Question:
A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
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In perfectly inelastic collisions, momentum is conserved but maximum kinetic energy is lost.
Updated On: Mar 19, 2026
- 56%
- 62%
- 44%
- 50%
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The Correct Option is B
Solution and Explanation
Step 1: Initial kinetic energy:
Kᵢ = (1)/(2)m(2v)² + (1)/(2)(2m)v²
Kᵢ = 2mv² + mv² = 3mv²
Step 2: Total momentum before collision:
vecp = (2mv)hati + (2mv)hatj
Step 3: After perfectly inelastic collision, combined mass = 3m
Velocity after collision:
vf = frac√((2mv)² + (2mv)²)3m = \frac2√(2)3v
Step 4: Final kinetic energy:
Kf = (1)/(2)(3m)(\frac2√(2)3v)²
= (4)/(3)mv²
Step 5: Percentage loss in energy:
% loss = (Kᵢ - Kf)/(Kᵢ) × 100
= (3 - (4)/(3))/(3) × 100 ≈ 62%
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