Is This Equation Correct? p² = p₀² + 2mΔk

[Superstring]

$$p^2=p_0^2+2m \Delta k$$

I derived it using the following:

$$k=\frac{1}{2}mv^2$$

$$p=mv$$

$$\frac{dk}{dv}=mv=p$$

$$\frac{dp}{dv}=m$$

$$\frac{dk}{p}=dv$$

$$\frac{dp}{m}=dv$$

$$\frac{dk}{p}=\frac{dp}{m}$$

$$mdk=pdp$$

$$\int_{k_0}^{k}mdk=\int_{p_0}^{p}pdp$$

$$m\Delta k=\frac{1}{2}(p^2-p_0^2)$$

$$p^2=p_0^2+2m \Delta k$$

As far as I can tell it's correct, but I though I'd get a second opinion.

 

[Bob S]

Since p² = 2mk, where k = ½mv²

mk = ½p²

m (k₂ - k₁) = ½(p₂² - p₁²)

Bob S

 

[K^2]

Isn't it easier to just note this?

$$\Delta k = k - k_0 = \frac{p^2}{2m} - \frac{p_0^2}{2m}$$

Edit: Ninja'd.

 

[kiwakwok]

As well as what Bob S mentioned, I would say your equation is dimensionally valid.
By the way, why you need such an equation? Or what is it mainly used for?

 

[harrylin]

It's correct for speeds much smaller than that of light.

 

[Superstring]

Thanks for the responses :).

By the way, why you need such an equation? Or what is it mainly used for?

As far as uses, it's a great deal simpler to solve for the mass than if you were to use a 3-equation system with p=mv, p₀=mv₀, and Δk=(m/2)(v²-v₀²) if you were only given p, p₀ and Δk.