InterviewSolution
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InterviewSolution
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Two balls having masses m and 2m are fastened to two light strings of same length l figure. The other ends of the strings ar fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is elastic. a. Find velocities of the balls just after their collision. b. How high will the balls rise after the collision. |
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Answer» Let the velocilty of m reaching at lower end `=v_1` From work energy principle `:.(1/2)xxmv_1^2-(1/2)xxm(0)^2=mgl` `rarr v_1=sqrt(2gl)` similarly velocity of heavy block will be ltbr. `v_2=sqt(2gl)` `:. V_1=v_2=u(say)` Let fiN/A,l velocity of m and 2m are `v_1 and v_2` respectively. According to law of conservation of momenum `rarr mxxu-2mu=mv_1+2mv_2` `rarr v_1+2v_2=-u` `Again v_1-v-2=[(u-v)]=-2u` Substracting `3v_2=u` `rarr v_2=u/3=sqrt(2gl)/3` substituting in ii ltbr. `v_1-v_2=-2u` `rarr v_1=-2u+v_2` `=-2u+(u/3)` `=(-5)/3u=(-5)/3xxsqrt(2gl)` `(sqrt(50gl))/3` b. putting the work energy principle (1/2)xx2mxx(0)^2-(1/2)xx2mxx(v_2)^2` `=-2mxxgxxh` `[hrarr` height reached by heavy ball]` `rarr h=(L/9)` similarly `(1/2)xxmx(0)^2-(1/2)xxmv_1^2=mxxgxxh_2` [height reached by small ball] ltbr. `(1/2)xx(50gL)/9 gxxh_2` `h_2=(0.25L)/9` some `h_2` is more than 2L the velocity at highest point will not be zero. |
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