Question

A ball of mass m is released from A inside a smooth wedge of mass m What is the speed of the wedge when the ball reaches point B. Please ans this question

Answer 1

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Answer 2

Answer:  The speed is $\sqrt g r / 3 \sqrt 2.$

Given data states the mass m released inside wedge from A to B. therefore the law of conservation of energy be conserved, thereby loss in Potential energy = Gain in Kinetic energy

            => Potential energy at B = Kinetic energy before sliding from A + Sliding Kinetic energy (vertical and horizontal component of velocity

            => $mgr.cos45^{\circ} = \frac{1}{2} mv^2+\frac{1}{2} m(v cos45^{\circ}-v)^2+\frac{1}{2} m(v.sin45^{\circ})^2$

            => $g r / \sqrt 2 = v^2 / 2 + \frac {1} {2} ( v_1 ^2 / 2 -2 v v_1 / \sqrt 2 - v_1 ^2) + 1 / 2 \sqrt 2 v_1^2 .$

And, by linear momentum conservation at B,

                   $mv = m ( v_1 cos 45 - v )$

                    => $v_1^2 = 3 \sqrt 2.$

Putting the value in earlier equation, we get,

                   $v = \sqrt g r / 3 \sqrt 2.$