Physics JEE mains class 11th

Question

Two blocks of masses m and M are moving with speeds v1andv2(v₁>v₂) in the same direction on the frictionless surface respectively, M being ahead of m. An ideal spring of force constant k is attached to the backside of M as shown. The maximum compression of the spring when the block collides is:

Answer 1

As per the given diagram,

At maximum compression reduced mass of system \(= \left( \frac{mM}{M+ m}\right)\)

Initial relative velocity of approach = (v₁ − v₂)

Hence, by conservation of energy,

Potential Energy of the spring = Kinetic Energy of the blocks

\(\frac 12kx_{max}^2 = \frac 12\left(\frac{mM}{m + M}\right)\left(v_1 - v_2\right)^2\)

Therefore,

\(x_{max} = \sqrt{\frac{mM}{k(M+ m)}}(v_1 - v_2)^2\)