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

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 C , immediately before it leaves the second incline is

Answer: (B)

Solution

After the collision at B, the block has speed

45

m/s. It then slides down 3 m along the 30° incline. The vertical drop is

3 sin (30°) = 1.5

m. Using energy conservation:

\frac{1}{2} m v_{C}^{2} = \frac{1}{2} m (45) + m g (1.5) = \frac{1}{2} m (45) + 15 m

, giving

v_{C}^{2} = 45 + 30 = 75

. However, the solution states

v_{C} = 105

m/s, suggesting the vertical drop is 3 m:

v_{C}^{2} = 45 + 60 = 105

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