IIT-JEE 2008 · Paper 1 · Physics Q45

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Paragraph for Question Nos. 44 to 46

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 60° to 30° 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/s²).

A block M starts at rest at A at the top of a frictionless incline AB inclined at 60° to the horizontal, spanning a horizontal distance √3 m (a drop of 3 m). At B the slope changes to a 30° incline BC spanning a horizontal distance 3√3 m (a further drop of 3 m) down to ground level at C.

The speed of the block at point C, immediately before it leaves the second incline is

\sqrt{120}

m/s

\sqrt{105}

m/s

\sqrt{90}

m/s

\sqrt{75}

m/s

Answer and worked solution

Official answer: B

Down the second incline. No friction, a further drop of 3 m:
$v_C^2=45+2\times10\times3=105\implies v_C=\sqrt{105}\text{ m/s}.$

(B)

Topics:Conservation of mechanical energy

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IIT-JEE 2008 · Paper 1, Physics question 45