A small block of mass `M` moves on a friction-less surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from `60^(@)` to `30^(@)` at point `B`. The block is initially at rest at `A`. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the blocks at point `B`, immediately after it strikes the second incline is
A
`sqrt(30) m//s`
B
`sqrt(15) m//s`
C
0
D
`-sqrt(15) m//s`
Text Solution
Verified by Experts
The correct Answer is:
C
In elastic collision, component of `v_(1)` parallel to `BC` will remailn unchanged, while component perpendicular to `BC` will remain unchanged in magnitude but its direction will be reversed. `v_(||) =v_(1)cos30^(@)=(sqrt60)(1/2)=sqrt(15)m//s` Now the vertical component of velocity of the block `v'=v_(_|_)cos30^(@)-v_(||)cos60^(@)` `=(sqrt(15))((sqrt(3))/2)-(sqrt(45))(1/2)=0`
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