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 #65498, Physics / Mechanics | Relativity

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.

Solution:

Before collision

After collision

Kinetic energy of the proton before collision $E_1^k = \frac{m_p v_1^2}{2}$ , where $m_p$ is proton's mass.

Kinetic energy of the proton after collision $E_2^k = \frac{m_p v_2^2}{2} = \frac{16}{25} E_1^k$ .

So $\frac{m_p v_2^2}{2} = \frac{16}{25} \cdot \frac{m_p v_1^2}{2}$ , and we may evaluate $v_2 = \frac{4}{5} v_1$ .

Assuming that the system is closed we may use the momentum conservation law:

$m_{p}\vec{v}_{1} + \vec{0} = m_{p}\vec{v}_{2} + m_{x}\vec{v}_{x}$ , where $m_{x}$ is unknown particle's mass.

Or in scalar form, $m_p v_1 = -m_p v_2 + m_x v_x$ .

$$m _ {p} v _ {1} = - m _ {p} \frac {4}{5} v _ {1} + m _ {x} v _ {x} \Rightarrow \frac {9}{5} m _ {p} v _ {1} = m _ {x} v _ {x} \Rightarrow v _ {x} = \frac {9}{5} \frac {m _ {p}}{m _ {x}} v _ {1}$$

Next, according to the law of conservation of energy, $E_1^k = E_2^k + E_x^k$ .

$$\frac {m _ {p} v _ {1} ^ {2}}{2} = \frac {m _ {p} v _ {2} ^ {2}}{2} + \frac {m _ {x} v _ {2} ^ {2}}{2} \Rightarrow \frac {m _ {p} v _ {1} ^ {2}}{2} = \frac {m _ {p} \left(\frac {4}{5} v _ {1}\right) ^ {2}}{2} + \frac {m _ {x} \left(\frac {9}{5} \frac {m _ {p}}{m _ {x}} v _ {1}\right) ^ {2}}{2} \Rightarrow$$ $$m _ {p} v _ {1} ^ {2} = \frac {1 6}{2 5} m _ {p} v _ {1} ^ {2} + \frac {8 1}{2 5} m _ {x} \left(\frac {m _ {p}}{m _ {x}}\right) ^ {2} v _ {1} ^ {2} \Rightarrow \frac {9}{2 5} m _ {p} = \frac {8 1}{2 5} m _ {x} \left(\frac {m _ {p}}{m _ {x}}\right) ^ {2} \Rightarrow$$ $$\frac {m _ {p}}{m _ {x}} = 9 \left(\frac {m _ {p}}{m _ {x}}\right) ^ {2} \Rightarrow \frac {m _ {p}}{m _ {x}} = \frac {1}{9} \Rightarrow \frac {m _ {x}}{m _ {p}} = 9$$

Answer:

$$\frac {m _ {x}}{m _ {p}} = 9$$

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