Question

When a body is projected vertically up from the ground with certain velocity, its potential energy and kinetic energy at a point A are in the ratio 2 : 3. If the same body is projected with double the previous velocity, then at the same point A the ratio of its potential energy to kinetic energy is

• A. 9 : 1
• B. 2 : 9
• C. 1 : 9
• D. 9 : 2
• E. 3 : 2

Hint:

At a fixed height, potential energy remains constant, but total energy changes with initial velocity.

Answer 1

The Correct Option is B

Concept: Total mechanical energy is conserved: \[ E = KE + PE \] At a fixed height, potential energy remains same but total energy changes if initial velocity changes.

Step 1: Let initial velocity be \(u\). Total energy: \[ E = \frac{1}{2}mu^2 \]

Step 2: Given ratio at point A. \[ PE : KE = 2 : 3 \] Let: \[ PE = 2x, \quad KE = 3x \Rightarrow E = 5x \]

Step 3: Now double velocity \(2u\). New total energy: \[ E' = \frac{1}{2}m(2u)^2 = 2mu^2 = 4E = 20x \]

Step 4: At same point A, PE remains same. \[ PE = 2x \]

Step 5: Find new KE. \[ KE = E' - PE = 20x - 2x = 18x \]

Step 6: New ratio. \[ PE : KE = 2x : 18x = 1 : 9 \]

Step 7: Conclusion. \[ \boxed{1 : 9} \]