<p>A uniform rod of length \"L\" rests on a frictionless horizontal surface. The

Question

<p>A uniform rod of length "L" rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed "v" strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fifth the mass of the rod. Given [L, v] Determine:</p>
<p>a. The final angular speed of the rod.</p>
<p>b. What fraction of the original energy was lost in the collision?</p>
<p>&#160;</p>
<p>I don't even know how to picturize this question...</p>
<p>thanks for any help!!! : )</p>

Explanation / Answer

length of the rod L =L mass of the rod M = M mass of bullet m = M / 5 = 0.2 M initial speed of rod U = 0 initial speed of bullet  u = v velocity of the system after collision be V from law of conservation of energy ,total energy before collision = total energy after collision                              ( 1/ 2) mv^ 2 + ( 1/ 2) MU^ 2 = ( 1/ 2) I w^ 2                              ( 1/ 2)*0.2M *v^ 2 + 0 = ( 1/ 2) Iw^ 2        ----( 1) where I = moment of inertia of the bullet -rod system               = moment of inertia of bullet + moment of inertia of the rod               = m(L/2)^ 2 + (1/3) ML^ 2               = (0.2 M *0.25L^ 2) +(0.3333ML^ 2)               = 0.3833 ML^ 2 plug this in eq( 1) we get 0.1 M v^ 2 = 0.5*0.3833 ML^ 2 w^ 2                                        0.1 v^ 2 = 0.1916 L^ 2w^ 2                                              v^ 2 = 1.916 L^ 2w^ 2                                           w^ 2 = 0.5217 v^ 2/ L^ 2                                              w = 0.7223 v / L   angular speed w = 0.7223 v / L (b) from law of conservation of momentum mv + MU = ( m+ M) V from this V = mv / ( m+ M)    since U = 0                   = 0.2M v / ( 1.2M )
                  = 0.166666 v initial total mechanical energy E = ( 1/ 2) mv^ 2 + ( 1/ 2) MU^ 2                                      = ( 1/ 2)*0.2M *v^ 2 + 0                                      = 0.1 Mv^ 2 final total mechanical energy E ' = ( 1/ 2) (m+ M) V^ 2                                                 = 0.5 *1.2M V^ 2                                                 = 0.5*1.2M *(0.16666v)^ 2                                                 = 0.01666 v^ 2 required fraction = ( E - E ' ) / E                           = 0.8333   i.e., 83.33 %

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