Q.Show that the total mechanical energy of a system is conserved if the forces doing work on it are conservative. OR Two bodies A & B having masses m1 & m2 moving with v1 & v2 velocities undergoes completely inelastic collision. Show that the heavier mass (m2) is undisturbed while the lighter mass (m1) reverses its velocity.

(a) For conservative forces, work done equals minus the change in potential energy, so kinetic + potential energy stays constant. (b) The described outcome — heavier mass undisturbed, lighter mass reverses — is the textbook result for an elastic collision in the limit $m_2\gg m_1$; a genuinely perfectly inelastic collision instead makes the two bodies stick together and move with one common velocity, which would not show this reversal, so this derivation uses the elastic-collision equations that actually produce the stated behaviour.

(a) Conservation of mechanical energy: By the work-energy theorem, the net work done on a particle equals its change in kinetic energy: $W = \Delta KE = KE_f - KE_i$. For a conservative force, the work done is defined to equal the negative of the change in potential energy: $W = -\Delta PE = -(PE_f - PE_i)$ (this is exactly how potential energy is defined for such forces). Equating the two expressions for $W$:

$KE_f - KE_i = -(PE_f - PE_i) \;\Rightarrow\; KE_f + PE_f = KE_i + PE_i.$

So the total mechanical energy $E = KE + PE$ has the same value at every instant — it is conserved — whenever only conservative forces do work on the system.

(b) Collision result — heavier mass undisturbed, lighter mass reverses: A quick honesty note first: the behaviour described (one mass bouncing back while the other stays undisturbed) is the classic result for a perfectly elastic collision when $m_2 \gg m_1$, not a perfectly inelastic one — in a truly perfectly inelastic collision the two bodies stick together and move off with one common final velocity, so no reversal is possible. The derivation below uses the elastic-collision equations, which are what actually produce the stated result; this is worked out for completeness, taking body B (mass $m_2$) as initially at rest ($v_2=0$) for simplicity, which is the standard form of this classic problem.

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