Chapter 8, Problem 89P

To determine

The Speed of $2.0\text{ kg}$ ball and the direction of the $3.0\text{ kg}$ ball after collision.

Answer to Problem 89P

The speed of $2.0\text{ kg}$ ball after collision is $5.3434\text{m}/\text{s}$ and the direction of $3.0\text{ kg}$ ball after collision from incident direction is $26.43°$ .

Explanation of Solution

Given:

The mass of ball 1 is $m_1=2.0\text{kg}$ .

The mass of ball 2 is $m_2=3.0\text{kg}$ .

The speed of ball 1 before collision is $u_1=10\text{m}/\text{s}$ .

The speed of ball 2 before collision is $u_2=0\text{m}/\text{s}$ .

The speed of ball 2 after collision is $v_1=4.0\text{m}/\text{s}$ .

The direction of ball 1 after collision from incident direction is ${\theta }_1=30°$ .

Formula Used:

The expression for conservation of momentum is X-direction is given by,

  $m_1{u}_1\cos \left(0°\right)+m_2{u}_2\cos \left(0°\right)=m_1{v}_1\cos \left({\theta }_1\right)+m_2{v}_2\cos \left({\theta }_2\right)$

The expression for conservation of momentum is Y-direction is given by,

  $m_1{u}_1\sin \left(0°\right)+m_2{u}_2\sin \left(0°\right)=m_1{v}_1\sin \left({\theta }_1\right)+m_2{v}_2\sin \left(-{\theta }_2\right)$

The expression for kinetic energy after collision is,

  $K_2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

The expression for kinetic energy before collision is,

  $K_1=\frac{1}{2}{m}_1{u}_1{}^2+\frac{1}{2}{m}_2{u}_2{}^2$

Calculation:

The expression for conservation of momentum is X-direction is calculated as,

  $\begin{array}{c}{m}_1{u}_1\cos \left(0°\right)+m_2{u}_2\cos \left(0°\right)=m_1{v}_1\cos \left({\theta }_1\right)+m_2{v}_2\cos \left({\theta }_2\right)\\ 2\cdot 10+3\cdot 0=2\cdot v_1\cdot \cos \left({30}^{\circ }\right)+3\cdot 4\cdot \cos \left({\theta }_2\right)\\ 20=2v_1\cdot \frac{\sqrt{3}}{2}+12\cos \left({\theta }_2\right)\\ \cos \left({\theta }_2\right)=\frac{20-\sqrt{3}{v}_1}{12}\end{array}$ ....... (1)

The expression for conservation of momentum is Y-direction is calculated as,

  $\begin{array}{c}{m}_1{u}_1\sin \left(0°\right)+m_2{u}_2\sin \left(0°\right)=m_1{v}_1\sin \left({\theta }_1\right)+m_2{v}_2\sin \left(-{\theta }_2\right)\\ 2\cdot 0.+3\cdot 0\cdot 0=2\cdot v_1\cdot \sin \left({30}^{\circ }\right)-3\cdot 4\sin \left({\theta }_2\right)\\ 2v_1\cdot \frac{1}{2}=12\sin \left({\theta }_2\right)\\ \sin \left({\theta }_2\right)=\frac{v_1}{12}\end{array}$ ....... (2)

Square Equation (1) and (2) and add them

  $\begin{array}{c}{\left(\cos \left({\theta }_2\right)=\frac{20-\sqrt{3}{v}_1}{12}\right)}^2\\ {\left(\sin \left({\theta }_2\right)=\frac{v_1}{12}\right)}^2\\ {\cos }^2\left({\theta }_2\right)+{\sin }^2\left({\theta }_2\right)={\left(\frac{20-\sqrt{3}{v}_1}{12}\right)}^2+{\left(\frac{v_1}{12}\right)}^2\\ {\left(\frac{20-\sqrt{3}{v}_1}{12}\right)}^2+{\left(\frac{v_1}{12}\right)}^2=1\end{array}$

Further simplify the above,

  $\begin{array}{c}{\left(\frac{20-\sqrt{3}{v}_1}{12}\right)}^2+{\left(\frac{v_1}{12}\right)}^2=1\\ \frac{400+3v_1^2-40\sqrt{3}{v}_1}{{\left(12\right)}^2}+\frac{v_1^2}{{\left(12\right)}^2}=1\\ 400+3v_1^2-40\sqrt{3}{v}_1+v_1^2=144\\ v_1^2-10\sqrt{3}{v}_1+100-36=0\end{array}$

Further simplify the above,

  $\begin{array}{c}{v}_1^2-10\sqrt{3}{v}_1+64=0\\ v_1=\frac{-\left(-10\sqrt{3}\right)\pm \sqrt{\left({\left(10\sqrt{3}\right)}^2-4\cdot 64\cdot 1\right)}}{2\cdot 1}\\ v_1=\frac{10\sqrt{3}\pm \sqrt{\left(4\left(25\cdot 3\right)-4\cdot 64\right)}}{2}\\ v_1=\frac{10\sqrt{3}\pm 2\sqrt{\left(75-64\right)}}{2}\end{array}$

This implies,

  $v_1=5\sqrt{3}\pm \sqrt{11}$

So,

  $v_1=11.9766\text{m}/\text{s}$ and $v_1=5.3434\text{m}/\text{s}$

The value of total kinetic energy ffter collision when $v_1=11.9766\text{m}/\text{s}$ is calculated as,

  $\begin{array}{c}{K}_2=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2\\ =\frac{1}{2}\cdot 2\cdot {\left(11.9766\right)}^2+\frac{1}{2}\cdot 3\cdot {\left(4\right)}^2\\ =143.43+24\\ =167.43\text{ J}\end{array}$

The value of total kinetic energy after collision when $v_1=5.3434\text{m}/\text{s}$ is calculated as,

  $\begin{array}{c}{K}_2=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2\\ =\frac{1}{2}\cdot 2\cdot {\left(5.3434\right)}^2+\frac{1}{2}\cdot 3\cdot {\left(4\right)}^2\\ =28.55+24.00\\ =52.55\text{ J}\end{array}$

The value of total kinetic energy before collision is calculated as,

  $\begin{array}{c}{K}_1=\frac{1}{2}{m}_1{u}_1{}^2+\frac{1}{2}{m}_2{u}_2{}^2\\ =\frac{1}{2}\cdot 2\cdot {\left(10\right)}^2+\frac{1}{2}\cdot 3\cdot 0\\ =100\text{ J}\end{array}$

When $v_1=11.9766\text{m}/\text{s}$ kinetic energy after collision increases, this is not possible. So $v_1=5.3434\text{m}/\text{s}$ is correct answer.

From Equation (2),

  $\begin{array}{c}\sin \left({\theta }_2\right)=\frac{v_1}{12}\\ =\frac{5.3434}{12}\\ {\theta }_2={\sin }^{-1}\left(\frac{5.3434}{12}\right)\\ =26.43°\end{array}$

Conclusion:

Therefore, the speed of $2.0\text{ kg}$ ball after collision is $5.3434\text{m}/\text{s}$ and the direction of $3.0\text{ kg}$ ball after collision from incident direction is $26.43°$ .