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/s 2 )
(i) The speed of the block at point B immediately after it strikes the second incline is
Text Solution
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(i) At point B there is perfectly inelastic collision so component of velocity ⊥ to incline plane becomes zero and component parallel to second surface is retained
velocity immediately after it strikes second incline
V = 
= 
× 
= 
V = 
m/s
(ii) At point ‘C’
V C 2 = 45 + 2 × 10 × 3
V C = 
m/s
(iii) The block coming down from incline AB makes an angle 30° with incline BC. If the block collides with incline BC elastically, the angle of block after collision with the incline shall be 30°. Hence just after collision with incline BC the velocity of block shall be horizontal. So immediately after the block strikes second inclined, its vertical component of velocity will be zero.
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