Question

A neutron of mass 1.67 ×10^-27 kg moving with a velocity 1.2×10^7 ms^-1 collides head-on with a deuteron of mass 3.34×10^-27 kg initially at rest. If the collision were perfectly elastic, what would be the speed of deuteron after the collision?

Answer 1

Answer on Question #76287, Physics / Mechanics | Relativity

A neutron of mass $1.67 \times 10^{\wedge} - 27 \, \mathrm{kg}$ moving with a velocity $1.2 \times 10^{\wedge} 7 \, \mathrm{ms}^{\wedge} - 1$ collides head-on with a deuteron of mass $3.34 \times 10^{\wedge} - 27 \, \mathrm{kg}$ initially at rest. If the collision were perfectly elastic, what would be the speed of deuteron after the collision?

Solution

$$m_n = 1.67 \cdot 10^{-27} \, \mathrm{kg}$$ $$v_{1n} = 1.2 \cdot 10^7 \, \mathrm{m/s}$$ $$m_d = 3.34 \cdot 10^{-27} \, \mathrm{kg}$$ $$v_{1d} = 0$$ $$v_{2d} - ?$$

To answer this question we should use two laws:

1) The law of conservation of momentum:

$$m_n \times v_{1n} + m_d \times v_{1d} = m_n \times v_{2n} + m_d \times v_{2d}$$

as $v_{1d} = 0$ (deuteron was initially at rest.)

then $m_d \times v_{1d} = 0$

and the equation of the law of conservation of momentum is:

$$m_n \times v_{1n} = m_n \times v_{2n} + m_d \times v_{2d}$$

2) Energy conservation law:

$$\frac{m_n \times v_{1n}^2}{2} + \frac{m_d \times v_{1d}^2}{2} = \frac{m_n \times v_{2n}^2}{2} + \frac{m_d \times v_{2d}^2}{2}$$

as $v_{1d} = 0$, then $m_d \times v_{1d}^2 = 0$

and the equation of energy conservation law is:

$$\frac{m_n \times v_{1n}^2}{2} = \frac{m_n \times v_{2n}^2}{2} + \frac{m_d \times v_{2d}^2}{2}$$

Solve the system of two equations:

$$\left\{ \begin{array}{l} 1.67 \times 10^{-27} \times 1.2 \times 10^7 = 1.67 \times 10^{-27} \times v_{2n} + 3.34 \times 10^{-27} \times v_{2d} \\ \frac{1.67 \times 10^{-27} \times 1.2 \times 10^7}{2} = \frac{1.67 \times 10^{-27} \times v_{2n}}{2} + \frac{3.34 \times 10^{-27} \times v_{2d}}{2} \end{array} \right.$$ $$\left\{ \begin{array}{l} 1.2 \times 10^7 = v_{2n} + 2 \times v_{2d} \\ (1.2 \times 10^7)^2 = v_{2n}^2 + 2 \times v_{2d}^2 \end{array} \right.$$ $$\left\{ \begin{array}{l} v_{2n} = 1.2 \times 10^7 - 2 \times v_{2d} \\ (1.2 \times 10^7)^2 = (1.2 \times 10^7 - 2 \times v_{2d})^2 + 2 \times v_{2d} \end{array} \right.$$ $$\left\{ \begin{array}{l} v_{2n} = 1.2 \times 10^7 - 2 \times v_{2d} \\ (1.2 \times 10^7)^2 = (1.2 \times 10^7)^2 - 2 \times 2 \times 1.2 \times 10^7 \times v_{2d}^2 + 4 \times v_{2d}^2 + 2 \times v_{2d} \end{array} \right.$$ $$\left\{ \begin{array}{l} v_{2n} = 1.2 \times 10^7 - 2 \times v_{2d} \\ 6 \times v_{2d}^2 = 2 \times 2 \times 1.2 \times 10^7 \times v_{2d} \end{array} \right.$$ $$\left\{ \begin{array}{l} v_{2n} = 1.2 \times 10^7 - 2 \times v_{2d} \\ v_{2d} = 8 \times 10^6 \end{array} \right.$$ $$\left\{ \begin{array}{l} v_{2n} = -4 \times 10^6 \\ v_{2d} = 8 \times 10^6 \end{array} \right.$$ $$v_{2d} = 8 \times 10^6 \text{ m/s}$$

$v_{2n} = -4 \times 10^6 \text{ m/s}$ (minus before value of speed means that a neutron changed its direction on the opposite after perfectly elastic collision).

Answer: $8 \times 10^6 \text{ m/s}$