Question

Consider a box that explodes into two pieces while moving with a constant positive velocity along an x-axis. If one piece, with mass , ends up with positive velocity ${\mathit{v}}_{\mathbf{1}}$,then the second piece, with mass ${\mathit{m}}_{\mathbf{2}}$, could end up with (a) a positive velocity ${\mathit{v}}_{\mathbf{2}}$(Fig. 9-25a), (b) a negative velocity ${\mathit{v}}_{\mathbf{2}}$(Fig. 9-25b), or (c) zero velocity (Fig. 9-25c). Rank those three possible results for the second piece according to the corresponding magnitude of ${\mathit{v}}_{\mathbf{1}}$, the greatest first.

Short answer

The rank of these three possible results for the second piece according to the corresponding magnitude of $\overrightarrow{v_1}$ is (b) > (c) > (a) .

Step-by-step solution

The given data

For case (a),$\overrightarrow{v_2}=\mathrm{positive}$

For case (b),$\overrightarrow{{\mathrm{v}}_2}=\mathrm{negative}$

For case (c),$\overrightarrow{{\mathrm{v}}_2}=\mathrm{zero}$

Understanding the concept of the conservation of momentum

As they explode and the box is broken into pieces, the pieces move with different velocities. Using the conservation of momentum, we can find the equation for the velocity of the second box and using that equation, we can find the velocity for the different conditions.

Formula:

According to the conservation of the momentum,$\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}={\mathrm{m}}_1+{\mathrm{v}}_1+{\mathrm{m}}_2+{\mathrm{v}}_2$ (1)

Calculation of the rank according to the velocities

From equation (1), we can get the value of the velocity of the second piece as follows:

$$\begin{gathered} {\mathrm{m}}_2{\mathrm{v}}_2=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}-{\mathrm{m}}_1{\mathrm{v}}_1 \\ {\mathrm{v}}_2=\frac{\left(\left(\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}\right)-{\mathrm{m}}_1{\mathrm{v}}_1\right)}{{\mathrm{m}}_2}...........\left(2\right) \end{gathered}$$

For case (a)

If${\mathrm{v}}_2$ is positive, then in the above equation (2)$\left(\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}\right)$ ,must be greater than${\mathrm{m}}_1{\mathrm{v}}_1$

As the masses are constant, there${\mathrm{v}}_1$ must be less.

For case (b)

If${\mathrm{v}}_2$is negative, then in the above equation (2)$\left(\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}\right)$ ,must be smaller than${\mathrm{m}}_1{\mathrm{v}}_1$

As the masses are constant, there${\mathrm{v}}_1$must be greater.

For case (c)

Ifis zero, then in the above equation (2)$\left(\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{v}\right)$ ,must be equal to${\mathrm{m}}_1{\mathrm{v}}_1$

As masses are constant, there will be greater as compared to case (2), and they will be smaller than case (b).

So we can say that the rank of three possible results for the second piece, according to the corresponding magnitude of$\overrightarrow{{\mathrm{v}}_1}$ , as (b) > (c) > (a) , greatest first.