Laws of MotionHard
Question
A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60o with the ground is:
Options
A.√3 m/s
B.2 m/s
C.2√3 m/s
D.3 m/s
Solution
Let AB = l , B = (x , y)
= vxî + vyĵ
= √3î + vyĵ → (i)
x2 + y2 = l2
2x vx + 2y vy = 0 ⇒ √3 +
vy = 0
⇒ √3 + (tan60o) vy = 0
⇒ vy = − 1
Hence from (i)
= √3 î − ĵ
Hence
vB = 2 m/s
= vxî + vyĵ= √3î + vyĵ → (i)x2 + y2 = l2
2x vx + 2y vy = 0 ⇒ √3 +
vy = 0 ⇒ √3 + (tan60o) vy = 0
⇒ vy = − 1
Hence from (i)
= √3 î − ĵ Hence
vB = 2 m/s
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