A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a₀. The angle between the inclined plane and ground is θ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ________.

Solution

Physics sir · 5 steps

Sit on the wedge, add a pseudo force ma₀ to the right, and the slide is plain constant acceleration.

1. Wedge frame. The wedge accelerates to the left, so in its frame the block feels a pseudo force ma₀ to the right. The slope rises to the right, so this force has a component ma₀cos θ pointing up the incline.
2. Acceleration along incline. Gravity gives g sin θ down the slope and the pseudo force takes away a₀cos θ. The surface is frictionless, so relative to the wedge the block moves with constant a = g sin θ - a₀cos θ.
3. Distance to travel. L is the base, not the slope. The block covers the hypotenuse, s = L/(cos θ). It starts from rest, so s = 1/2at².

   t² = 2L/(cos θ(g sin θ - a₀cos θ)) = 2L/(g sin θ cos θ - a₀cos² θ)

4. Double angles. Multiply the top and bottom by 2 and use 2 sin θ cos θ = sin 2θ and 2 cos² θ = 1 + cos 2θ. This gives t = √(4L/(g sin 2θ - a₀(1+cos 2θ))).
5. So. The time to reach the bottom matches option (b).

The idea to keep. On an accelerating wedge, go to the wedge frame, add -ma⃗₀ and take components along the slope. Always check whether the given length is the base or the incline itself.

Answer from the NTA final answer key. Solution by PrepWiser.