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
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A small both A of mass m is attached to a massless rigid rod of length 1m pivoted at point P and kept at an angle of 60° with vertical as shown in figure. At distance of 1m below point P, an identical bob B is kept at rest on a smooth horizontal surface that extends to a circular track of radius R as shown in figure. If bob B just manages to complete the circular path of radius R upto a point Q after being hit elastically by bob A, then radius R is
Vertical Circular Motion diagram for Q27 - JEE Main 2026 Morning
Displays bob A on a rod hitting bob B which enters a circular track.

Solution & Explanation

Related Formula
V = √(2gl(1 - θ)) Vmin = √(5gR)
Core Logic

Bob A gains kinetic energy converting from potential energy when it falls from angle 60°. Since the collision is elastic and masses are identical, Bob A will transfer its entire momentum to Bob B. Bob B will then use this velocity to complete the vertical circular loop.

Step 1: Velocity of Bob A before collision

Velocity of Bob A at the lowest point:

VA = √(2gl(1 - θ)) VA = √(2 × 10 × 1 (1 - (1)/(2))) = √(10) m/s
Step 2: Velocity of Bob B after collision

Due to elastic collision between identical masses, velocities are exchanged.

VB = VA = √(10) m/s
Step 3: Condition for completing loop

To just complete the circular track of radius R, the minimum velocity required at the lowest point is:

VB = √(5gR) √(10) = √(5 × 10 × R)

10 = 50R

R = (1)/(5) m
Pattern Recognition

Sees: "elastic collision" + "identical masses" → velocities swap instantly. "just completes circular path" → velocity is exactly √(5gR) at the bottom.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Systems of Particles and Rotational Motion

Vertical Circular Motion diagram for Q27 - JEE Main 2026 Morning
Displays bob A on a rod hitting bob B which enters a circular track.

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

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