91Ó°ÊÓ

i) Mass of running shoe box, $m_{1}is3.2kg$.

ii) Mass of ballet slipper box, $m_{2}is2.0kg$.

iii) Height of the table, $h_{i}is0.40m$.

iv) Speed of the mass $m_1$, $V_{i1}is3.0m/s$.

You can use the concept of conservation of momentum and conservation of mechanical energy. Using conservation of momentum, you find the initial velocity of the boxes at the edge of the table. Using conservation of mechanical energy, you can find the final velocity at the ground

$$\begin{gathered} m_1{V}_{i1}+m_2{V}_{i2}=m_1{V}_{f1}+m_2{V}_{f2} \\ \frac{1}{2}m{v}_i^2+mg{h}_1=\frac{1}{2}m{v}_f^2+mg{h}_2 \end{gathered}$$ $$\begin{gathered} m_1{V}_{i1}+m_2{V}_{12}=m_1{V}_{f1}+m_2{V}_{f2} \\ \frac{1}{2}m{v}_i^2+mg{h}_1=\frac{1}{2}m{v}_f^2+mg{h}_2 \end{gathered}$$

First, we find the velocity of theboxes at the edge of the table; both move together, so they have thesame velocity, v; we can say that v can be calculated using momentum conservation law as,

$$\begin{gathered} \left(3.2\right)\left(3.0\right)+\left(20\right)\left(0\right)=\left(3.2\right)v+\left(2.0\right)v \\ v=\frac{9.6}{5.2} \\ =1.846m/s \end{gathered}$$

Now, this is the initial velocity for thenext projectile motion. We know that both boxes initially have kinetic energy and potential energy, which will convert to kinetic energy at the ground.

You get,

$$\begin{gathered} K{E}_i+P{E}_i=K{E}_f+P{E}_f \\ \frac{1}{2}m{v}^2+mg{h}_1=K{E}_f+mg{h}_f \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \frac{1}{2}\left(5.2kg\right)\left(1.846m/s^2\right)+\left(5.2kg\right)\left(9.8m/s^2\right)\left(0.40m\right)=K{E}_f+\left(5.2\right)\left(9.8\right)\left(0\right) \\ K{E}_f=29.24\left(1kg.m^2/s^2×\frac{1J}{1kg.m^2/s^2}\right) \\ =29.24J \\ =29J \end{gathered}$$

The kinetic energy of the boxes just before they strike the floor, $K{E}_{f}is29J$.