JEE Main · Physics ↓ Falling

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

Year 2026 2025 2024 Total
Questions 8 15 8 31

In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg, moving in opposite directions with speeds of 10 m/s and 30 m/s, respectively, strike each other and stick together. The rise in temperature (in °C), if all the heat produced during the collision is retained by these spheres, is : (specific heat of sphere material 31 cal/kg·°C and 1 cal = 4.2 J)

Solution & Explanation

Related Formula
Δ K = (1)/(2) m₁m₂m₁ + m₂ urel² (1 - e²)

Q = msΔ T

Step 1: Loss of Kinetic Energy

For a perfectly inelastic collision, e = 0. Relative speed urel = v₁ + v₂ (since moving in opposite directions). urel = 10 + 30 = 40 m/s.

Δ K = (1)/(2)((15 × 25)/(15 + 25))(40)² Δ K = (1)/(2)((375)/(40))(1600) Δ K = (1)/(2) × 375 × 40 = 7500 J
Step 2: Heat Conversion

This lost energy turns into heat Q. The combined mass is m = 15 + 25 = 40 kg. Specific heat s = 31 cal/kg·°C = 31 × 4.2 J/kg·°C = 130.2 J/kg·°C.

Q = mtotal · s · Δ T 7500 = 40 × 130.2 × Δ T
Step 3: Calculate Temperature Rise
Δ T = (7500)/(40 × 130.2) Δ T = (750)/(520.8) ≈ 1.44°C
Pattern Recognition

Sees: "inelastic collision" + "rise in temperature" → Lost kinetic energy = Heat gained. Use reduced mass relative velocity shortcut to find Δ K instantly instead of conserving momentum first.

Chapter Mix

Class 11 Physics: Work, Energy and Power Class 11 Physics: Thermal Properties of Matter

Collisions diagram for Q43 - JEE Main 2026 Morning
Collisions diagram for Q43 - JEE Main 2026 Morning

Reference Study Guides

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_2026_23_january_morning

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

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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%)