Question

5. Derive the expression for the loss of kinetic energy in completely inelastic collision of two bodies.​

Answer 1

Answer:

$\rm \Delta K = \dfrac{m_1 m_2 (u_1 - u_2)^2}{2(m_1 + m_2)}$

Explanation:

Let mass of two bodies be $\sf m_1$ and $\sf m_2$ and initial velocity of two bodies be $\sf v_1$ and $\sf v_2$.

In completely inelastic collision two bodies stick together after collision.

So, body bodies will have same final velocity, let it be v.

By conservation of momentum, we get:

$\rm m_1 u_1 + m_2 u_2 = m_1 v + m_2 v \\ \\ \rm v = \dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Loss of Kinetic Energy:

$\rm \Delta K = K_i - K_f \\ \\ \rm = \bigg( \dfrac{1}{2} m_1 {u_1}^2 + \dfrac{1}{2} m_2 {u_2}^2 \bigg) - \dfrac{1}{2} (m_1 + m_2) {v}^2$

By substituting value of v, we get:

$\rm \Delta K = \bigg( \dfrac{1}{2} m_1 {u_1}^2 + \dfrac{1}{2} m_2 {u_2}^2 \bigg) - \dfrac{1}{2} (m_1 + m_2) {\bigg(\dfrac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \bigg)}^2$

After solving the equation, we get:

$\rm \Delta K = \dfrac{m_1 m_2 (u_1 - u_2)^2}{2(m_1 + m_2)}$