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:

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

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

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

Answer 2

The problem asks for the speed of a particle at point B after it starts from rest at point A and moves along a frictionless track. We are given the initial and final heights of the particle and the acceleration due to gravity.

Concept Used:

Since the track is frictionless and the only force doing work is gravity (which is a conservative force), the total mechanical energy of the particle is conserved. The principle of conservation of mechanical energy states that the sum of the kinetic energy (KE) and potential energy (PE) of the particle remains constant throughout its motion.

The formula for conservation of energy is:

\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]

where Kinetic Energy \(KE = \frac{1}{2}mv^2\) and Gravitational Potential Energy \(PE = mgh\).

Step-by-Step Solution:

Step 1: Identify the initial and final states and list the given values.

Let's set the reference level for potential energy at the bottom of the track (h=0).

• Initial State (at point A):
  • Initial velocity, \(v_A = 0\) (since the particle is "gently pushed").
  • Initial height, \(h_A = 1 \, \text{m}\).
• Final State (at point B):
  • Final velocity, \(v_B = ?\)
  • Final height, \(h_B = 0.5 \, \text{m}\).
• Given:
  • Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\).

Step 2: Apply the principle of conservation of mechanical energy.

The total mechanical energy at point A must be equal to the total mechanical energy at point B.

\[ \frac{1}{2}mv_A^2 + mgh_A = \frac{1}{2}mv_B^2 + mgh_B \]

Step 3: Substitute the known values into the conservation of energy equation.

\[ \frac{1}{2}m(0)^2 + m(10)(1) = \frac{1}{2}mv_B^2 + m(10)(0.5) \]

Step 4: Simplify the equation.

The mass \(m\) is a common factor in all terms and can be canceled out.

\[ 0 + 10(1) = \frac{1}{2}v_B^2 + 10(0.5) \] \[ 10 = \frac{1}{2}v_B^2 + 5 \]

Final Computation & Result:

Step 5: Solve the equation for the final speed \(v_B\).

Rearrange the equation to isolate the term with \(v_B^2\):

\[ \frac{1}{2}v_B^2 = 10 - 5 \] \[ \frac{1}{2}v_B^2 = 5 \] \[ v_B^2 = 10 \] \[ v_B = \sqrt{10} \, \text{m/s} \]

The speed of the particle when it reaches point B is \(\sqrt{10} \, \text{m/s}\). This corresponds to option (B).