A block of mass m = 1 kg slides with velocity υ=6m/s 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 l = 1 m, the value of θ is approximately: take g=10m/s2
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a
69º
b
55º
c
63º
d
49º
answer is C.
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Detailed Solution
Initial angular momentum of block about hinge before Collison = Final angular momentum of combine system after collision ⇒mvl=L ---(1)Now, using energy conservation Loss in rotational kinetic energy of combine system = Gain in potential energy of combine system⇒L22I=mgl1−cosθ+Mgl21−cosθ ∵cosθ=l−hl⇒h=l1−cosθ ⇒mvl22ml2+Ml23=m+M2gl1−cosθ ∵I=Iblock+Irod=ml2+Ml23& from 1⇒1×6×1221×12+2×123=1+2210×11−cosθ⇒cosθ=2350∵m=1kg, l=1m and M=2kg⇒θ=630
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A block of mass m = 1 kg slides with velocity υ=6m/s 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 l = 1 m, the value of θ is approximately: take g=10m/s2