Chapter 7, Problem 29P

To determine

The landing distance of each cube.

Answer to Problem 29P

The landing distance of cube is $s_1=34.24\text{ cm}$ .

The landing distance of small cube is $s_2=138.24\text{ cm}$ .

Explanation of Solution

Given:

The mass of the cube is $m_1$ .

The mass of small cube is $m_2=\frac{m_1}{2}$ .

The height of the incline is $h=30\text{ cm}$ .

The height of the table from the floor is $H=90\text{ cm}$ .

Formula used:

The expression for the speed of the sliding cube is,

  $u_1=\sqrt{2gh}$

Here, g is the acceleration due to gravity and h is the height of sliding.

The expression for relative velocity is,

  $u_1-u_2=-\left(v_1-v_2\right)$

Here, $u_1$ is the initial velocity of cube, $u_2$ is the initial velocity of the small cube, $v_1$ is the final velocity of cube, and $v_2$ is the final velocity of the small cube.

The expression for law of conservation of momentum is,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Here, $m_1$ is the mass of the cube, $m_2$ is the mass of small cube, $u_1$ is the initial velocity of cube, $u_2$ is the initial velocity of the small cube, $v_1$ is the final velocity of cube, and $v_2$ is the final velocity of the small cube.

The expression for falling time is,

  $t=\sqrt{\frac{2H}{g}}$

Here, $H$ is the falling height and g is the acceleration due to gravity.

The expression for landing distance is,

  $s=vt$

Here, v is the speed and t is the time.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$

The initial speed of the cube is,

  $\begin{array}{c}{u}_1=\sqrt{2gh}\\ =\sqrt{2\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(30\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)}\\ =2.42\text{m}/\text{s}\end{array}$

The initial speed of the small cube is,

  $\begin{array}{c}{u}_2=\sqrt{2gh}\\ =\sqrt{2\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(0\right)}\\ =0\end{array}$

The relative velocity is,

  $\begin{array}{l}{u}_1-u_2=-\left(v_1-v_2\right)\\ \left(2.42\text{m}/\text{s}\right)-\left(0\right)=v_2-v_1\\ v_1=v_2-2.42\end{array}$ …… (1)

The speed of each object using the relation of conservation of momentum is,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ m_1\left(2.42\text{m}/\text{s}\right)+\frac{m_1}{2}\left(0\right)=m_1{v}_1+\frac{m_1}{2}{v}_2\\ 2.42=v_1+\frac{v_2}{2}\\ 2.42=\left(v_2-2.42\right)+\frac{v_2}{2}\end{array}$

  $\begin{array}{l}4.84=\frac{3v_2}{2}\\ v_2=3.23\text{m}/\text{s}\end{array}$

The speed of the cube after sliding using equation (1) is,

  $\begin{array}{l}{v}_1=v_2-2.42\\ v_1=\left(3.23\text{m}/\text{s}\right)-2.43\\ v_1=0.8\text{m}/\text{s}\end{array}$

The time of fall is,

  $\begin{array}{l}t=\sqrt{\frac{2H}{g}}\\ t=\sqrt{\frac{2\times \left(90\text{ cm}\times \frac{1\text{ m}}{100\text{ cm}}\right)}{9.81\text{m}/{\text{s}}^{\text{2}}}}\\ t=0.428\text{ s}\end{array}$

The landing distance of cube is,

  $\begin{array}{c}{s}_1=v_{1}t\\ s_1=\left(0.8\text{m}/\text{s}\right)\left(0.428\text{ s}\right)\\ s_1=0.3424\text{ m}\times \frac{100\text{ cm}}{1\text{ m}}\\ s_1=34.24\text{ cm}\end{array}$

The landing distance of small cube is,

  $\begin{array}{c}{s}_2=v_{2}t\\ s_2=\left(\text{3}\text{.23 }\text{m}/\text{s}\right)\left(0.428\text{ s}\right)\\ s_2=1.3824\text{ m}\times \frac{100\text{ cm}}{1\text{ m}}\\ s_2=138.24\text{ cm}\end{array}$

Conclusion:

Thus, the landing distance of cube is $s_1=34.24\text{ cm}$ and the landing distance of small cube is $s_2=138.24\text{ cm}$ .