JEE Main · Physics ↓ Falling

Work, Energy and Power appeared 31 times across 3 years — 3.6% of Physics. This question is from Conservation of Mechanical Energy.

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Questions 8 15 8 31
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A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

Solution & Explanation

Related Formula

Conservation of Mechanical Energy:

Etotal = K + U = constant

At the initial height S (velocity v = 0):

Etotal = mgS

At any height x above the ground:

U = mgx and K = (1)/(2)mv²
Core Logic

Let the height of the particle at that instant be x. We are given: K = 3U

Substitute the energy terms:

(1)/(2)mv² = 3mgx

By energy conservation:

K + U = Etotal 3U + U = mgS 4U = mgS 4(mgx) = mgS x = (S)/(4)
Step 1: Calculating the Speed

Now find the speed v at this height x = S/4. Since K = 3U:

(1)/(2)mv² = 3mgx (1)/(2)mv² = 3mg((S)/(4)) v² = (6gS)/(4) = (3gS)/(2) v = √((3gS)/(2))
Pattern Recognition

Standard ratio trick: If K = n U, then by energy conservation (n+1)U = Etotal. This immediately yields:

x = (S)/(n+1)

Here n = 3, so x = S/4. This rapid shortcut lets you find the height in a split second!

Chapter Mix

Class 11 Physics: Work, Energy and Power

Reference Study Guides

Work Energy And Power

More Work, Energy and Power Previous-Year Questions — Page 7

Q49 jee_main_2024_31_jan_morning Conservation Of Momentum
An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is :
  • A. M₁ / (M₁ + M₂)
  • B. (M₂)/(M₁)
  • C. M₂ / (M₁ + M₂)
  • D. (M₁)/(M₂)

Solution

Related Formula
KE = (p²)/(2m)
Core Logic

By conservation of linear momentum (since no external horizontal force acts on the system):

0 = M₁ v₁ + M₂ v₂ | p₁| = | p₂| = p

Both the artillery and the shell acquire the exact same magnitude of momentum during firing.

Step 2: Kinetic Energy Ratio

The kinetic energy is related to momentum by KE = (p²)/(2m). Since p is identical for both bodies:

KE ∝ (1)/(m)

Therefore, the ratio of kinetic energy of the artillery (M₁) to the shell (M₂) is:

KE₁KE₂ = ((p²)/(2M₁))/((p²)/(2M₂)) = (M₂)/(M₁)
Chapter Mix

Class 11 Physics: Work, Energy And Power

More Work, Energy and Power Questions — jee_main_2025_03_april_morning

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Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)