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. $\frac{4}{9}$
• B. $\frac{5}{9}$
• C. $\frac{1}{9}$
• D. $\frac{8}{9}$

Answer 1

The Correct Option is D

To solve the problem of determining the fraction of energy lost by body A after the collision, we need to apply concepts from elastic collision theory.

1. In an elastic collision, both momentum and kinetic energy are conserved.
2. Consider two bodies A and B. Initially, body A has mass 4m and velocity u, while body B has mass 2m and is at rest.
3. The initial momentum of the system is given by: \( p_{\text{initial}} = (4m)u + (2m)(0) = 4mu \).
4. Let the final velocities of bodies A and B be v_1 and v_2 respectively.
5. The conservation of momentum gives us: \( 4mu = 4mv_1 + 2mv_2 \) Simplifying, we get: \( 2u = 2v_1 + v_2 \) (Equation 1).
6. The conservation of kinetic energy gives us: \( \frac{1}{2}(4m)u^2 = \frac{1}{2}(4m)v_1^2 + \frac{1}{2}(2m)v_2^2 \) Simplifying, we get: \( 2u^2 = 2v_1^2 + v_2^2 \) (Equation 2).
7. From Equations 1 and 2, we can solve for v_1 and v_2.
8. Solving these simultaneous equations, we find: v_1 = \frac{u}{3} and v_2 = \frac{8u}{3}.
9. The initial kinetic energy of body A: \( KE_{\text{initial}} = \frac{1}{2}(4m)u^2 = 2mu^2 \).
10. The final kinetic energy of body A: \( KE_{\text{final}} = \frac{1}{2}(4m)\left(\frac{u}{3}\right)^2 = \frac{4m}{2} \cdot \frac{u^2}{9} = \frac{2mu^2}{9} \).
11. The energy lost by body A: \( \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 2mu^2 - \frac{2mu^2}{9} = \frac{18mu^2}{9} - \frac{2mu^2}{9} = \frac{16mu^2}{9} \).
12. The fraction of the energy lost by body A is: \( \frac{\Delta KE}{KE_{\text{initial}}} = \frac{\frac{16mu^2}{9}}{2mu^2} = \frac{16}{18} = \frac{8}{9} \).

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