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A ball of mass m moving with velocity v_0 experiences a head-on elastic collision with one of the spheres of a stationary rigid dumbbell as whown in figure. The mass of each sphere equals m//2, and the distance between them is l. Disregarding the size of the spheres, find the proper angular momentum overset~M of the dumbbell after the collision, i.e. the angular momentum in the reference frame moving translationally and fixed to the dumbbell's centre of inertia. |
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Answer» Solution :From conservation of linear momentum along the DIRECTION of incident ball for the system consists with colliding ball and phhere `mv_0=mv^'+m/2v_1` (1) where `v^'` and `v_1` are the velocities of ball and SPHERE 1 respectively after collision. (Remember that the collision is head on). As the collision is perfectly elastic, from the definition of co-efficient of restitution, `1=(v^'-v_1)/(0-v_0)` or, `v^'-v_1=-v_0` (2) Solving (1) and (2), we get, `v_1=(4v_0)/(3)`, DIRECTED towards right. In the C.M. frame of spheres 1 and 2 (FIGURE) `overset~vecp_1=-overset~vecp_2` and `|overset~vecp_1|=|overset~vecp_2|=mu|vecv_1-vecv_2|` Also, `vecr_(1C)=-vecr_(2C)`, thus `overset~vecM=2[l/2(m//2)/(2)(4v_0)/(3)hatn]` (where `hatn` is the unit vector in the sense of `vecr_(1C)xxoverset~vecp_1`) Hence `overset~M=(mv_0l)/(3)` 
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