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

Answer: (B)

$\sqrt{10} \, \text{m/s}$

Answer 2

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 :

$mg = \frac{1}{2}mv^2 + \frac{mg}{2}$

$\frac{mg}{2} = \frac{1}{2}mv^2$

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