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 $$g=10m/s2$$

Question

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 $$g=10m/s2$$

Infinity Learn · Accepted Answer

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​⇒$$L22I=mgl1$$−$$cos$$θ$$+Mgl21$$−$$cos$$θ  ∵$$cos$$θ$$=l$$−hl⇒$$h=l1$$−$$cos$$θ ⇒$$mvl22ml2+Ml23=m+M2gl1$$−$$cos$$θ ∵$$I=Iblock+Irod=ml2+Ml23$$& from $$1$$​​⇒$$1$$×$$6$$×$$1221$$×$$12+2$$×$$123=1+2210$$×$$11$$−$$cos$$θ⇒$$cos$$θ$$=2350$$∵$$m=1kg$$, $$l=1m$$ and $$M=2kg$$​⇒θ$$=630$$

Detailed Solution

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