JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Work, Power & Energy

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A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from

6 0^{\circ}

3 0^{\circ}

at point B . The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic

(g = 10 m / s^{2})

.

The speed of the block at point B immediately after it strikes the second incline is

(A) $60 m / s$
(B) $45 m / s$ ✓ Correct
(C) $30 m / s$
(D) $15 m / s$

Answer: (B)

Solution

The block slides down 3 m along the 60° incline. Using energy conservation,

v = 2 g h = 2 \times 10 \times 3 = 60

m/s just before collision. In a totally inelastic collision, the component perpendicular to the new incline is lost. The component along the 30° incline is

v cos (30°) = 60 \times \frac{3}{2} = 45

m/s.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Work, Power & Energy

Solution

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