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Physics · LAWS OF MOTION
हिन्दीA ball is released from rest at a height \( h = 0.45\text{ m} \) on a smooth inclined plane 1 inclined at \( 30^\circ \) to the horizontal. After reaching the bottom, it travels up a smooth inclined plane 2 for which \( \sin\theta_2 = \frac{1}{3} \). Assuming \( g = 10\text{ m/s}^2 \), what is the total time taken by the ball from release until it comes to instantaneous rest on plane 2?
View Answer & Solution
Correct answer · Option D
1.5 s
Explanation
On plane 1, acceleration \( a_1 = g\sin 30^\circ = 5\text{ m/s}^2 \), distance \( s_1 = \frac{0.45}{\sin 30^\circ} = 0.9\text{ m} \), giving time \( t_1 = \sqrt{\frac{2 s_1}{a_1}} = 0.6\text{ s} \) and bottom speed \( v = \sqrt{2gh} = 3\text{ m/s} \). On plane 2, acceleration \( a_2 = g \sin\theta_2 = \frac{10}{3}\text{ m/s}^2 \), so ascent time \( t_2 = \frac{v}{a_2} = 0.9\text{ s} \). Total time \( t = 0.6 + 0.9 = 1.5\text{ s} \).
