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

Since the track is frictionless, we can use the principle of conservation of mechanical energy. At point A, the particle has potential energy and no kinetic energy, while at point B, it will have both kinetic and potential energy.

Calculate Potential Energy Difference Between Points A and B:

• The height of A is 1 m, and the height of B is 0.5 m.
• The difference in height, 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 \( U_B = mg \times 0.5 \).

Setting up the equation:

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

Simplify and solve for v:

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

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

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