Two inclined planes OA and OB having inclination (with horizontal) `30^(@)` and `60^(@)`, respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity `u = 10(sqrt3) ms^(-1)` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate

The velocity with which particle strikes the plane OB,

Options

A. `15ms^(-1)`
B. `30 ms^(-1)`
C. `20 ms^(-1)`
D. `10ms^(-1)`

Answer

D

Explanation

d. Two given planes are mutually perpendicular and the particle
is projected perpendicularly from plane OA. It means `vecu` is
parallel to plane OB.
At the instant of collision of the particle with OB, its velocity
is perpendicular to OB or velocity component parallel to OB
is zero.
First considering motion of particle parallel to plane OB,
`u = 10sqrt3 ms^(-1)`,
Acceleration `= -g sin `60^(@) = - 5sqrt(3) ms^(-2)`
`v = 0 , t= ? s = ?`
Using ` v = u + at rArr 0 = 10 sqrt(3) - 5 (sqrt3) t rArr t = 2s.`
`s = ut + 1/2 at^2 rArr OQ = 10sqrt3 xx 2 - 1/2 5sqrt3 (2)^2`
`rArr OQ = 10sqrt3m`.
Now considering motion of the particle normal to plane OB.
Initial, velocity = 0, acceleration `= g cos 60^@ = 5 ms^(-2)`
` t = 2 s , v = ? s = PQ = ? `
Using `v = u + at , we get v = 10ms^(-1)`
`s = it + 1/2 at^2 or PO = 10m`
`h = PO sin 30^@ = 10 xx sin 30^@ = 5,`
Inclination of velocity at point P with the vertical is `30^@`.
therefore its vertical component is ,
` u cos `30^(@) = 15ms^(-1)` (upwards)
Considering vertically upward motion of the particle from P.
Initial velocity `= 15ms^(-1)`,
Acceleration `= -g = -10 ms^(-2) , v= 0 , s = H = ?`
Using `v^2 = u^2 + 2as,` we get `H = 11.25m`
Maximum height reached by particle above O is
h + H = 16.25m
Distance PQ` = sqrt((PO)^2+(OQ)^2) = sqrt((10^2) + (10sqrt(3))^2) = 20 m` .