InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two blocks `A` and `H`. each of mass `m`, are connected by a massless spring of natural length `I`. and spring constant `K`. The blocks are initially resting in a smooth horizontal floor with the spring at its natural length, as shown in Fig. A third identical block `C`, also of mass `m`, moves on the floor with a speed `v` along the line joining `A` and `B`. and collides elastically with `A`. Then A. (a) The KE of the AB system at maximum compression of the spring is zero.B. (b) The KE of the AB system at maximum compression of the spring is `(1//4)mv^2`.C. (c) The maximum compression of the spring is `vsqrt(m/k)`.D. (d) The maximum compression of the spring is `vsqrt((m)/(2k))` |
|
Answer» Correct Answer - B::D Initially, there will be collision between C and A which is elastic. So by conservation of momentum, we have `mv=mv_A+mv_C` `v=v_A+v_C` (i) And as in elastic collision, KE after collision is same as before collision, hence `1/2mv^2=1/2mv_A^2+1/2mv_C^2` i.e., `v^2=v_A^2+v_C^2` Subtracting Eq. (ii) from the square of Eq. (i), we have `2v_Av_C=0` So, either `v_A=0` or `v_C=0` `v_A=0` corresponds to no interaction between A and C, so the only physically possible solution is `v_C=0`, which is the light of Eq. (i) gives `v_A=v`, i.e., after collision C stops and A starts moving with velocity v. Now A will move and compress the spring which in turn accelerates B and retards A and finally both A and B will move with same velocity (say V). In this situation, compression of the spring will be maximum. As external force is zero, momentum of the system `(A+B+spri ng)` is conserved, i.e., `mv=(m+m)VimpliesV=v/2` By conservation of mechanical energy, `1/2mv^2=1/2(m+m)V^2+1/2kx_0^(2_2)` or `mv^2-2m(v_0/2)^2=kx^2` i.e., `k=(mv^2)/(2x_0^2)` or `x_0=vsqrt((m)/(2k))` |
|