91Ó°ÊÓ

The graph of displacement versus time for the two bodies which undergo an elastic one-dimensional collision is given.

From the slope of the lines, we can guess about velocity, and hence, also the motion of the bodies and center of mass. Using the concept of velocity of the center of mass, we can find the relation between the masses of two bodies.

Formula:

The velocity of the center of the mass of a body, ${\mathrm{v}}_{\mathrm{com}}=\frac{{\mathrm{m}}_1{\mathrm{v}}_1+{\mathrm{m}}_2{\mathrm{v}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$ (1)

Velocity is a slope of the x vs t graph

The slope of the line1 is zero. This represents that one of the bodies is stationary.

The slope of line 3 is non-zero. This represents that the other body is moving.

Since the line representing the center of mass always lies between the lines representing corresponding masses, Line segment 2 corresponds to the motion of the center of mass before the collision.

Similarly, Line segment 5 corresponds to the motion of the center of mass after collision considering the value of the center of the mass.

Let the body which was initially stationary have mass m₁anditsinitial velocity be ${\mathrm{v}}_{1\mathrm{i}}$and final velocity ${\mathrm{v}}_{1\mathrm{f}}$. Let the mass of the other body be m₂anditsinitial velocity is ${\mathrm{v}}_{2\mathrm{i}}$and final velocity ${\mathrm{v}}_{2\mathrm{f}}$. Let the initial velocity of their center of mass be ${\mathrm{v}}_{\mathrm{i}}$and its final velocity ${\mathrm{v}}_{\mathrm{f}}$. Then, the initial and the final velocities of the center of mass are given using equation (1) as follows:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{i}}=\frac{{\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{i}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{i}}}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ {\mathrm{v}}_{\mathrm{f}}=\frac{{\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{f}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{f}}}{{\mathrm{m}}_1+{\mathrm{m}}_2} \end{gathered}$$

Since the velocity of the center of mass is the same after collision. So, using the above equations, we get

$$\begin{gathered} \frac{{\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{i}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{i}}}{{\mathrm{m}}_1+{\mathrm{m}}_2}=\frac{{\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{f}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{f}}}{{\mathrm{m}}_1+{\mathrm{m}}_2} \\ {\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{i}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{i}}={\mathrm{m}}_1{\mathrm{v}}_{1\mathrm{f}}+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{f}} \end{gathered}$$

From the given graph we can write that

$$\begin{gathered} {\mathrm{v}}_{1\mathrm{i}}=0 \\ {\mathrm{v}}_{2\mathrm{f}}=0 \\ {\mathrm{v}}_{2\mathrm{i}}={\mathrm{v}}_{1\mathrm{f}} \end{gathered}$$

Thus, the above equation after substituting the values can be given as:

$$\begin{gathered} {\mathrm{m}}_1\left(0\right)+{\mathrm{m}}_2{\mathrm{v}}_{2\mathrm{i}}={\mathrm{m}}_1{\mathrm{v}}_{2\mathrm{i}}+{\mathrm{m}}_2\left(0\right) \\ {\mathrm{m}}_1={\mathrm{m}}_2 \end{gathered}$$

Therefore, the mass of the body that was moving faster before the collision is equal to that of the other body.