Chapter 9, Problem 77AP

To determine

The maximum height attained by $m_1$ and $m_2$ after collision.

Answer to Problem 77AP

The maximum height attained by $m_1$ and $m_2$ after collision is $13.9\text{m}$ and $0.556\text{m}$ respectively.

Explanation of Solution

Write the expression for the conservation of energy for $m_1$ .

  $K_i+U_i=K_f+U_f$                                                                                         (I)

Here $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy, $U_f$ is the final potential energy and $m_1$ is the mass.

Substitute $m_{1}gh$ for $U_i$, $0$ for $K_i$,0 for  $U_f$ and $\frac{1}{2}{m}_1{v}_{1i}^2$ for $K_f$ in equation (I).

    $0+m_{1}gh=\frac{1}{2}{m}_1{v}_{1i}^2+0$

Simplify the above expression for $v_{1i}$.

    $v_{1i}=\sqrt{2gh}$                                                                                                    (II)

Here $m_1$ is the mass sliding down, $g$ is the acceleration due to gravity, $v_{1i}$ is the initial velocity of the mass $m_1$ and $h$ is initial height.

Write the expression for the conservation of energy for mass $m_2$ .

  $K_i+U_i=K_f+U_f$                                                                                      (III)

Here $K_i$ is the initial kinetic energy, $U_i$ is the initial potential energy, $K_f$ is the final kinetic energy, $U_f$ is the final potential energy and $m_2$ is the mass.

Substitute $m_{2}gh$ for $U_i$, $0$ for $K_i$, 0 for $U_f$ and $\frac{1}{2}{m}_2{v}_{2i}^2$ for $K_f$ In equation (III)

    $0+m_{2}gh=\frac{1}{2}{m}_2{v}_{2i}^2+0$

Simplify the above expression for $v_{2i}$.

  $v_{2i}=\sqrt{2gh}$                                                                                                  (IV)

Here $m_2$ is the mass sliding down, $g$ is the acceleration due to gravity, $v_{2i}$ is the initial velocity of the mass $m_2$ and $h$ is initial height.

Write the expression for the velocity perfectly elastic one dimensional collision for $m_1$ mass.

    $v_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right)v_{1i}+\left(\frac{2m_2}{m_1+m_2}\right)v_{2i}$                                                               (V)   

Here, $v_{1f}$ is the final velocity mass $m_1$ and $m_2$ is the mass of the second block.

Substitute $\sqrt{2g{h}_i}$ for $v_{1i}$ and $-\sqrt{2g{h}_i}$ for $v_{2i}$  in equation (V).

    $v_{1f}=\left(\frac{m_1-m_2}{m_1+m_2}\right)\sqrt{2gh}+\left(\frac{2m_2}{m_1+m_2}\right)\left(-\sqrt{2gh}\right)$                                             (VI)

After the collision of the block mass $m_1$ and $m_2$ slides back up to some certain height. Hence, kinetic energy of each block converts to potential energy to which helps in achieving that height.

Write the expression for the conservation of energy for mass $m_1$.

    $\frac{1}{2}{m}_1{v}_{1f}^2=m_{1}g{h}_1$                                                                                         (VII)

Here, $h_1$ is the final height of mass $m_1$.

Substitute $\left(\frac{m_1-m_2}{m_1+m_2}\right)\sqrt{2g{h}_i}+\left(\frac{2m_2}{m_1+m_2}\right)\left(-\sqrt{2g{h}_i}\right)$ for $v_{1f}$ in above equation and simplify for $h_1$.

    $h_1=\frac{{\left(\left(\frac{m_1-m_2}{m_1+m_2}\right)\sqrt{2gh}+\left(\frac{2m_2}{m_1+m_2}\right)\left(-\sqrt{2gh}\right)\right)}^2}{2g}$                                     (VIII)

Write the expression for the velocity perfectly elastic one dimensional collision for $m_2$ mass.

    $v_{2f}=\left(\frac{2m_1}{m_1+m_2}\right)v_{1i}+\left(\frac{m_2-m_1}{m_1+m_2}\right)v_{2i}$                                                            (IX)   

Here, $v_{2f}$ is the final velocity mass $m_2$.

Substitute $\sqrt{2g{h}_i}$ for $v_{1i}$ and $-\sqrt{2g{h}_i}$ for $v_{2i}$  in equation (V) and solve for $v_{1f}$.

    $v_{2f}=\left(\frac{2m_1}{m_1+m_2}\right)\sqrt{2gh}+\left(\frac{m_2-m_1}{m_1+m_2}\right)\left(-\sqrt{2gh}\right)$                                              (X)

After the collision of the block mass $m_1$ and $m_2$ both will slides back. Hence, kinetic energy of $m_2$ block converts to potential energy to help in achieving that height.

Write the expression for the conservation of energy for mass $m_2$ .

    $\frac{1}{2}{m}_2{v}_{2f}^2=m_{2}g{h}_2$                                                                                         (XI)

Here, $h_2$ is the final height of mass $m_2$.

Substitute $\left(\frac{2m_1}{m_1+m_2}\right)\sqrt{2g{h}_i}+\left(\frac{m_2-m_1}{m_1+m_2}\right)\left(-\sqrt{2g{h}_i}\right)$ for $v_{2f}$. Simplify the above equation for the value of $h_2$.

    $h_2=\frac{{\left(\left(\frac{2m_1}{m_1+m_2}\right)\sqrt{2gh}+\left(\frac{m_2-m_1}{m_1+m_2}\right)\left(-\sqrt{2gh}\right)\right)}^2}{2g}$                                       (XII)

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $5\text{m}$ for $h_i$ in equation (II)

    $\begin{array}{c}{v}_{1i}=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\\ =9.90\text{m/s}\end{array}$

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $5\text{m}$ for $h_i$ in equation (IV)

    $\begin{array}{c}{v}_{2i}=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\\ =9.90\text{m/s}\end{array}$

Substitute $2\text{kg}$ for $m_1$ , $4\text{kg}$ for $m_2$ , $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $5\text{m}$ for $h_i$ in equation (VIII).

    $\begin{array}{c}{h}_1=\frac{{\left(\left(\frac{2\text{kg}-4\text{kg}}{2\text{kg}+4\text{kg}}\right)\left(\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\right)+\left(\frac{2\left(4\text{kg}\right)}{2\text{kg}+4\text{kg}}\right)\left(-\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\right)\right)}^2}{2\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ =\frac{{\left(-3.299\text{m}/\text{s}-13.199\text{m}/\text{s}\right)}^2}{19.6\text{m}/\text{s}}\\ =13.9\text{m}\end{array}$

Substitute $2\text{kg}$ for $m_1$ , $4\text{kg}$ for $m_2$ , $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $5\text{m}$ for $h_i$ in equation (XII).

    $\begin{array}{c}{h}_2=\frac{{\left(\left(\frac{2\left(2\text{kg}\right)}{2\text{kg}+4\text{kg}}\right)\left(\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\right)+\left(\frac{4\text{kg}-2\text{kg}}{2\text{kg}+4\text{kg}}\right)\left(-\left(\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(5\text{m}\right)}\right)\right)\right)}^2}{2\left(9.8{\text{m/s}}^{\text{2}}\right)}\\ =\frac{{\left(6.599\text{m}/\text{s}-3.299\text{}\text{m}/\text{s}\right)}^2}{19.6\text{m}/\text{s}}\\ =0.556\text{m}\end{array}$

Thus, the maximum height attained by $m_1$ and $m_2$ after collision is $13.9\text{m}$ and $0.556\text{m}$ respectively.