Question

Develop the theory of one-dimensional elastic collision.

Answer 1

Step 1: Defining elastic collision in one dimension:

1. Elastic collision in one dimension.
2. The collision in which both the momentum and kinetic energy are conserved and colliding bodies continue to move along the same straight line after the collision.

Consider two perfectly elastic bodies A and B of masses M₁ and M₂ moving along the same straight line with velocities u₁ and u₂.

From the figure, the two bodies will collide if $u_1>u_2$

The two bodies undergo a head-on collision and continue moving along the same straight line with velocities v₁ and v₂.

The two bodies will separate if $v_2>v_1$

Step 2: Deriving the theory

Applying momentum conservation:

$M_1{u}_1+M_2{u}_2=M_1{v}_1+M_2{v}_2..........\left(1\right)$

Since kinetic energy is also conserved, we have:

$\frac{1}{2}{M}_1{u_1}^2+\frac{1}{2}{M}_2{u_2}^2=\frac{1}{2}{M}_1{v_1}^2+\frac{1}{2}{M}_2{v_2}^2.............\left(2\right)$

From equation (1) we get:

$M_1\left(u_1-v_1\right)=M_2\left(v_2-u_2\right).........\left(3\right)$

From equation (2) we get:

$M_1\left({u_1}^2-{v_1}^2\right)=M_2\left({v_2}^2-{u_2}^2\right)............\left(4\right)$

Dividing (4) by (3)

$$\begin{gathered} u_1+v_1=v_2+u_2 \\ {u}_1-u_2=v_2-v_1............\left(5\right) \end{gathered}$$

This implies that the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision.

From equation (5), we get:

$v_2=u_1-u_2+v_1$

Substituting the value of v₂ in (1), we get the value of v₁ as:

$v_1=\frac{\left(M_1-M_2\right)u_1+2M_2{u}_2}{M_1+M_2}..........\left(6\right)$

From equation (5), we get:

$v_1=v_2-u_1+u_2$

In equation (6) put the value of v₁, we get the value of v₂

$v_2=\frac{\left(M_2-M_1\right)u_2+2M_1{u}_1}{M_1+M_2}$