Question

A particle is placed at the point \( A \) of a frictionless track \( ABC \) as shown in the figure. It is gently pushed toward the right. The speed of the particle when it reaches the point \( B \) is: (Take \( g = 10 \, \text{m/s}^2 \)).

• A. 20 m/s
• B. \( \sqrt{10} \, \text{m/s} \)
• C. \( 2 \sqrt{10} \, \text{m/s} \)
• D. 10 m/s

Answer 1

The Correct Option is B

Given the frictionless track, mechanical energy is conserved. At point A, only potential energy exists, whereas at point B, both kinetic and potential energy are present.

Potential Energy Difference Between Points A and B:

• Height at A is 1 m, and at B is 0.5 m.
• The height difference, h, is \( 1 - 0.5 = 0.5 \, \text{m} \).Apply Conservation of Mechanical Energy:

\( U_A + KE_A = U_B + KE_B \)

At point A, \( KE_A = 0 \) and \( U_A = mgh = mg \times 1 \). At point B, \( KE_B = \frac{1}{2}mv^2 \) and \( U_B = mg \times 0.5 \).

Equation Setup:

\( mg \times 1 = \frac{1}{2}mv^2 + mg \times 0.5 \)

Simplify and solve for v:

\[ mg = \frac{1}{2}mv^2 + \frac{mg}{2} \]

\[ \frac{mg}{2} = \frac{1}{2}mv^2 \]

\[ v = \sqrt{g} = \sqrt{10} \, \text{m/s} \]