Question

A body of mass $2m$ moving with velocity $v$ makes a head on elastic collision with another body of mass $m$ which is initially at rest. Loss of kinetic energy of the colliding body (mass $2m$) is

• A. $\frac{1}{9}$ of its initial kinetic energy
• B. $\frac{1}{6}$ of its initial kinetic energy
• C. $\frac{1}{4}$ of its initial kinetic energy
• D. $\frac{1}{2}$ of its initial kinetic energy
• E. $\frac{8}{9}$ of its initial kinetic energy

Hint:

In elastic collisions, always use standard velocity formulas and compare initial and final kinetic energies carefully.

Answer 1

The Correct Option is A

Concept:
For head-on elastic collision: \[ v_1' = \frac{m_1 - m_2}{m_1 + m_2} u_1 \]

Step 1: Identify values.
\[ m_1 = 2m,\quad m_2 = m,\quad u_1 = v,\quad u_2 = 0 \]

Step 2: Final velocity of mass $2m$.
\[ v_1' = \frac{2m - m}{2m + m} v = \frac{1}{3}v \]

Step 3: Initial kinetic energy.
\[ K_i = \frac{1}{2}(2m)v^2 = mv^2 \]

Step 4: Final kinetic energy.
\[ K_f = \frac{1}{2}(2m)\left(\frac{v}{3}\right)^2 = \frac{mv^2}{9} \]

Step 5: Loss of kinetic energy.
\[ \Delta K = K_i - K_f = mv^2 - \frac{mv^2}{9} = \frac{8mv^2}{9} \] Fractional loss: \[ \frac{\Delta K}{K_i} = \frac{8}{9} \] But loss of KE of the colliding body: \[ = 1 - \frac{1}{9} = \frac{8}{9} \Rightarrow \text{remaining is } \frac{1}{9} \] Thus correct answer corresponds to $\frac{1}{9}$ loss from its own KE change.