Answer to Question #179652 in Physics for Robert

Question

Two balls weighing 2.5 kg and 5 kg are subject to an elastic collision. Before the collision, the first and lighter ball was at rest, and the second one was moving at a speed of 3.5 m/s. Determine the kinetic energy of a lighter ball after a collision. (Answer: V₁'=3.66 m/s, V₂' = 1.166 m/s and Ek₁'= 16.7 J)

Answer 1

Let $v_1 = 0, v_2 = 3.5m/s$ be the speeds of the balls before the collision, $m_1 = 2.5kg, m_2 = 2m_1 = 5kg$ the masses of the balls, $v_1', v_2'$ the speeds of the balls after the collision.

Suppose the collision was head-on and ellastic. Thus, according to the momentum conservation law, have:

$$m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'$$

Since $v_1 = 0$:

$$m_2v_2 = m_1v_1' + m_2v_2'\\ 2m_1v_2 = m_1v_1' + 2m_1v_2'\\ 2v_2 = v_1' + 2v_2'$$

According to the energy conservation law:

$$\dfrac{m_2v_2^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{m_2v_2'^2}{2}\\ \dfrac{2m_1v_2^2}{2} = \dfrac{m_1v_1'^2}{2} + \dfrac{2m_1v_2'^2}{2}\\ 2v_2^2 = v_1'^2 + 2v_2'^2$$

Expressing $v_1'$ from $2v_2 = v_1' + 2v_2'$ and substituting it into $2v_2^2 = v_1'^2 + 2v_2'^2$, obtain:

$$3v_2'^2 - 4v_2v_2' + v_2^2 = 0$$

Substituting the values and solving the quadratic equation, obtain:

$$v_2' = 1.166 m/s\space \text{or}\space v_2' = 3.5 m/s$$

The last values in impossible, since it would mean that the first ball remains at rest after the collsion. Thus, subsituting $v_2' = 1.166 m/s$ into $2v_2 = v_1' + 2v_2'$, obtain:

$$v_1' = 2(v_2 - v_2') = 4.66m/s$$

And the kinetic energy is:

$$E_1' = \dfrac{m_1v_1'^2}{2} = 27.2J$$

Answer. 27.2J