Question

Two blocks 1 and 2 of masses m and 2m respectively are connected by a spring of force constant k. The masses are moving to the right with uniform velocity v each, the heavier mass, leading the lighter one. The spring is of natural length in the motion.Block 2 collides head on with a third block 3 of mass m, at rest, the collision being completely inelastic, Determine the velocity of blocks at the instant of maximum compression of the spring.

• A. $\frac{5v}{4}m/sec$
• B. $\frac{5v}{2}m/sec$
• C. $\frac{5v}{3}m/sec$
• D. $\frac{3v}{4}m/sec$

Answer 1

The correct option is D $\frac{3v}{4}m/sec$

$Collisionb/w\text{blocks}2\&3$

$2mv=(2m+m)V^1$

$V^1=\frac{2v}{3}$

Whey both the blocks have equal velocity the spring will has may compression.

$\text{Let velocity}{v}_0$

Law of conservation of Momentum

$\mathrm{m/}v+3\mathrm{m/}\frac{2v}{3}=\mathrm{m/}{v}_0+3\mathrm{m/}{v}_0$

$So,V_0=\frac{3v}{4}$