Answer to Question #125622 in Physics for Mwansa Kunda

Question

.Two identical ball-head on. The initial velocity of one is 0.75m/s ⁄ –East , while that of the other one is 0.43m/s ⁄ -West. If the collision is perfectly elastic. Determine the velocity of each ball.

Answer 1

According to the law of conservation of momentum (in projection on the x-axis directed to the East):

$$mv_1 -mv_2 = mv_1' + mv_2'\\ v_1 - v_2 = v_1' + v_2'$$

where $v_1$ and $v_2$ are velocities of the first and second ball respectively before the collision, $v_1'$ and $v_2'$ - after the collision. Since masses are identical they can be canceled.

Since the collision is perfectly elastic, the energy conservation law holds:

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

Thus, have two equations for two variables: $v_1'$ and $v_2'$.

$$v_1 - v_2 = v_1' + v_2'\\ v_1^2 + v_2^2 = v_1'^2 + v_2'^2$$

Substituting the values, obtain:

$$v_1' + v_2' = 0.75-0.43=0.32\\ v_1^2 + v_2^2 = 0.75^2 + 0.43^2=0.7474$$

Expressing $v_1'$ from the first equation and substituting it to the second one, get:

$$v_1' = 0.32-v_2'\\ ( 0.32-v_2')^2 + v_2'^2 = 0.7474\\ 0.1024-0.64v_2' + 2v_2'^2 = 0.7474\\ 2v_2'^2-0.64v_2'-0.645 = 0$$

The solutions are:

$$v_2' =- 0.43m/s \\ v_2' =0.75m/s$$

We should chose the last one, for the ball cannot not to change its velocity after the collision.

Then:

$$v_1' = 0.32-0.75 = -0.43m/s$$

Answer. The velocities after the collision: 1st ball 0.43 m/s to the West, 2nd ball 0.75 m/s to the East.