# Conservation or energy problem. What am I doing wrong? [closed]

For a given ion with a charge of $$+1$$ being accelerated by $$U$$ volts the distance it travels in a time $$t$$ is $$d$$ given by: $$d = v_0 t + \frac{1}{2} a t^2$$ for $$v_0=0$$ $$a = 2\frac{d}{t^2}$$ and for $$a=0$$ $$v = \frac{d}{t}$$ therefore $$F = m a = m (2\frac{d}{t^2}) = 2 m \frac{d}{t^2}$$ For the energy of the ion $$E_i$$ $$E_i = F_i d = (2 m_i \frac{d}{t^2}) d = 2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$ Since $$E_v = (1 q_e) U$$ where $$q_e$$ is the universal charge constant it follow that: $$E_i = E_v\ \ =>\ \ \ 2 m_i v_i^2 = q_e U\ \ =>\ \ \ v_i^2 = q_e \frac{U}{2m_i}$$ Therefore: $$\biggl[\ \ \ v_i = \sqrt{ \frac{1}{2} \frac{q_e U}{m_i}}\ \ \ \ \Biggl]$$ However for $$v_0 = 0$$ $$\operatorname{KE}_i = \frac{1}{2} m_i v_i^2 \ \ \ \text{and} \ \ \ \operatorname{KE}_v = q_e U$$ and since $$\operatorname{KE}_i = \operatorname{KE}_v$$ is given per definition, since we are talking about the same ion in both cases it follows that

$$\frac{1}{2} m_i v_i^2 = q_e U \ \ \ \ => \ \ \ \ v_i^2 = q_e \frac{U}{\frac{1}2 m_i} = 2 q_e \frac{U}{m_i}$$ therefore

$$\biggl[\ \ \ v_i = \sqrt{ 2 \frac{q_e U}{m_i}}\ \ \ \ \Biggl]$$

What is up with the factor of $$\ \sqrt{2}\ \$$ and $$\ \ \sqrt{\frac{1}{2}}\ \$$ ???

What am I doing wrong?

Thanks!

• Your first method is incorrect, because you assumed both a zero and nonzero acceleration at the same time. Sep 14 at 21:40

This part is incorrect:

$$E_i = F_i d = (2 m_i \frac{d}{t^2}) d = 2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$

Specifically $$2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$

is wrong because by your own omission $$v_i \neq \frac{d}{t}$$

• Therefore $v_i = \sqrt{2 \frac{q_e U}{m_i}}\ \ \$ Thanks! Sep 14 at 22:34
• and $d = v_0 t + \frac{1}{2} a t^2\ \$ should be $d = \frac{1}{2} t ( 2 v_0 + a t)\ \$ and $v = \frac{2 v_0 + a t}{2}\ \$ and if I substitute $a = \frac{d}{t^2}\ \$ then I get $2v = v\ \$ as $v_0\ \$ approaches $0\ \$, so clearly wrong. Thanks! Sep 14 at 22:56

The reason is the way you're making the assumptions of a and v0 to be zero. Either of those is useful on its own but when you do both, you're essentially saying that a=0 and v0=0 at the same time. So then d=0 and F=0 too, so it's a trivial situation where the ion isn't moving and nothing is happening. The odd factors don't actually mean much because they're being multiplied by 0.

• Very helpful. Thanks! Sep 14 at 22:04