A particle is placed at point A on a frictionless track ABC as shown. It is gently pushed to the right. The speed of the

Question

A particle is placed at point A on a frictionless track ABC as shown. It is gently pushed to the right. The speed of the particle when it reaches point B is: (Take g = 10 m/s^2)

Accepted Answer

Given: The track is frictionless, the particle starts from rest at point A, $h_A = 1 \, \text{m}$ , $h_B = 0.5 \, \text{m}$ , and $g = 10 \, \text{m/s}^2$ .

Find: The speed at point B.

Since the track is frictionless, the mechanical energy is conserved.

U_A + KE_A = U_B + KE_B

At point A, kinetic energy is zero and potential energy is $mg \times 1$ . At point B, kinetic energy is $\frac{1}{2}mv^2$ and potential energy is $mg \times 0.5$ .

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

Simplifying,

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

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

So,

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

Therefore, the speed of the particle at point B is $\sqrt{10} \, \text{m/s}$ . The correct option is B.

Given: The particle starts from rest at point A on a frictionless track. $v_A = 0$ , $h_A = 1 \, \text{m}$ , $h_B = 0.5 \, \text{m}$ , and $g = 10 \, \text{m/s}^2$ .

Find: The final speed $v_B$ at point B.

Use conservation of mechanical energy:

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

Substitute the known values:

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

Cancel the common factor $m$ and simplify:

10 = \frac{1}{2}v_B^2 + 5

Now solve for $v_B$ :

\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}

Hence, the speed of the particle when it reaches point B is $\sqrt{10} \, \text{m/s}$ , which corresponds to option B.

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