Question

Two inclined planes OA and OB having inclination (with horizontal) ${30}^{\circ }$ and ${60}^{\circ }$ respectively , intersect each other at O as shown in figure.A particle is projected from point P with velocity along a direction perpendicular to plane OA. if the particle strickes planne OB perpendicularly at Q, calculate ,

(i) Velocity with which particle strikes the plane OB. (u) 16.25

(ii) Time of flight (v) 20

(iii) Verticla height h of P from O (w) 10

(iv) Maximum height from the ground (x)5

(v) Distance PQ (y)2

• A. (i) - (w), (ii) - (y), (iii) - (x), (iv) - (u), (v) - (v)
• B. (i) - (x), (ii) - (u), (iii) - (v), (iv) - (w), (v) - (y)
• C. (i) - (y), (ii) - (w), (iii) - (x), (iv) - (u), (v) - (v)
• D. None of these

Answer 1

The correct option is A

(i) - (w), (ii) - (y), (iii) - (x), (iv) - (u), (v) - (v)

Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane OA. It means $\overrightarrow{u}$ is parallel to plane OB.

At the instant of collision of the particle with OB , its velocity is perpendicular to OB or velocity compoment parallel to OB is zero.

For considering motion of particle parallel to plane OB , $\overrightarrow{u}$ = $10\sqrt{3}\frac{m}{s^{-1}}$

Acceleration = - g sin ${60}^{\circ }$ =$-5\sqrt{3}\frac{m}{s^{-2}}$ = 0 , t = ? S = ?

Using v = u + at, t = 2 seconds s = ut + $\frac{1}{-2}$ a $t^{-2}$ or OQ = 10 $\sqrt{3}$ meter.

Now considering motion of the particle normal of the particle normal to plane OB,

Intial velocity = 0 , acceleration = h cos ${60}^{\circ }$ = $5\sqrt{3}\frac{m}{s^{-2}}$

T = 2 second , v = ? ,s = PO = ?

Using v = u + at, v = 10$\frac{m}{s^{-}1}$

s = ut + $\frac{1}{2}$ a $t^2$ or OP = 10 m

h = PO sin ${30}^{\circ }$ = 10 * sin ${30}^{\circ }$ = 5 m

Inclination of $\overrightarrow{u}$ with the vertical is ${30}^{\circ }$, therefore its vertical component is u cos ${30}^{\circ }$ = $15\frac{m}{s^{-1}}$ (upward)

Considering vertically upward motion of the particle from P, Intial velocity = 15 $\frac{m}{{}^{-}1}$ acceleration.

= g - 10 $10\frac{m}{s^{-2}}$ v = 0 , S = H = ?

Using $v^2$ = $u^2$ + 2as , H = 11.25 m.

$\therefore$ Maximum height reached by particle by above O = h + H = 16.25 m.

Distance PQ = $\sqrt{{PQ}^2+{OQ}^2}$ = $\sqrt{{\left(10\right)}^2+{\left(10\sqrt{3}\right)}^2}$ = 20 m