Motion of a charged particle in an electric field

[Nemo's]

Homework Statement

Each of the nuclei below are accelerated from rest through the same potential difference.Which one completes the acceleration with the lowest speed?
¹H₁ ⁴He₂ ⁷Li₃
⁹Be₄

Homework Equations

F=ma=QE

The Attempt at a Solution

v∝a∝Q/m so for lowest v Q/m lowest is 1 so answer is ¹H₁

Homework Statement

The following particles are each accelerated from rest through the same potential difference. Which one completes acceleration with the greatest momentum?
α β proron neutron

Homework Equations

F=ma=QE

The Attempt at a Solution

mv∝QE so biggest mass and biggest charge give biggest momentum so the answer is α

 

[mfb]

¹H₁ has the largest Q/m.
While a larger Q/m leads to a larger velocity, those are not proportional to each other.

mv∝QE is not true.

 

[technician]

The energy gained by each particle = q x V
They all pass through the same V so the energy gained for each in order is 1, 2, 3 and 4 units.
This is KE so for each use 0.5mv^2 with the appropriate masses to find the values of velocity, v.

 

[Nemo's]

¹H₁ has the largest Q/m.

Yes that's right I don't know why I reversed it. My bad. So it must be the Lithium.

While a larger Q/m leads to a larger velocity, those are not proportional to each other. mv∝QE is not true.

Why's that?

 

[Nemo's]

The energy gained by each particle = q x V
They all pass through the same V so the energy gained for each in order is 1, 2, 3 and 4 units.
This is KE so for each use 0.5mv^2 with the appropriate masses to find the values of velocity, v.

Thank you so much the energy approach is a lot
easier

 

[mfb]

Why's that?

Why should the equation be true?
Energy and momentum are not the same.

 

[Nemo's]

O.K so force (EQ) is directly proportional to the rate of change of momentum(Δmv/t) This one is true right?

 

[mfb]

$F=\frac{dp}{dt} = m \frac{dv}{dt} = ma$ (for constant m), indeed. As time is not given, I don't think this helps here.

 

[Nemo's]

I see. So using proportionality here can't solve this problem.
O.K so by using the energy method I can get the speed of each particle then I could multiply each speed by the corresponding mass to get the momentum?
The problem is it's a multiple choice question and I shouldn't be wasting so much time on a single question in an exam. Isn't there a faster way ?

 

[mfb]

O.K so by using the energy method I can get the speed of each particle then I could multiply each speed by the corresponding mass to get the momentum?

You can do this, or just compare the different particles ("larger mass leads to whatever, that leads to ..., ...")

 

[technician]

The charge of the particles is1, 2, 3 4...so the energy gained is also 1, 2 3 and 4 (relatively!)
The mass of the particles is 1, 4, 7 and 9.
so using energy =1/2 m v² gives
1 = 1/2 x 1 x v²
2 = 1/2 x 4 x v²
3 = 1/2 x 7 x v²
4 = 1/2 x 9 x v²

if you sort this out you should see that for 1 v² = 2
for 2 v² = 1
for 3 v² = 6/7
for 4 v² = 8/9

 

[Nemo's]

O.K Thank you so much :D