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IIT-JEE 2008 · Paper 1, Physics question 46

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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=10g=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.

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

Choose one option
Answer and worked solution

Official answer: C

Elastic impact. The velocity component along the second incline is kept and the normal component is reversed: the velocity is reflected in the incline's surface.

Angles. The incoming velocity points 60° below the horizontal; the incline slopes 30° below the horizontal, so the velocity is 30° below the incline. After reflection it is 30° above the incline, i.e. at −30°+30°=0°-30°+30°=0°: horizontal.

Vertical component: 0.

(C)

Topics:Elastic and inelastic collisions
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