Q.Prove that there is a loss of kinetic energy in a perfectly (completely) inelastic collision of two particles. OR If, on increasing the speed of a vehicle by 2 m/sec, its kinetic energy gets doubled, then what would be its initial speed?
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Start your 14-day free trial to unlock the full solution →In a perfectly inelastic collision, deltaKE = [m1m2/(2(m1+m2))]*(u1-u2)^2 > 0, proving kinetic energy is always lost (except in the trivial case u1 = u2).
Consider two particles of masses m1 and m2 moving along the same straight line with initial velocities u1 and u2 (u1 not equal to u2), which collide and stick together (perfectly inelastic collision), moving afterward with a common velocity v.
Step 1 - Find v using conservation of momentum (momentum is always conserved in a collision, elastic or not):
m1u1 + m2u2 = (m1+m2)v v = (m1u1 + m2*u2)/(m1+m2)
Step 2 - Kinetic energy before collision:
KE_i = (1/2)m1u1^2 + (1/2)m2u2^2
Step 3 - Kinetic energy after collision:
KE_f = (1/2)*(m1+m2)*v^2
Step 4 - Loss in kinetic energy, deltaKE = KE_i - KE_f. Substituting v from Step 1 and simplifying (standard algebra) gives the compact result:
deltaKE = [m1m2 / (2(m1+m2))] * (u1-u2)^2
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