Question

An unstable atomic nucleus of mass $\mathbf{17}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$ initially at rest disintegrates into three particles. One of the particles, of mass $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$, moves in the y direction with a speed of $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Another particle, of mass $\mathbf{8}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$, moves in the x direction with a speed of $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Find (a) the velocity of the third particle and (b) the total kinetic energy increase in the process.

Short answer

1. The velocity of third particle is $\left(9.33\hat{i}-8.33\hat{j}\right)\times {10}^6\mathrm{m}/\mathrm{s}$.
2. The total increase in kinetic energy in process is $439\mathrm{fJ}$.

Step-by-step solution

Identification of given data

The mass of unstable nucleus is $M=17\times {10}^{-27}\mathrm{kg}$

The mass of particle along y direction is $m_1=5\times {10}^{-27}\mathrm{kg}$

The speed of particle along y direction is $v_y=6\times {10}^6\mathrm{m}/\mathrm{s}$

The mass of particle along x direction is$m_2=8.40\times {10}^{-27}\mathrm{kg}$

The speed of particle along x direction is $v_x=4\times {10}^6\mathrm{m}/\mathrm{s}$

The conservation of momentum is used to find the final speed of each disk.

Determination of velocity of third particle

(a)

Apply the conservation of momentum to find the velocity of third particle.

$m_2{v}_x-m_1{v}_y-\left(M-m_1-m_2\right)V=0$

Substitute all the values in the above equation.

$$\begin{gathered} \left[\begin{array}{c}\left(8.40\times {10}^{-27}kg\right)\\ \left(4\times {10}^{6}m/s\right)\hat{i}\end{array}\right]-\left[\begin{array}{c}\left(5\times {10}^{-27}kg\right)\\ \left(6\times {10}^{6}m/s\right)\hat{j}\end{array}\right]-\left[\left(\begin{array}{c}17\times {10}^{-27}\mathrm{kg}\\ -5\times {10}^{-27}\mathrm{kg}\\ -8.40\times {10}^{-27}\mathrm{kg}\end{array}\right)\right]V=0 \\ V=\left(9.33\hat{i}-8.33\hat{j}\right)\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the velocity of third particle is $\left(9.33\hat{i}-8.33\hat{j}\right)\times {10}^6\mathrm{m}/\mathrm{s}$.

Determination of total kinetic energy increase in the process

(b)

The speed of the third particle is given as:

$$\begin{gathered} V=\sqrt{{\left(9.33\times {10}^6\mathrm{m}/\mathrm{s}\right)}^2+{\left(-8.33\times {10}^6\mathrm{m}/\mathrm{s}\right)}^2} \\ V=12.51\times {10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

The total increase in kinetic energy in process is given as:

$K=\frac{1}{2}{m}_1{v}_y^2+\frac{1}{2}{m}_2{v}_x^2+\frac{1}{2}\left(M-m_1-m_2\right)V^2$

Substitute all the values in the above equation.

Therefore, the total increase in kinetic energy in process is $439\mathrm{fJ}$.