Question 4

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

Since the track $$ABC$$ is completely frictionless, the total mechanical energy of the particle remains conserved throughout its motion.

$$U_A + K_A = U_B + K_B$$

$$\implies mgh_A + 0 = mgh_B + \frac{1}{2}mv_B^2$$

$$\implies v_B = \sqrt{2g(h_A - h_B)}$$

$$\implies v_B = \sqrt{2 \times 10 \times (1 - 0.5)} = \sqrt{10}\text{ m s}^{-1}$$

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