Chapter 8, Problem 64P

To determine

The proof that final velocities are $v_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}$ and $v_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}$ .

Answer to Problem 64P

It is proved that final velocities are $v_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}$ and $v_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}$ .

Explanation of Solution

Formula used:

The expression for the conservation of momentum given by,

  $m_1{v}_{\text{1i}}+m_2{v}_{\text{2i}}=m_1{v}_{\text{1f}}+m_2{v}_{\text{2f}}$

The expression for the conservation of kinetic energy is given by,

  $\frac{1}{2}{m}_1{v}_{\text{1i}}^2+\frac{1}{2}{m}_2{v}_{\text{2i}}^2=\frac{1}{2}{m}_1{v}_{\text{1f}}^2+\frac{1}{2}{m}_2{v}_{\text{2f}}^2$

Calculation:

The expression for the conservation of momentum is calculated as,

  $\begin{array}{c}{m}_1{v}_{\text{1i}}+m_2{v}_{\text{2i}}=m_1{v}_{\text{1f}}+m_2{v}_{\text{2f}}\\ m_1\left(v_{\text{1i}}-v_{\text{1f}}\right)=m_2\left(v_{\text{2f}}-v_{\text{2i}}\right)\end{array}$ …… (1)

The expression for the conservation of kinetic energy is calculated as,

  $\begin{array}{c}\frac{1}{2}{m}_1{v}_{\text{1i}}^2+\frac{1}{2}{m}_2{v}_{\text{2i}}^2=\frac{1}{2}{m}_1{v}_{\text{1f}}^2+\frac{1}{2}{m}_2{v}_{\text{2f}}^2\\ m_1\left(v_{\text{1i}}^2-v_{\text{1f}}^2\right)=m_2\left(v_{\text{2f}}^2-v_{\text{2i}}^2\right)\\ m_1\left(v_{\text{1i}}-v_{\text{1f}}\right)\left(v_{\text{1i}}+v_{\text{1f}}\right)=m_2\left(v_{\text{2f}}-v_{\text{2i}}\right)\left(v_{\text{2f}}+v_{\text{2i}}\right)\end{array}$

This implies,

  $\begin{array}{c}\frac{m_1\left(v_{\text{1i}}-v_{\text{1f}}\right)\left(v_{\text{1i}}+v_{\text{1f}}\right)}{m_1\left(v_{\text{1i}}-v_{\text{1f}}\right)}=\frac{m_2\left(v_{\text{2f}}-v_{\text{2i}}\right)\left(v_{\text{2f}}+v_{\text{2i}}\right)}{m_2\left(v_{\text{2f}}-v_{\text{2i}}\right)}\\ \left(v_{\text{1i}}+v_{\text{1f}}\right)=\left(v_{\text{2f}}+v_{\text{2i}}\right)\\ v_{\text{1i}}-v_{\text{2i}}=v_{\text{2f}}-v_{\text{1f}}\\ v_{\text{2f}}-v_{\text{1f}}=-\left(v_{\text{2i}}-v_{\text{1i}}\right)\end{array}$ …… (2)

Further simplify above,

  $\begin{array}{c}{m}_2\left(v_{\text{2f}}-v_{\text{1f}}\right)=-\left(m_2\right)\left(v_{\text{2i}}-v_{\text{1i}}\right)\\ m_2{v}_{\text{2f}}-m_2{v}_{\text{1f}}=-m_2{v}_{\text{2i}}+m_2{v}_{\text{1i}}\end{array}$ …… (3)

Add equation (1) and (3).

  $\begin{array}{c}{m}_1{v}_{\text{1i}}+m_2{v}_{\text{2i}}+m_2{v}_{\text{2f}}-m_2{v}_{\text{1f}}=m_1{v}_{\text{1f}}+m_2{v}_{\text{2f}}+\left(-m_2{v}_{\text{2i}}+m_2{v}_{\text{1i}}\right)\\ \left(m_1-m_2\right)v_{\text{1i}}+2m_2{v}_{\text{2i}}=\left(m_1+m_2\right)v_{\text{1f}}\\ v_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}\end{array}$

Multiply equation (2) by $m_1$ .

  $\begin{array}{c}{m}_1\left(v_{\text{2f}}-v_{\text{1f}}\right)=-m_1\left(v_{\text{2i}}-v_{\text{1i}}\right)\\ m_{1}v-m_1{v}_{\text{1f}}=-m_1{v}_{\text{2i}}+m_1+v_{\text{1i}}\end{array}$ …… (4)

Add equation (1) and (4).

  $\begin{array}{r}{m}_1{v}_{\text{1i}}+m_2{v}_{\text{2i}}-m_1{v}_{\text{2f}}+m_1{v}_{\text{1f}}=m_1{v}_{\text{1f}}+m_2{v}_{\text{2f}}+m_1{v}_{\text{2i}}-m_1{v}_{\text{1i}}\\ \left(m_1-m_2\right)v_{\text{1i}}+2m_2{v}_{\text{2i}}=\left(m_1+m_2\right)v_{\text{1f}}\\ 2m_1{v}_{\text{1i}}+\left(m_2-m_1\right)v_{\text{2i}}=\left(m_1+m_2\right)v_{\text{2f}}\\ v_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}\end{array}$

Conclusion:

Therefore, It is proved that final velocities are $v_{\text{1f}}=\frac{m_1-m_2}{m_1+m_2}{v}_{\text{1i}}+\frac{2m_2}{m_1+m_2}{v}_{\text{2i}}$ and $v_{\text{2f}}=\frac{2m_1}{m_1+m_2}{v}_{\text{1i}}+\frac{m_2-m_1}{m_1+m_2}{v}_{\text{2i}}$ .