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Question

In the given figure, a particle A moves along the line y=30 m with a constant velocity of magnitude 3 m/s and parallel to the x axis. At the instant particle A passes the y axis, the particle B leaves the origin with a zero initial speed and a constant acceleration of magnitude 0.40 m/s2. What angle θ betweenaB and the positive direction of the yaxis would result in a collision?


A
30
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B
15
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C
60
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D
45
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Solution

The correct option is C 60
Given that:
Vertical distance between A and B, (y)=30 m
Velocity of A, (uA)=3 m/s
Initial velocity of B, (uB)=0 m/s
Acceleration of B, (aB)=0.40 m/s2


Let the particles collide at point P.
In vertical direction for B:
y=12ayt2
30=12(0.40)cosθ t2
150=cosθ t2 ...........(i)
In horizontal direction for A and B:
uAt=12axt2
t=2uAax
t=2×3(0.40)sinθ
t=15sinθ
Substituting t in equation (i),
150=cosθ×(15sinθ)2
150=cosθ×2251cos2θ
1cos2θ=3cosθ2
22cos2θ=3cosθ
2cos2θ+3cosθ2=0

Solving this, we get
cosθ=12 or cosθ=2 [neglect]
cosθ=12
θ=60
​​​​​

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