InterviewSolution
Saved Bookmarks
| 1. |
A stationary body explodes into four identical fragments such that three of them fly off mutuallyperpendicular to each other, each with same K.E., Eo. The minimum energy of explosion will be(A) 6Eo15.4Eo3(C) 4Eo(D) 8Eo |
|
Answer» Let the three equal mass fragments move along x, y, z direction with velocity vₒ (each have same k.e = E = ½mvₒ²) . conservation of momentum: before = after => 0 = mv̂ + m[vₒî + vₒĵ + vₒk̂] => v̂ = ivth fragment`s velocity = - [vₒî + vₒĵ + vₒk̂] ---------------------> (i) energy of explosion = k.e(f) - k.e(i) = k.e(f) - 0 = ½m(√3vₒ)² + 3(½mvₒ²) = 6½mvₒ² = 6E. |
|