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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A particle is placed at the point A of a frictionless track ABC as shown in figure. It is gently pushed toward right. The speed of the particle when it reaches the point B is: (Take g = 10 ~m/s²).
Conservation of Mechanical Energy diagram for Q37 - JEE Main 2024 Morning
A particle on a frictionless track moving from height 1m at point A to 0.5m at point B.

Solution & Explanation

Related Formula
Kᵢ + Uᵢ = Kf + Uf (1)/(2) m u² + mghᵢ = (1)/(2) m v² + mghf
Core Logic

Since the track is frictionless, mechanical energy is conserved. We can apply the Principle of Conservation of Mechanical Energy (COME) between point A and point B.

Step 1: Apply Conservation of Energy

At point A (initially pushed gently, u ≈ 0):

KEA + UA = KEB + UB 0 + mg(hA) = (1)/(2) mv² + mg(hB)

Substitute the given values (hA = 1 ~m, hB = 0.5 ~m):

mg(1) = (1)/(2) mv² + mg(0.5) mg(0.5) = (1)/(2) mv² v² = 2g(0.5) = g
Step 2: Calculate Velocity

Given g = 10 ~m/s²:

v = √(g) = √(10) ~m/s
Pattern Recognition

For a mass sliding down a frictionless slope, its speed relies only on the vertical height dropped: v = √(2gΔ h).

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_30_jan_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%)