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Q.State the principal of conservation of mechanical energy. Prove the principle of conservation of mechanical energy in the case of a freely falling body. (1+4=5) OR What is an elastic collision ? Obtain the expression for velocities after collision when two bodies undergo perfectly elastic collision in one dimension. (1+4=5)

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Subjective· 5mImportance★★★★★
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The total mechanical energy (KE + PE) of a system stays constant under conservative forces alone; for a body falling freely from height hh, KE + PE = mghmgh at every point of the fall, proving the principle.

Statement: In the absence of non-conservative (dissipative) forces such as friction or air resistance, the total mechanical energy of a system — the sum of its kinetic energy (KE) and potential energy (PE) — remains constant throughout its motion, even though KE and PE may individually change (converting into each other).

Proof for a freely falling body:

Consider a body of mass mm released from rest from a height hh above the ground (point A). Let it fall freely (only gravity acting, air resistance neglected) and consider an intermediate point B at height xx above the ground, and the ground itself as point C.

At point A (top, start of fall):

  • Height above ground = hh, speed v=0v = 0
  • KEA=0KE_A = 0
  • PEA=mghPE_A = mgh
  • TEA=KEA+PEA=0+mgh=mghTE_A = KE_A + PE_A = 0 + mgh = mgh

At point B (height xx, having fallen a distance h−xh-x):

Using v2=2g(h−x)v^2 = 2g(h-x) (from v2=u2+2asv^2=u^2+2as, u=0u=0):

  • KEB=12mv2=12m×2g(h−x)=mg(h−x)KE_B = \tfrac12 m v^2 = \tfrac12 m \times 2g(h-x) = mg(h-x)
  • PEB=mgxPE_B = mgx
  • TEB=mg(h−x)+mgx=mgh−mgx+mgx=mghTE_B = mg(h-x) + mgx = mgh - mgx + mgx = mgh

At point C (ground, height = 0):

Using v2=2ghv^2 = 2gh:

  • KEC=12m(2gh)=mghKE_C = \tfrac12 m(2gh) = mgh
  • PEC=0PE_C = 0
  • TEC=mgh+0=mghTE_C = mgh + 0 = mgh …

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