Question

Two blocks A and B of masses 2m and 3m placed on smooth horizontal surface are connected with a light spring. Two blocks are given velocities as shown when spring is at natural length.

$\begin{array}{cccc}& Column-I& & Column-II\\ \left(A\right)& Minimummagnitudeofvelocityof& \left(p\right)& v\\ & A\left(V_{A_{min}}\right)duringmotion& & \\ \left(B\right)& Minimummagnitudeofvelocityof& \left(q\right)& \frac{v}{5}\\ & A\left(V_{A_{max}}\right)duringmotion& & \\ \left(C\right)& Minimummagnitudeofvelocityof& \left(r\right)& 0\\ & A\left(V_{B_{max}}\right)duringmotion& & \\ \left(D\right)& Velocityofcenterofmass& \left(s\right)& \frac{7v}{5}\\ & \left(V_{cm}\right)ofthesystemcomprisingof& & \\ & blocksA,Bandspring& & \end{array}$

• A. (A)-p; (B)-s; (C)-pr; (D)-q
• B. (A)-s; (B)-p; (C)-pr; (D)-q
• C. (A)-p; (B)-pr; (C)-s; (D)-q
• D. (A)-p; (B)-s; (C)-q; (D)-pr

Answer 1

The correct option is A (A)-p; (B)-s; (C)-pr; (D)-q

Step I : $V_{cm}=\frac{3mv-2mv}{5m}=\frac{v}{5}$
Step II : In COM frame
Initial velocity of $A=(-V-\frac{v}{5})=-\frac{6v}{5}$ to left
Initial velocity of $B=v-\frac{v}{5}=\frac{4}{5}v$
$=\frac{4}{5}v$ to right
Blocks are executing SHM in CM frame with initial position as equilibrium position.
Step III : Velocity variation of B in ground frame, considering right as + ve
From $(\frac{4v}{5}+\frac{v}{5})=vto\frac{4v}{5}+\frac{v}{5}=-\frac{3v}{5}$
So, $|v_{B_{max}}|=vand|v_{B_{min}}|=0$
Velocity variation A in ground frame
From $(\frac{6v}{5}+\frac{v}{5})=\frac{7v}{5}to\frac{-6v}{5}+\frac{v}{5}=-v$
Thus minimum velocity of A is $\frac{-v}{5}$ when spring is at maximum extension.