Question

Two identical blocks A and B, each of mass 'm' resting on a smooth floor are connected by a light spring of natural length L and spring constant K, with the spring at its natural length. A third identical block 'C' (mass m) moving with a speed v along the line joining A and B collides with A. the maximum compression in the spring is

• A. $v\sqrt{\frac{m}{2k}}$
• B. $m\sqrt{\frac{v}{2k}}$
• C. $\sqrt{\frac{mv}{2k}}$
• D. $\frac{mv}{2k}$

Answer 1

Answer: (A)

$v\sqrt{\frac{m}{2k}}$

Answer 2

The correct option is A)
On compression, consider collision to be instantaneous and use linear momentum conservation since no external force is acting.

⇒ Velocity of A after impact VA​=v
Now spring will compress until VA​>VB​ and maximum compression will be when VA​=VB.

So using moment conservation on system "A+B+spring",

⇒ Total linear momentum of the system will remain constant.
mv=m(VA​when max comp+VB​when max compress)

∴ mv=m(Vmax compress​+Vmax compress​)
⇒ Vmax compress​ = 2v​

Total K.E of block A & B at max compression K.E total​=2×21​m(Vmax compress​)2=$\frac{mv^2}{4}$
Initial K.E (after collision) K.Ei​=$\frac{1}{2}$​mv2
The difference in kinetic energy =$\frac{1}{2}$​mv2−$\frac{1}{4}$​mv2=$\frac{1}{4}$​mv2
The difference will be converted to P.E. of spring.
⇒ $\frac{mv^2}{4}=\frac{1}{2}kx^2$
⇒$x=v\sqrt{\frac{m}{2k}}$