A ball of mass m is attached to cord of length L, pivoted at point O, as shown. The ball is released from rest at point A, swings down and makes an inelastic collision with a block of mass $$2m$$ kept on a rough horizontal floor. The coefficient of restitution of collision is $$e=23$$ , and coefficient of friction between block and surface is μ . After collision the ball comes momentarily to rest at C when cord makes an angle θ with the vertical and block comes to rest after travelling a distance of $$3L2$$ on the horizontal surface the value of μ and θ are respectively.

Question

A ball of mass m is attached to cord of length L, pivoted at point O, as shown. The ball is released from rest at point A, swings down and makes an inelastic collision with a block of mass $$2m$$ kept on a rough horizontal floor. The coefficient of restitution of collision is $$e=23$$ , and coefficient of friction between block and surface is μ . After collision the ball comes momentarily to rest at C when cord makes an angle θ  with the vertical and block comes to rest after travelling a distance of $$3L2$$  on the horizontal surface the value of μ  and θ  are respectively.

Infinity Learn · Accepted Answer

Let $$v1$$, $$v2$$  are speeds of ball & block, $$v1=(m1$$−$$em2m1+m2)u$$             $$v2=m1(1+e)um1+m2$$  [since block is at rest]            Here e $$=$$ $$2/3$$, $$m1=m$$ , $$m2=2m$$            Applying conservation energy for ball            ∴ $$mgL=12mu2$$            ⇒ $$u=2gL$$            ∴  $$v1=(m$$−$$43m)3m2gL=$$−$$192gL$$            Applying conservation of energy for ball            $$12m(181$$ $$2gL)=mgL(1$$−$$cos$$θ$$)$$            $$181=1$$−$$cos$$θ⇒ $$cos$$θ$$=8081$$            For block,            $$V2=m(5/3)2gL3m=592gL$$            ∴ $$02$$−$$2581(2gL)=2($$−μ$$g)3L2$$            $$5081$$×$$3=$$μ

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