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A wedge of mass M = 4 kg with a smooth inclined face making 45…

Question

A wedge of mass M = 4 kg with a smooth inclined face making 45 degrees with the horizontal rests on a frictionless horizontal floor. A small block of mass m = 2 kg is placed on the smooth inclined face and released from rest, so the block slides down while the wedge recoils horizontally. Taking g = 10 m/s^2, find the magnitude of the horizontal acceleration of the wedge (in m/s^2).

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Step-by-step solution

Let A be the horizontal acceleration of the wedge and a_r the acceleration of the block relative to the wedge along the incline.

Using a non-inertial frame or Newton's laws with the constraint that the block stays on the face, the standard result for a frictionless block on a frictionless wedge is a_wedge = m g sin(theta) cos(theta) / (M + m sin^2(theta)).

With theta = 45 deg, sin(theta)cos(theta) = 1/2 and sin^2(theta) = 1/2, so a_wedge = m g (1/2) / (M + m/2) = (2)(10)(1/2)/(4 + 1) = 10/5 = 2 m/s^2.

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