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Trying to do a physics problem but I can't understand the mathematical logic on why the term m cancels out completely in this equation: $\frac{1}{2}mv_0^2=\frac{1}{2}mv_1^2+mg\cdot 2r$

I understand canceling the first terms where m is multiplied my half, but I dont understand how the whole m can also get cancelled out so that the final result is: $\frac{1}{2}v_0^2=\frac{1}{2}v_1+g\cdot 2r$

CyclotomicField
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bontu
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1 Answers1

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It's because you can factor out the $m$ from the right side of the equation:

$$\frac{1}{2}mv_0^2=m\left(\frac{1}{2}v_1^2+2gr\right)$$

Now, when both sides are divided by $m$, they cancel.

Gordon
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