Question

A body of A moves an initial velocity of 2.5 m/s and collides with an identical body of which is at rest. Find the final velocity of A and B after collision if this collision is perfectly elastic

Answer 1

Let's consider the momentum conservation law:

$$mv_1 = mv_1' + mv_2'$$

where $m$ is the mass of the bodies, $v_1 = 2.5m/s$ is the velocity of the first body before the collistion and $v_1', v_2'$ are velocities of the first and second bodies respectively. Factoring out $m$, obtain equation (1):

$$v_1 = v_1' + v_2'$$

Since the collision is perfectly elastic, let's consider the energy conservation law and obtain equation (2):

$$\dfrac{mv_1^2}{2} = \dfrac{mv_1'^2}{2} +\dfrac{mv_2'^2}{2} \\ v_1^2 = v_1'^2 + v_2'^2$$

Expressing $v_1'$ from the equation (1) and substituting it into the equation (2), obtain:

$$v_1' = v_1-v_2'\\ v_1^2 = ( v_1-v_2')^2 + v_2'^2\\ v_1^2 = v_1^2 - 2v_1v_2' + v_2'^2 + v_2'^2\\ 2v_2'^2 - 2v_1v_2' = 0\\ v_2'(v_2' - v_1) = 0$$

There are two possible solutions:

$$v_2' = 0, \space v_2' = v_1$$

The first one doesn't fit, for it means that no collision occured. Thus, the answer is:

$$v_2' = v_1 = 2.5m/s$$

The velocity of the first object after the collision is then:

$$v_1' = v_1 - v_2' = 0m/s$$

Answer. 0 m/s, 2.5 m/s.