Question

A particle of charge q and mass m travels through a potential difference of v from rest the final momentum of the particle is

Answer 1

Hello Dear.

Here is the answer---

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Given Conditions ⇒ 

Charge = q

Potential Difference = v

Let the Final Velocity of the Charge be x.

Initial Velocity (u) = 0
[Since the body is at rest]


Using the Formula,

 Potential = Work Done/Charge
⇒ Work Done = v × q ------eq(i)

Now, By the Work -Energy Theorem,

Work done = Change in Kinetic Energy
                  = 1/2 × m × (x² - u²)
                  =(1/2) × m × (x² - 0)
              W = (x²m)/2  ----eq(ii)


From eq(i) and eq(ii)

   v × q = (x²m)/2
     x² = $\frac{2vq}{m}$
    x = $\sqrt{ \frac{2vq}{m} }$
∴ Velocity = $\sqrt{ \frac{2vq}{m} }$


Now, Using the Formula,

 Momentum(p) = Mass × Velocity
                    p = m × $\sqrt{ \frac{2vq}{m} }$
                    p = $\sqrt{2vqm}$

Thus, the Final Momentum of the Particle is √2vqm.


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Hope it helps.

Have a Good Day.