91Ó°ÊÓ

1. $M_1=16.7×{10}^{-27}\mathrm{kg}\mathrm{and}{\mathrm{v}}_1=\left(6.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}$
2. $M_2=8.35×{10}^{-27}\mathrm{kg}\mathrm{and}{\mathrm{v}}_2=\left(-8.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$
3. $M_3=11.7×{10}^{-27}\mathrm{kg}$
   

Using the formula of conservation of momentum, find the linear momentum of the third particle of mass$\mathbf{11}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$. From this momentum, find the velocity of the third particle. Then, the amount of kinetic energy appeared in transformation by adding K.E of all particles.

Formulae are as follow:

$$\begin{gathered} m_i{v}_i=m_f{v}_f \\ k=\frac{1}{2}m{v}^2 \end{gathered}$$

Here, m is mass, v is velocity and k is kinetic energy.

Mass of the nucleus is,

$m_i=M_1+M_2+M_3$

Substitute the values and solve as:

$$\begin{gathered} m_i=16.7×{10}^{-27}\mathrm{kg}+8.35×{10}^{-27}\mathrm{kg}+11.7×{10}^{-27}\mathrm{kg} \\ {m}_i=36.75×{10}^{-27}\mathrm{kg} \end{gathered}$$

As the nucleus is at rest before the transformation, the velocity of nucleus is,

$v_i=0\frac{\mathrm{m}}{\mathrm{s}}$

Now, according to conservation of momentum,

$$\begin{gathered} m_i{v}_i=m_f{v}_f \\ {m}_i{v}_i=m_1{v}_1+m_2{v}_2+m_3{v}_3 \\ 0=m_1{v}_1+m_2{v}_2+m_3{v}_3 \\ {m}_3{v}_3=-m_1{v}_1+m_2{v}_2 \end{gathered}$$

Substitute the values and solve as:

role="math" localid="1661313325635" $$\begin{gathered} m_3{v}_3=-\left(16.7×{10}^{-27}\mathrm{kg}\right)\left(6.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}-\left(8.35×{10}^{-27}\mathrm{kg}\right)\left(-8.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j} \\ {m}_3{v}_3=\left(-1.0×{10}^{-19}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}+\left(0.67×{10}^{-27}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j} \end{gathered}$$

Therefore, the linear momentum of the third particle of mass role="math" localid="1661313375641" $11.7×{10}^{-27}\mathrm{kg}$is $m_3{v}_3=\left(-1.0×{10}^{-19}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}+\left(0.67×{10}^{-27}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$.

The linear momentum of the third particle of mass$11.7×{10}^{-27}\mathrm{kg}$is,

$m_3{v}_3=\left(-1.0×{10}^{-19}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}+\left(0.67×{10}^{-27}\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}$

$v_3=\frac{\left(-1.0×{10}^{-19}\mathrm{kg}{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)\hat{i}+\left(0.67×{10}^{-19}\mathrm{kg}{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)\hat{j}}{\left(11.7×{10}^{-27}\mathrm{kg}\right)}$

Finding resultant of the numerator and dividing it by mass,

$v_3=1.03×{10}^7\frac{m}{s}$

Therefore,

$$\begin{gathered} KE=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2+\frac{1}{2}{m}_3{v}_3^2 \\ KE=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2+\frac{1}{2}{m}_3 \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} KE=\left\{\begin{array}{c}\frac{1}{2}\left(16×{10}^{-27}\mathrm{kg}\right){\left[\left(6.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{i}\right]}^2+\frac{1}{2}\left(8.35×{10}^{-27}\mathrm{kg}\right){\left[\left(-8.00×{10}^6\frac{\mathrm{m}}{\mathrm{s}}\right)\hat{j}\right]}^2\\ \frac{1}{2}\left(11.7×{10}^{-27}\mathrm{kg}\right){\left(1.03×{10}^7\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\end{array}\right\} \\ KE=1.188×{10}^{-12}~1.19×{10}^{-12}\mathrm{J} \end{gathered}$$.

Therefore, the amount of kinetic energy appeared in transformation is $1.19×{10}^{-12}\mathrm{J}$.

Using the conservation of momentum, the velocity of one of the objects from the velocities of the other objects of the system can be found. From this, the kinetic energy of the system is found.