Chapter 7, Problem 45P

To determine

The mass of the unknown atom.

Answer to Problem 45P

The mass of the unknown atom is $40.75\text{ u}$ .

Explanation of Solution

Given:

The mass of neon atom is $m=20.0\text{ u}$ .

Formula used:

The expression for conservation of energy is,

  $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

Here, $m_1$ is the mass of the object 1, $m_2$ is the mass of object 2, $u_1$ is the initial velocity of object 1, $u_2$ is the initial velocity of the object 2, $v_1$ is the final velocity of object 1, and $v_2$ is the final velocity of the object 2.

Calculation:

The representation of the given situation is shown below.

  

Sketch the triangle for the direction of momentum of the ball as shown below.

  

The angle $\beta$ is,

  $\begin{array}{l}\beta =180°-\theta -\alpha \\ \beta =180°-55.6°-50°\\ \beta =74.4°\end{array}$

Apply law of sines for the triangle.

  $\begin{array}{l}\frac{m_1{v}_1}{\sin \alpha }=\frac{m_1{u}_1}{\sin \beta }=\frac{m_2{v}_2}{\sin \theta }\\ \frac{m_1{v}_1}{\sin \alpha }=\frac{m_1{u}_1}{\sin \beta }\\ v_1=\frac{\sin \alpha }{\sin \beta }{u}_1\\ v_1=\frac{\sin 50°}{\sin 74.4°}{u}_1\\ v_1=0.80u_1\end{array}$

  $\begin{array}{l}\frac{m_1{u}_1}{\sin \beta }=\frac{m_2{v}_2}{\sin \theta }\\ v_2=\frac{\sin \theta }{\sin \beta }\frac{m_1{u}_1}{m_2}\\ v_2=\frac{\sin 55.6°}{\sin 74.4°}\frac{\left(20.0\text{ u}\right)u_1}{m_2}\\ v_2=17.13\frac{u_1}{m_2}\end{array}$

Apply the law of conservation of energy.

  $\begin{array}{l}\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ \frac{1}{2}\left(20.0\text{ u}\right)u_1^2+\frac{1}{2}{m}_2\left(0\right)=\frac{1}{2}\left(20.0\text{ u}\right){\left(0.80u_1\right)}^2+\frac{1}{2}{m}_2{\left(17.13\frac{u_1}{m_2}\right)}^2\\ \left(10.0\text{ u}\right)u_1^2=\left(6.4\text{ u}\right)u_1^2+146.7\frac{u_1^2}{m_2}\\ \frac{146.7}{m_2}=3.6\text{ u}\\ m_2=40.75\text{ u}\end{array}$

Conclusion:

Thus, the mass of the unknown atom is $m_2=40.75\text{ u}$ .