Question

Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:

• A. \(\frac19\)
• B. \(\frac89\)
• C. \(\frac49\)
• D. \(\frac59\)

Answer 1

The Correct Option is B

To solve the problem, we need to determine the fraction of energy lost by body A after a head-on elastic collision with body B. Let's approach this step by step:

1. Before the collision, body A has a mass of \(4m\) and moves with a speed \(u\). Body B has a mass of \(2m\) and is initially at rest. The initial kinetic energies are:
   • Kinetic energy of A: \(\frac{1}{2} \times 4m \times u^2 = 2mu^2\)
   • Kinetic energy of B: \(0\) (as B is at rest)
2. In a head-on elastic collision, both momentum and kinetic energy are conserved. Conservation of momentum gives: 4mu = 4mv + 2mv_b\), where \(v\) and \(v_b\) are the final velocities of A and B, respectively.
3. Conservation of kinetic energy gives: \(\frac{1}{2} \cdot 4m \cdot u^2 = \frac{1}{2} \cdot 4m \cdot v^2 + \frac{1}{2} \cdot 2m \cdot v_b^2\)
4. From the conservation equations, the final velocity expressions for A and B in an elastic collision are:
   • \(v = \frac{(4m - 2m)}{4m + 2m} \cdot u = \frac{1}{3}u\)
   • \(v_b = \frac{2 \times 4m}{4m + 2m} \cdot u = \frac{4}{3}u\)
5. The final kinetic energy of A is: \(\frac{1}{2} \cdot 4m \cdot \left(\frac{1}{3}u\right)^2 = \frac{2}{9}mu^2\)
6. The energy lost by A can be calculated as the initial energy minus the final energy: 2mu^2 - \frac{2}{9}mu^2 = \frac{16}{9}mu^2\)
7. Thus, the fraction of energy lost by A is: \(\frac{\frac{16}{9}mu^2}{2mu^2} = \frac{8}{9}\)

Thus, the fraction of energy lost by the colliding body A is \(\frac{8}{9}\), which matches the correct answer.