Angular momentum
mvl=(ml2+32ml2)ω
mvl=35ml2ω
ω=5l3v
21Iω2=2mg2l(1−cosθ)+mgl(1−cosθ)
21(35ml2)25l29v2=2mgl(1−cosθ)
5×23mv2=2mgl(1−cosθ)
103×2×1036=1−cosθ
1−5027=cosθ
cosθ=5023=0.46
θ=cos−1(0.46)=63∘
A block of mass m=1kg slides with velocity v=6ms−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=2kg, and length ℓ=1m, the value of θ is approximately(takeg=10ms−2)

Held on 3 Sept 2020 · Verified 6 Jul 2026.
63∘
55∘
69∘
49∘
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