JEE Main · Physics ↓ Falling

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

Year 2026 2025 2024 Total
Questions 8 15 8 31
Practice all Work, Energy and Power questions → View full JEE Main weightage & trends →

A bob of mass 'm' is suspended by a light string of length 'L'. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B. The ratio of kinetic energies (K.E.)A(K.E.)B is:
Vertical circular motion path of a pendulum bob for Q42 - JEE Main 2024 29 January Shift 2
The diagram displays a bob of mass m in vertical circular motion with velocity indicators at points A, B, and C.

Solution & Explanation

Related Formula

For a body to just complete a vertical loop of radius L:

  • Speed at the lowest point A: vA = √(5gL)
  • Speed at the highest point B: vB = √(gL)
Core Logic

The kinetic energy at any point is given by:

K.E. = (1)/(2)mv²

Thus:

(K.E.)A = (1)/(2)m vA² = (1)/(2)m(5gL) (K.E.)B = (1)/(2)m vB² = (1)/(2)m(gL)
Step 1: Calculate the Ratio

Taking the ratio of the kinetic energies at A and B:

(K.E.)A(K.E.)B = ((1)/(2)m(5gL))/((1)/(2)m(gL)) = (5)/(1) = 5:1
Pattern Recognition

Since K.E. ∝ v², the ratio of kinetic energies is simply the ratio of the squares of the critical velocities at the bottom and top of the vertical loop: (√(5gL))² : (√(gL))² = 5:1.

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_2024_29_january_evening

Practice all Work, Energy and Power previous-year questions →

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%)