InterviewSolution
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InterviewSolution
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A shell flying with velocity `v=500m//s` burts into three identical fragements so that the kinetic energy of the system increases `eta=1.5` times. What maximum velocity can one of the fragments obtain? |
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Answer» From the conservation of linear momentum of the shell just before and after its fragementation `3vecv=vecv_1+vecv_2+vecv_3` (1) where `vecv_1`, `vecv_2` and `vecv_3` are the velocities of its fragments. From the energy conservation `3etav^2=v_1^2+v_2^2+v_3^2` (2) Now `overset~vecv_i` or `vecv_(iC)=vecv_i-vecv_C=vecv_i-vecv` (3) where `vecv_C=vecv`=velocity of the C.M. of the fragements the velocity of the shell. Obviously in the C.M. frame the linear momentum of a system is equal to zero, so `overset~vecv_1+overset~vecv_2+overset~vecv_3=0` (4) Using (3) and (4) in (2), we get `3etav^2=(vecv+overset~vecv_1)^2+(vecv+overset~vecv_2)^2+(vecv-overset~vecv_1-overset~vecv_2)^2=3v^2+2overset~v_1^2+2overset~v_2^2+2overset~vecv_1*overset~vecv_2` or, `2overset~v_1^2+2overset~v_1overset~v_2costheta+2overset~v_2^2+3(1-eta)v^2=0` (5) If we have had used `overset~vecv_2=-overset~vecv_1-overset~vecv_3`, then Eq. 5 were contain `overset~v_3` instead of `overset~v_2` and so on. The problem being symmetrical we can look for the maximum of any one. Obviously it will be the same for each. For `overset~v_1` to be real in Eq. (5) `4overset~v_2^2cos^2thetage8(2oversetv_2^2+3(1-eta)v^2)` or `6(eta-1)v^2ge(4-cos^2theta)overset~v_2^2` So, `overset~v_2levsqrt((6(eta-1))/(4-cos^2theta))` or `overset~v_(2(max))=sqrt(2(eta-1))v` Hence `v_(2(max))=|vecv+overset~vecv_2|_(max)=v+sqrt(2(eta-1))v=v(1+sqrt(2(eta-1)))=1km//s` Thus owing to the symmetry `v_(1(max))=v_(2(max))=v_(3(max))=v(1+sqrt(2(eta-1)))=1km//s` |
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