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Question

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 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

 



Correct Answer
A

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

Your Answer
B

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

Your Answer
C

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

Your Answer
D

None of these


Solution

The correct option is B

(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 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 , u = 103ms1

Acceleration = - g sin  60 =53ms2  = 0 , t = ? S = ?

Using v = u + at, t = 2 seconds s = ut +  12 a   t2 or OQ = 10  3 meter.

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

Intial velocity = 0 , acceleration = h cos 60 = 53ms2

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

Using v = u + at, v = 10ms1

s = ut + 12t2 or OP = 10 m

h = PO sin 30 = 10 * sin 30 = 5 m

Inclination of u with the vertical is 30, therefore its vertical component is u cos 30 = 15ms1 (upward)

Considering vertically upward motion of the particle from P, Intial velocity = 15 m1 acceleration.

= g - 10 10ms2 v = 0 , S = H = ?

Using  v2u2 + 2as , H = 11.25 m.

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

Distance PQ = PQ2+OQ2(10)2+(103)2 = 20 m

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