Two particles of mass 3 kg & 2 kg collide head on. Their relative velocity before the collision is 15 m/s and after the collision is 5 m/s. The loss of kinetic energy of the system is _____

Engineering

Physics

Conservation of Energy

Collision

Momentum and Energy Conservation for System of Particles

Question

Two particles of mass 3 kg & 2 kg collide head on. Their relative velocity before the collision is 15 m/s and after the collision is 5 m/s. The loss of kinetic energy of the system is ______ J .

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In a head-on collision, the loss of kinetic energy is calculated using the formula for inelastic collisions. The formula is derived from the conservation of momentum and the definition of the coefficient of restitution (e). The loss of kinetic energy (ΔK) is given by:

$Δ K = \frac{1}{2} μ u_{r e l}^{2} (1 - e^{2})$

where μ is the reduced mass, $μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} k g$ , u_rel is the initial relative velocity (15 m/s), and e is the coefficient of restitution, $e = \frac{v_{r e l}}{u_{r e l}} = \frac{5}{15} = \frac{1}{3}$ . Substituting the values:

$Δ K = \frac{1}{2} \times \frac{6}{5} \times 15^{2} \times (1 - {(\frac{1}{3})}^{2}) = \frac{1}{2} \times \frac{6}{5} \times 225 \times \frac{8}{9} = 120$

Final Answer: 120 J