InterviewSolution
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InterviewSolution
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Two identical blocks `A` and `B` connected by massless string, are placed on a frictionless horizontal plane. A bullet havig the same mass, moving with speed `u` strickes block `B` from behind as shown. If the bullet gets embedded into block `B` then finda. the velocity of `A,B,C` after collision. b. impulse on `A` due to tension in the string,c. impulse on `C` due to normal force of collision, d. impulse on `B` due to normal force of collision. |
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Answer» (a) After collision, the blocks & the bullet will move together with same velocity, say `v` By conservation of linear momentum `m u=3m u` `v=(u)/(3)` (b) Net impulse on `A` is due to tension force, Impulse on `A=P_(f)-P_(i)` `intTdt=(m u)/(3)-0` (c) On the bullet `C`, net impulse is due to `N` `-intNdt=P_(f)-P_(i)` `=(m u)/(3)-m u` `=(-2m)/(3)` (d) On `B` two impulsive forces act i.e. Normal & Tension. `vec"J"=int(N-T)dt=intN" "dt-intT" "dt=(m u)/(3)` `rArr" "intN" "dt=(2m u)/(3)` The impulse due to normal force on both the colliding bodies is equal. Thus we can directly say impulse on `B` due to normal is same as impulse on `C` due to normal. |
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