Question

Two bodies A and B of masses \( 2m \) and \( m \) are projected vertically upwards from the ground with velocities \( u \) and \( 2u \) respectively. The ratio of the kinetic energy of body A and the potential energy of body B at a height equal to half of the maximum height reached by body A is:

• A. 8 : 1
• B. 1 : 1
• C. 4 : 1
• D. 2 : 1

Hint:

When dealing with energy conservation problems, always consider the total energy at different points, and apply conservation of mechanical energy to find the velocities and energies at other points.

Answer 1

The Correct Option is D

To solve the problem, we need to find the ratio of the kinetic energy of body A and the potential energy of body B at a particular height.

Step 1: Determine Maximum Height of Body A

The maximum height \( H_A \) reached by body A is calculated using the equation of motion:

\( H_A = \frac{u^2}{2g} \).
At half the maximum height, the height \( h \) is:

\( h = \frac{H_A}{2} = \frac{u^2}{4g} \).

Step 2: Calculate Kinetic Energy of Body A at Height \( h \)

The kinetic energy \( KE_A \) is:

\( KE_A = \frac{1}{2} \times 2m \times v_A^2 \),
where \( v_A \) is the velocity of body A at height \( h \). Using energy conservation:

\( \frac{1}{2} \times 2m \times u^2 = \frac{1}{2} \times 2m \times v_A^2 + 2m \times g \times \frac{u^2}{4g} \),

\( 2mu^2 = 2mv_A^2 + mu^2 \),
\( v_A^2 = \frac{u^2}{2} \).

Thus,

\( KE_A = 2m \times \frac{u^2}{4} = \frac{mu^2}{2} \).

Step 3: Determine Maximum Height of Body B

The maximum height \( H_B \) reached by body B is:

\( H_B = \frac{(2u)^2}{2g} = \frac{4u^2}{2g} = \frac{2u^2}{g} \).

At height \( h = \frac{u^2}{4g} \), the potential energy \( PE_B \) is:

\( PE_B = m \times g \times \frac{u^2}{4g} = \frac{mu^2}{4} \).

Step 4: Calculate the Ratio

The ratio of kinetic energy of body A to potential energy of body B is:

\( \text{Ratio} = \frac{KE_A}{PE_B} = \frac{\frac{mu^2}{2}}{\frac{mu^2}{4}} = 2 \).

Therefore, the ratio is 2 : 1.

Answer: 2 : 1

Answer 2

Step 1: Maximum Height Reached by Body A 

Using the formula for maximum height: \[ H_A = \frac{u^2}{2g} \]

Step 2: Kinetic Energy of Body A at Half of Maximum Height

At height \( \frac{H_A}{2} \), we apply conservation of mechanical energy: 
Total energy at launch: \[ E = \frac{1}{2} (2m) u^2 = mu^2 \] Potential energy at \( \frac{H_A}{2} \): \[ PE = 2m \cdot g \cdot \frac{H_A}{2} = m g H_A \] Remaining energy is kinetic energy: \[ K_A = mu^2 - m g H_A \] But from earlier, \( H_A = \frac{u^2}{2g} \), so: \[ m g H_A = m g \cdot \frac{u^2}{2g} = \frac{1}{2} mu^2 \] Thus: \[ K_A = mu^2 - \frac{1}{2} mu^2 = \frac{1}{2} mu^2 \]

Step 3: Potential Energy of Body B at \( \frac{H_A}{2} \)

\[ P_B = m g \cdot \frac{H_A}{2} = m g \cdot \frac{u^2}{4g} = \frac{1}{4} mu^2 \]

Step 4: Ratio of Kinetic Energy to Potential Energy

\[ \frac{K_A}{P_B} = \frac{\frac{1}{2} mu^2}{\frac{1}{4} mu^2} = \frac{1}{2} \cdot \frac{4}{1} = 2 \] 
So, the required ratio is: \[ \boxed{2 : 1} \]

Final Answer:

Option (4)