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(B) When the body hits the smooth inclined plane,the impulse acts perpendicular to the surface of the plane. Therefore,the component of velocity paral

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$\sqrt{3}\, m/s$

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(B) When the body hits the smooth inclined plane,the impulse acts perpendicular to the surface of the plane. Therefore,the component of velocity parallel to the inclined plane remains unchanged.Let the initial velocity be $v = 3\, m/s$ directed vertically downward.The angle of the inclined plane with the horizontal is $	heta = 30^{\circ}$.The component of the initial velocity $v$ parallel to the inclined plane is $v_{\parallel} = v \sin 30^{\circ}$.After the collision,the body moves horizontally with velocity $v_f$. The component of this final velocity $v_f$ parallel to the inclined plane is $v_{f\parallel} = v_f \cos 30^{\circ}$.Since the surface is smooth,there is no force parallel to the plane,so the velocity component parallel to the plane is conserved:$v \sin 30^{\circ} = v_f \cos 30^{\circ}$$v_f = v 	an 30^{\circ}$Substituting the values $v = 3\, m/s$ and $	an 30^{\circ} = 1/\sqrt{3}$:$v_f = 3 	imes (1/\sqrt{3}) = \sqrt{3}\, m/s$.

As shown in the figure,a body of mass $m$ moving vertically downward with a speed of $3\, m/s$ hits a smooth fixed inclined plane and rebounds with a velocity $v_f$ in the horizontal direction. If the angle of the inclined plane is $30^{\circ}$,the velocity $v_f$ will be:

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