Chapter 8, Problem 88P

(a)

To determine

The speed $v_2$ of larger mass after collision and angle ${\theta }_2$ .

Answer to Problem 88P

The speed of larger mass after collision is $\sqrt{2}{v}_0$ and the angle is $45°$

Explanation of Solution

Given:

The mass of small object is $m_1=m$ .

The mass of large object is $m_2=2m$ .

The speed of small object before collision is $u_1=3v_0$ .

The speed of large object before collision is $u_2=0$ .

The speed of small object after collision is $v_1=\sqrt{5}{v}_0$ .

Formula Used:

The expression for the vertical angle is given by,

  $\sin {\theta }_1=\sqrt{\frac{{\tan }^2{\theta }_1}{1+{\tan }^2{\theta }_1}}$

The expression for the horizontal angle is given by,

  $\cos {\theta }_1=\sqrt{\frac{1}{1+{\tan }^2{\theta }_1}}$

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

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

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

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

Calculation:

The vertical angle is calculated as,

  $\begin{array}{c}\sin {\theta }_1=\sqrt{\frac{{\tan }^2{\theta }_1}{1+{\tan }^2{\theta }_1}}\\ =\sqrt{\frac{2^2}{1+2^2}}\\ =\sqrt{\frac{4}{5}}\\ =\frac{2}{\sqrt{5}}\end{array}$

The horizontal angle is calculated as,

  $\begin{array}{c}\cos {\theta }_1=\sqrt{\frac{1}{1+{\tan }^2{\theta }_1}}\\ =\sqrt{\frac{1}{1+2^2}}\\ =\sqrt{\frac{1}{5}}\\ =\frac{1}{\sqrt{5}}\end{array}$

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

  $\begin{array}{c}{m}_1{u}_1\cos \left(0°\right)+m_2{u}_2\cos \left(0°\right)=m_1{v}_1\cos \left({\theta }_1\right)+m_2{v}_2\cos \left({\theta }_2\right)\\ m\cdot 3v_0+2m\cdot 0=m\cdot \sqrt{5}{v}_0\cdot \frac{1}{\sqrt{5}}+2m\cdot v_2\cos \left({\theta }_2\right)\\ 3m{v}_0=m{v}_0+2m{v}_2\cos \left({\theta }_2\right)\\ v_2\cos \left({\theta }_2\right)=v_0\end{array}$ ....... (1)

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

  $\begin{array}{c}{m}_1{u}_1\sin \left(0°\right)+m_2{u}_2\sin \left(0°\right)=m_1{v}_1\sin \left({\theta }_1\right)+m_2{v}_2\sin \left(-{\theta }_2\right)\\ m\cdot 0.+2m\cdot 0\cdot 0=m\cdot \sqrt{5}{v}_0\cdot \frac{2}{\sqrt{5}}-2m\cdot v_2\sin \left({\theta }_2\right)\\ 2m{v}_0-2m\cdot v_2\sin \left({\theta }_2\right)=0\\ v_2\sin \left({\theta }_2\right)=v_0\end{array}$ ........ (2)

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

  $\begin{array}{l}{\left(v_2\cos \left({\theta }_2\right)=v_0\right)}^2\\ {\left(v_2\sin \left({\theta }_2\right)=v_0\right)}^2\\ v_2^2{\cos }^2\left({\theta }_2\right)=v_0^2\\ v_2^2{\sin }^2\left({\theta }_2\right)=v_0^2\end{array}$

Further simplify the above,

  $\begin{array}{l}{v}_2^2=2v_0^2\\ v_2=\sqrt{2}{v}_0\end{array}$

Divide Equation (2) by (1).

  $\begin{array}{c}\frac{v_2\sin \left({\theta }_2\right)}{v_2\cos \left({\theta }_2\right)}=\frac{v_0}{v_0}\\ \tan \left({\theta }_2\right)=1\\ {\theta }_2={\tan }^{-1}\left(1\right)\\ {\theta }_2=45°\end{array}$

Conclusion:

Therefore, the speed of larger object after collision is $\sqrt{2}{v}_0$ and the angle is $45°$

(b)

To determine

The proof that collision is elastic.

Answer to Problem 88P

It is proved that collision is elastic.

Explanation of Solution

Formula used:

The expression for kinetic energy before collision is,

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

The expression for kinetic energy after collision is,

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

Calculation:

The expression for kinetic energy before collision is calculated as,

  $\begin{array}{c}{K}_1=\frac{1}{2}{m}_1{u}_1{}^2+\frac{1}{2}{m}_2{u}_2{}^2\\ =\frac{1}{2}m{\left(3v_0\right)}^2+\frac{1}{2}\cdot 2m{\left(0\right)}^2\\ =\frac{9}{2}m{v}_0{}^2\end{array}$

The expression for kinetic energy after collision is calculated as,

  $\begin{array}{c}{K}_2=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2\\ =\frac{1}{2}m{\left(\sqrt{5}{v}_0\right)}^2+\frac{1}{2}\cdot 2m{\left(\sqrt{2}{v}_0\right)}^2\\ =\frac{5}{2}\cdot m\cdot v_0{}^2+\frac{4}{2}\cdot m\cdot v_0{}^2\\ =\frac{9}{2}m{v}_0{}^2\end{array}$

Conclusion:

Since kinetic energy before collision is the same as kinetic energy after collision Therefore, it is proved that collision is elastic.