A small block of mass M moves on a frictionle | vedclass

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(B,B,C) $1.$ The height of $A$ above $B$ is $h_1 = \sqrt{3} 	an 60^{\circ} = \sqrt{3} \cdot \sqrt{3} = 3 \ m$. The velocity of the block just before

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(B,B,C) $1.$ The height of $A$ above $B$ is $h_1 = \sqrt{3} 	an 60^{\circ} = \sqrt{3} \cdot \sqrt{3} = 3 \ m$. The velocity of the block just before reaching $B$ is $v = \sqrt{2gh_1} = \sqrt{2 \cdot 10 \cdot 3} = \sqrt{60} \ m/s$. Since the collision is totally inelastic,the component of velocity perpendicular to the second incline is lost. The velocity along the second incline is $v_B = v \cos(60^{\circ}-30^{\circ}) = v \cos 30^{\circ} = \sqrt{60} \cdot \frac{\sqrt{3}}{2} = \sqrt{15 \cdot 3} = \sqrt{45} \ m/s$. Thus,option $(B)$ is correct.$2.$ The height of $B$ above $C$ is $h_2 = 3\sqrt{3} 	an 30^{\circ} = 3\sqrt{3} \cdot \frac{1}{\sqrt{3}} = 3 \ m$. Using the work-energy theorem from $B$ to $C$: $\frac{1}{2}Mv_C^2 - \frac{1}{2}Mv_B^2 = Mgh_2$. Substituting $v_B^2 = 45$ and $h_2 = 3$: $v_C^2 = v_B^2 + 2gh_2 = 45 + 2 \cdot 10 \cdot 3 = 45 + 60 = 105$. So,$v_C = \sqrt{105} \ m/s$. Thus,option $(B)$ is correct.$3.$ For an elastic collision,the component of velocity parallel to the incline remains unchanged,while the component perpendicular to the incline reverses direction. The velocity just before collision is $v = \sqrt{60} \ m/s$ at an angle of $60^{\circ}$ with the horizontal. The second incline is at $30^{\circ}$ to the horizontal. The angle between the velocity vector and the second incline is $60^{\circ}-30^{\circ} = 30^{\circ}$. The velocity components are $v_{\parallel} = v \cos 30^{\circ}$ and $v_{\perp} = v \sin 30^{\circ}$. After elastic collision,$v'_{\parallel} = v \cos 30^{\circ}$ and $v'_{\perp} = v \sin 30^{\circ}$ (directed away from the incline). The vertical component is $v_y = v'_{\parallel} \sin 30^{\circ} - v'_{\perp} \cos 30^{\circ} = (v \cos 30^{\circ}) \sin 30^{\circ} - (v \sin 30^{\circ}) \cos 30^{\circ} = 0$. Thus,option $(C)$ is correct.

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