Question

A proton undergoes a head on elastic collision with a particle of unknown mass which is initially at rest and rebound with 16/25 of its initial kinetic energy.Calculate the ratio of the unknown mass with respect to the mass of the proton.

Answer 1

Answer on Question #66741-Physics-Mechanics-Relativity

A proton undergoes a head on elastic collision with a particle of unknown mass which is initially at rest and rebounds with 16/25 of its initial kinetic energy. Calculate the ratio of the unknown mass with respect to the mass of the proton.

Solution

From the conservation of energy:

$$E _ {p} = E _ {p} ^ {\prime} + E$$ $$\frac {16}{25} E _ {p} = E _ {p} ^ {\prime}$$ $$\left(\frac {v ^ {\prime}}{v}\right) ^ {2} = \frac {16}{25} \rightarrow \frac {v ^ {\prime}}{v} = \frac {4}{5}.$$ $$\frac {25 - 16}{25} E _ {p} = E$$ $$\frac {m V ^ {2}}{m _ {p} v ^ {2}} = \frac {9}{25} \rightarrow \frac {m}{m _ {p}} = \frac {9}{25} \frac {v ^ {2}}{V ^ {2}}.$$

From the conservation of momentum:

$$m _ {p} v = m V - m _ {p} v ^ {\prime}$$ $$V = \frac {m _ {p}}{m} (v + v ^ {\prime}) = \frac {m _ {p}}{m} \left(v + \frac {4}{5} v\right) = \frac {9}{5} v \frac {m _ {p}}{m}$$

Thus,

$$\frac {m}{m _ {p}} = \frac {9}{25} \left(\frac {m}{m _ {p}}\right) ^ {2} \frac {25}{81} = 9 \left(\frac {m}{m _ {p}}\right) ^ {2}$$ $$\frac {m}{m _ {p}} = \frac {1}{9}$$

Answer: $\frac{1}{9}$

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