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Question: Answered & Verified by Expert

A block of mass m=1 kg slides with velocity v=6 m s-1 on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. if the rod has mass M=2 kg, and length =1 m, the value of θ is approximately(take g=10 m s-2

PhysicsRotational MotionJEE MainJEE Main 2020 (03 Sep Shift 1)
Options:
  • A 63°
  • B 55°
  • C 69°
  • D 49°
Solution:
1147 Upvotes Verified Answer
The correct answer is: 63°

Angular momentum

mvl=ml2+2ml23ω

mvl=53ml2ω

ω=3v5l

122=2mgl21-cos θ+mgl1-cos θ

1253ml29v225l2=2mgl 1-cos θ

35×2mv2=2mgl1-cosθ

310×362×10=1-cos θ

1-2750=cos θ

cos θ=2350=0.46

θ=cos-10.46=63°

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