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Two block of mass 2 kg and M kg are at rest on an inclined plane and are separated by a distance of 6.0 m as a showm in figure. The coefficient of friction between each of the blocks and the inclined plane is 0.25. The 2 kg block is given a velocity of 10.0 m/s up the inclined plans. It collides with M, comes back and has a velocity of 10 m/s when it reaches its initial position. The other block of mass M after the collision moves 0.5 m up and comes to rest. Calculate the coefficient of restitution between the 2 kg block and the mass M.["Take " sin theta ~~ tan theta = 0.5, cos theta = 1 " and " g = 10 m//sec^(2)] |
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Answer» Solution :For the block of mass 2 kg (=m) in going from the A to B `(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")` `(1)/(2)mv_(1)^(2)+0=(1)/(2)mv_(F)^(2)+mgh_(1)+F_(f)s` of `v_(f)^(2)=2[(1)/(2)v_(L)^(2)-gh_(1)-mu GS]` Substituting the given data, we get `v_(f)^(2)=2[(1)/(2)(10.0 m//s^(2))-(10m//s^(2))(0.30m)-(0.25)(10m//s^(2))(6.0m)]` Or, `v_(f)=8 m//s` For the block of mass 2 kg (=m) coming back from the position B to A `(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")` `(1)/(2)mu_(l)^(2)+mgh_(1)=(1)/(2)mu_(f)^(2)+0+F_(f)s` `u_(l)^(2)=2[(1)/(2)u_(f)^(2)+ mu gs-gh_(l)]` Substituting the given data, we get `u_(l)^(2)=2[(1)/(2)(-1.0m//s^(2))+(0.25)(10 m//s^(2))(6.0m)-(10m//s^(2))(0.30m)]` Or `u_(l)=-5.0 m//s`  For the block of Mass M in going from the position B to C `(KE+PE)_("initial")=(KE+PE)_("final")+W_("friction")` `(1)/(2)Mc_(1)^(2)+Mgh_(1)=(1)/(2)Mc_(r )^(2)+Mg(h_(1)+h_(2))+F_(f)S_(2)` or `(1)/(2)c_(1)^(2)=(1)/(2)c_(1)^(2)+gh_(2)+mu gs_(2)` `(1)/(2)c_(i)^(2)=(1)/(2)(0)+(10 m//s^(2))(0.25 m)+(0.25)(10 m//s^(2))(0.5m)` `c_(i)=sqrt(3)m//s` `e=(u_(i)-c_(i))/(v_(f)-c_(i))=((-5.0m//s)-(-sqrt(3)m//s))/((8.0 m//s)-0)=(5+sqrt(3))/(8)=0.84` 
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