Question

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.

Answer 1

Answer on Question #65602 – Physics - Mechanics - Relativity

Condition:

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:

The law of conservation of energy:

$$E_p = E_p' + E_x$$

where $E_p$ is energy of proton before the collision: $E_p = \frac{m_p V_p^2}{2}$,

$E_p'$ is energy of proton after the collision: $E_p' = \frac{m_p V_p'^2}{2}$,

$E_x$ is energy of particle after the collision

$$E_p' = \frac{16}{25} E_p \rightarrow \frac{m_p V_p'^2}{2} = \frac{16}{25} \frac{m_p V_p^2}{2} \rightarrow V_p'^2 = \frac{16}{25} V_p^2 \rightarrow V_p' = \frac{4}{5} V_p$$ $$E_x = E_p - E_p' = E_p - \frac{16}{25} E_p = \frac{9}{25} E_p \rightarrow \frac{m_x V_x^2}{2} = \frac{9}{25} \frac{m_p V_p^2}{2} \rightarrow m_x V_x^2 = \frac{9}{25} m_p V_p^2$$

The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision:

$$m_p V_p = m_p V_p' + m_x V_x \rightarrow m_x V_x = m_p (V_p - V_p') = m_p (V_p - \frac{4}{5} V_p) = \frac{1}{5} m_p V_p$$ $$\left\{ \begin{array}{l} (1) m_x V_x^2 = \frac{9}{25} m_p V_p^2 \\ (2) m_x V_x = \frac{1}{5} m_p V_p \end{array} \right. \rightarrow \begin{array}{c} (1) \\ (2) \end{array} \coloneqq V_x = \frac{9}{5} V_p$$

From (2): $\frac{m_x}{m_p} = \frac{V_p}{5 V_x} = \frac{V_p}{5 \cdot \frac{9}{5} \cdot V_p} = \frac{1}{9}$

Answer: $\frac{m_x}{m_p} = \frac{1}{9}$

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