# Using $E=mc²$ to simplify particle KE calculations?

In an old examination paper I found a question asking for the ratio between the KE of a nucleus and the KE of an alpha particle emitted from the nucleus.

For reference here is the question and answer (it's Question 1 (b) (iii)).

The answer can be found by using the exact values of the masses and velocities and $$KE = (1/2)mv²$$.

However, in the answer key, this (more simple) solution is also given: $$\frac{E_k\text{ of Ce}}{E_k\text{ of }\alpha}=\frac{m_dv_d^2}{m_\alpha v_\alpha^2}=\frac{m_d}{m_\alpha}\times\bigg(\frac{m_\alpha}{m_d}\bigg)^2=\frac{m_\alpha}{m_d}=\frac{1}{35}$$

I can't figure out what they are doing here.

In particular, I don't see where the substitution for $$v²$$ with the inverted masses comes from.

Is this some use of $$E=mc²$$?

I'm sure the answer is just some basic mathematics, but it escapes me.

Any help would be much appreciated.

• Please don't post formulas as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. – user191954 Oct 23 '18 at 13:29

This is just conservation of momentum, no need for $$E=mc^2$$. In the problem you have a stationary nucleus that decays, so initially the system has no momentum. After the decay you need the particles to have equal and opposite momentum, so considering magnitudes: $$p_{\alpha}=p_{d} \implies m_{\alpha}v_{\alpha}=m_{d}v_{d} \implies \frac {v_d^2}{v_{\alpha}^2}=\frac{m_{\alpha}^2}{m_d^2}$$
Conservation of Momentum $$p=mv$$ is the key for understanding this equation.
The Ce nucleus starts initially at rest, i.e. without any Momentum. After Decay of the particle, one has nonvanishing momenta for the nucleus; one of the Ce nucleus AFTER the Decay $$p_{Ce}$$ and one of the Alpha particle $$p_\alpha$$. Add These momenta up and set the total Momentum $$p_{tot} = p_{Ce}+p_{\alpha}$$ to Zero (total Momentum in the final decay is the same as the total Momentum initially). Then you will have the answer.