Question

A particle is released from height S above the surface of the earth. At certain height its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively.

• A. \( \frac{S}{2} \), \( \sqrt{\frac{3gS}{2}} \)
• B. \( \frac{S}{2} \), \( \frac{3gS}{2} \)
• C. \( \frac{S}{4} \), \( \sqrt{\frac{3gS}{2}} \)
• D. \( \frac{S}{4} \), \( \frac{3gS}{2} \)

Hint:

Use the conservation of energy principle to relate the potential and kinetic energies at different heights. Remember that the potential energy is proportional to the height above the surface.

Answer 1

The Correct Option is C

To solve this problem, we start by understanding the conditions given and relating them to the equations of motion and energy.

1. \(S\) is the height from which the particle is released. Initially, at height \(S\), the potential energy (PE) of the particle is maximum and the kinetic energy (KE) is zero.
2. As the particle falls, its potential energy decreases while its kinetic energy increases. At a certain height \(h\), the kinetic energy is given to be three times the potential energy.

Now, let's use the equations for potential energy and kinetic energy:

• Potential energy at height \(h\): \(PE = mgh\)
• Kinetic energy at height \(h\): \(KE = \frac{1}{2}mv^2\)

According to the problem, \(KE = 3 \times PE\):

\(\frac{1}{2}mv^2 = 3 \times mgh\)
\(v^2 = 6gh\) (Equation 1)

Using conservation of energy principle, the total mechanical energy at the initial point should equal the total mechanical energy at height \(h\):

• Total energy at height \(S\): \(mgS\) (since KE = 0 initially)
• Total energy at height \(h\): \(mgh + \frac{1}{2}mv^2\)

Setting the initial and current energies equal:

\(mgS = mgh + \frac{1}{2}mv^2\)
\(mgS = mgh + 3mgh\) (since KE = 3PE)
\(mgS = 4mgh\)
\(gS = 4gh\)
\(h = \frac{S}{4}\) (Equation 2)

Now substitute Equation 2 into Equation 1 to find the velocity:

\(v^2 = 6g \left(\frac{S}{4}\right)\)
\(v^2 = \frac{3gS}{2}\)
\(v = \sqrt{\frac{3gS}{2}}\)

Therefore, the height from the surface of the earth is \(\frac{S}{4}\) and the speed of the particle is \(\sqrt{\frac{3gS}{2}}\).

Answer 2

\( V^2 = 0 + 2g(S-x) \) \( V^2 = 2g(S-x) \) 

At B, Potential energy = mgx Kinetic energy 

= \( \frac{1}{2} mv^2 \) \( \frac{1}{2} mv^2 = 3mgx \) 

\( gx = \frac{1}{6} v^2 = \frac{1}{6} 2g(S-x) \) \( 4x = S \) 

\( x = \frac{S}{4} \) \( V = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} \)