A particle of mass `m` comes down on a smooth inclined plane from point `B` at a height of `h` from rest. The magni-tude of change in momentum of the particle between position `A` (just before arriving on horizontal surface) and `C` (assuming the angle of inclination of the plane as `theta` with respect to the horizontal) is

Options

A. `0`
B. `2msqrt((2gh))sintheta`
C. `2msqrt((2gh))sin(theta/2)`
D. `2msqrt((2gh))`

Answer

C

Explanation

The velocity of the particle when it reaches point `A` from `B` is
`v=sqrt((2gh))` (directed along `BA`)
`(p_(i))x=mvcostheta`………..i
`(p_(i))x=mvcostheta`……ii

When the particle reaches point `C`, the momentum is as shown in figure.
`(p_(f))_(x)=mv`..........iii
`(p_(f))_(h)=0`..........iv
Therefore `(/_\P)_(x)=(p_(f))_(x)-(p_(i))_(x)`
`=mv-mv costheta`..........v
`(/_\p)_(y)=(p_(f))y-(p_(i))_(y)=-mvsintheta`
Magnitude of change in momentum is
`sqrt([mv(1-costheta)]^(2)+(mvsintheta)^(2))=2m(sqrt(2gh))sin(theta/2)`