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A particle of mass m is dropped from a h...

A particle of mass m is dropped from a height h above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of`sqrt(2gh)`. If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

A

`(3 sqrt3)/(8) (u^(2))/(g)`

B

`2 sqrt2 (u^(2))/(g)`

C

`(5)/(8) (u^(2))/(g)`

D

`(3 sqrt2)/(4) (u^(2))/(g)`

Text Solution

Verified by Experts

The correct Answer is:
A

`P_(i)= P_(f)` [`because` by conservation of momentum]
`m u + m u cos theta= 2mv`
`rArr v= (u(1+ cos 60^(@)))/(2)= (3)/(4)u`
So horizontal range after collsion = vt
`=v sqrt((2H_("max"))/(g))= (3)/(4) u sqrt((2u^(2) sin^(2) (60^(@)))/(2g^(2))) = (3)/(4) u^(2) (sqrt((3)/(4)))/(g)= (3 sqrt3 u^(2))/(8g)`
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