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Question

In the diagram particle A moves along the line y = 253 m with a constant velocity v of magnitude 5 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with a zero initial speed and a constant acceleration a of magnitude 4 m/s2. What angle θ between v and the positive direction of the y axis would result in a collision?


A

30

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B

45

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C

60

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D

37

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Solution

The correct option is A

30


Resolving the velocity and acceleration of Ball B in x and y direction.

Ball B: vx = 0

ax = 4 sin θ

sy = 12(4 cos θ) t2

Where ay = 4 cos θ

Now ball A's position after time t

Velocity is constant and in x direction

s = 5t

Coordinates will be ( 5t, 25 3 )

Ball B should move 253 on y axis to catch up with A

sy = 2 cos θt2 = 253

t = 2532cos θ ............(1)

And also Ball B should move 5t on x axis to catch A 2 sin θt2 = 5t

Substituting the value of t from equation (1)

t=52 sin θ = 2532cos θ

Squaring both sides 254 sin2 θ = 2532 cos θ

2sin2θ = cosθ3

2 ( 1 - cos2θ) = cosθ3

2 cos2θ + cos θ3 - 2 = 0

cosθ=13±13+164

cosθ=634=32

θ=30


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