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Two blocks A and H. each of mass m, are ...

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))`

Text Solution

Verified by Experts

The correct Answer is:
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))`
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