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Step-by-Step Solution
Step 1: Identify the masses and velocities
Let the mass of body A be $4m$, moving with an initial speed $u$. The mass of body B is $2m$, initially at rest.
Step 2: Write down the formula for velocity after an elastic collision
In a one-dimensional elastic collision, the final velocity of body A ($v_A$) after collision is given by:
$ v_A = \dfrac{(m_A - m_B)}{(m_A + m_B)}\,u_A + \dfrac{2m_B}{(m_A + m_B)}\,u_B $
Here, $m_A = 4m$, $m_B = 2m$, $u_A = u$, and $u_B = 0$.
Step 3: Substitute the values
Since $u_B = 0$, the formula reduces to:
$ v_A = \dfrac{4m - 2m}{4m + 2m}\, u = \dfrac{2m}{6m}\, u = \dfrac{1}{3}\,u. $
Step 4: Calculate the initial kinetic energy of A
The initial kinetic energy ($KE_i$) of A is:
$ KE_i = \dfrac{1}{2}\,(4m)\,u^2 = 2\,m\,u^2. $
Step 5: Calculate the final kinetic energy of A
After collision, the speed of A is $v_A = \frac{1}{3}u$. Hence, the final kinetic energy ($KE_f$) of A is:
$ KE_f = \dfrac{1}{2}\,(4m)\left(\dfrac{1}{3}u\right)^2 = 2m \times \dfrac{u^2}{9} = \dfrac{2}{9} m\,u^2. $
Step 6: Find the fraction of energy lost by A
The fraction of initial energy lost by body A is:
$ \text{Fraction lost} = \dfrac{KE_i - KE_f}{KE_i}
= \dfrac{2\,m\,u^2 - \dfrac{2}{9}\,m\,u^2}{2\,m\,u^2}
= \dfrac{2\,m\,u^2\,(1 - \dfrac{1}{9})}{2\,m\,u^2}
= \dfrac{8}{9}. $
Answer
The fraction of energy lost by the colliding body A is $ \dfrac{8}{9} $.
