has a mass of 6.00 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00 10-3 kg particle is 130 m/s. Find the initial separation between the particles.

Kq^2/r = Kq^2/d + 1/2mV^2

I keep getting r = .177 but it is not correct :-(

 

[LowlyPion]

One particle has a mass of 3.00 10-3 kg and a charge of +7.80 µC. A second particle has a mass of 6.00 10-3 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and when the separation between them is 0.100 m, the speed of the 3.00 10-3 kg particle is 130 m/s. Find the initial separation between the particles.

Kq^2/r = Kq^2/d + 1/2mV^2

I keep getting r = .177 but it is not correct :-(

But both particles are free to move, and they both get accelerated in opposite directions by the same repulsive force no?

 

[nckaytee]

Would I not include the " + 1/2mV^2 " then?

 

[LowlyPion]

Would I not include the " + 1/2mV^2 " then?

But doesn't that include the MV²/2 for both particles?

 

[nckaytee]

so don't multiply it by 1/2?

 

[LowlyPion]

so don't multiply it by 1/2?

Not quite.

You need to treat the particles according to their mass. They have different mass no?

 

[nckaytee]

Kq^2/r = Kq^2/d + 1/2m1V^2 + 1/2m2V^2 ?

 

[LowlyPion]

Kq^2/r = Kq^2/d + 1/2m1V^2 + 1/2m2V^2 ?

I think so.

Now think in terms of F=ma to determine what the speeds would be. If the m is doubled then what is the acceleration for the same force on the doubled mass? If the Velocity over the same time is 130 on the smaller mass then what would the V of the more massive particle be?

 

[nckaytee]

So, the V would be halved! Thank you so much, Again! :-)

 

[LowlyPion]

So, the V would be halved! Thank you so much, Again! :-)

No problem.

Cheers.