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

The Correct Option is A

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 V_A​=v
Now spring will compress until V_A​>V_B​ and maximum compression will be when V_A​=V_B.

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

⇒ Total linear momentum of the system will remain constant. 
mv=m(V_A​when max comp+V_B​when max compress)
 
∴  mv=m(V_{max compress}​+V_{max compress​})
⇒ V_{max compress}​ = 2v​
 
Total K.E of block A & B at max compression K.E _total​=2×21​m(V_{max 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. 
⇒ \(\frac{mv^2}{4}=\frac{1}{2}kx^2\)
⇒\(x=v\sqrt{\frac{m}{2k}}\)